I have an input field that expects a 10 digit number. If the user enters and submits a number less than 10 digits, the function would simply add a "0" until the inputed value is 10 digits in length.
I haven't really used, or understand how recursive functions really work, but I'm basically looking at an efficient way of doing this. One minor issue I'm having is figuring out how to prepend the "0"s at the beginning of the string rather than appended to the end.
My thinking:
function lengthCheck(sQuery) {
for (var i = 0; i < sQuery.length; i++) {
if (sQuery.length !== 10) {
sQuery += "0";
//I'd like to add the 0s to the beggining of the sQuery string.
console.log(sQuery);
lengthCheck(sQuery);
} else return sQuery
}
}
Change:
sQuery += "0"; // added at end of string
to:
sQuery = "0" + sQuery; // added at start of string
To remove the for loop/recursion, you could slice out the desired length in one step:
function padZeros(sQuery) {
// the max amount of zeros you want to lead with
const maxLengthZeros = "0000000000";
// takes the 10 rightmost characters and outputs them in a new string
return (maxLengthZeros + sQuery).slice(-10);
}
Simple generic function using ES6 repeat:
// edge case constraints not implemented for brevity
function padZeros(sQuery = "", maxPadding = 10, outputLength = 10) {
// the max amount of zeros you want to lead with
const maxLengthZeros = "0".repeat(maxPadding);
// returns the "outputLength" rightmost characters
return (maxLengthZeros + sQuery).slice(-outputLength);
}
console.log('padZeros: ' + padZeros("1234567890"));
console.log('padZeros: ' + padZeros("123"));
console.log('padZeros: ' + padZeros(""));
Alternate version that doesn't affect strings over your set limit:
function padZerosIfShort(inputString = "", paddedOutputLength = 10) {
let inputLen = inputString.length;
// only padded if under set length, otherwise returned untouched
return (paddedOutputLength > inputLen)
? "0".repeat(paddedOutputLength - inputLen) + inputString
: inputString;
}
console.log('padZerosIfShort: ' + padZerosIfShort("1234567890", 5));
console.log('padZerosIfShort: ' + padZerosIfShort("123", 5));
console.log('padZerosIfShort: ' + padZerosIfShort("", 5));
It will ultimately depend on your needs how you want to implement this behavior.
The += operator adds things to the end of strings similar to:
sQuery=sQuery+"0"
You can add characters to the front of a string like this
sQuery="0"+sQuery
I also found something interesting here. it works like this:
("00000" + sQuery).slice(-5)
You would add zeros to the front then slice off everything except the last 5. so to get 10 characters you would use:
("0000000000" + n).slice(-10)
You don't need recursion to solve this, just a simple for loop should do the trick. Try this:
function lengthCheck (sQuery) {
for (var i = sQuery.length; i<10; i++) {
sQuery = "0" + sQuery;
}
return sQuery;
}
You're looking to pad the string with zeroes. This is an example I've used before from here and will shorten your code a little bit:
function lengthCheck (sQuery) {
while (sQuery.length < 10)
sQuery = 0 + sQuery;
return sQuery;
}
I believe this has already been answered here (or similar enough to provide you the solution): How to output integers with leading zeros in JavaScript
Related
Here is my question:
Given a string, which is made up of space separated words, how can I split that into N strings of (roughly) even length, only breaking on spaces?
Here is what I've gathered from research:
I started by researching word-wrapping algorithms, because it seems to me that this is basically a word-wrapping problem. However, the majority of what I've found so far (and there is A LOT out there about word wrapping) assumes that the width of the line is a known input, and the number of lines is an output. I want the opposite.
I have found a (very) few questions, such as this that seem to be helpful. However, they are all focused on the problem as one of optimization - e.g. how can I split a sentence into a given number of lines, while minimizing the raggedness of the lines, or the wasted whitespace, or whatever, and do it in linear (or NlogN, or whatever) time. These questions seem mostly to be unanswered, as the optimization part of the problem is relatively "hard".
However, I don't care that much about optimization. As long as the lines are (in most cases) roughly even, I'm fine if the solution doesn't work in every single edge case, or can't be proven to be the least time complexity. I just need a real world solution that can take a string, and a number of lines (greater than 2), and give me back an array of strings that will usually look pretty even.
Here is what I've come up with:
I think I have a workable method for the case when N=3. I start by putting the first word on the first line, the last word on the last line, and then iteratively putting another word on the first and last lines, until my total width (measured by the length of the longest line) stops getting shorter. This usually works, but it gets tripped up if your longest words are in the middle of the line, and it doesn't seem very generalizable to more than 3 lines.
var getLongestHeaderLine = function(headerText) {
//Utility function definitions
var getLongest = function(arrayOfArrays) {
return arrayOfArrays.reduce(function(a, b) {
return a.length > b.length ? a : b;
});
};
var sumOfLengths = function(arrayOfArrays) {
return arrayOfArrays.reduce(function(a, b) {
return a + b.length + 1;
}, 0);
};
var getLongestLine = function(lines) {
return lines.reduce(function(a, b) {
return sumOfLengths(a) > sumOfLengths(b) ? a : b;
});
};
var getHeaderLength = function(lines) {
return sumOfLengths(getLongestLine(lines));
}
//first, deal with the degenerate cases
if (!headerText)
return headerText;
headerText = headerText.trim();
var headerWords = headerText.split(" ");
if (headerWords.length === 1)
return headerText;
if (headerWords.length === 2)
return getLongest(headerWords);
//If we have more than 2 words in the header,
//we need to split them into 3 lines
var firstLine = headerWords.splice(0, 1);
var lastLine = headerWords.splice(-1, 1);
var lines = [firstLine, headerWords, lastLine];
//The header length is the length of the longest
//line in the header. We will keep iterating
//until the header length stops getting shorter.
var headerLength = getHeaderLength(lines);
var lastHeaderLength = headerLength;
while (true) {
//Take the first word from the middle line,
//and add it to the first line
firstLine.push(headerWords.shift());
headerLength = getHeaderLength(lines);
if (headerLength > lastHeaderLength || headerWords.length === 0) {
//If we stopped getting shorter, undo
headerWords.unshift(firstLine.pop());
break;
}
//Take the last word from the middle line,
//and add it to the last line
lastHeaderLength = headerLength;
lastLine.unshift(headerWords.pop());
headerLength = getHeaderLength(lines);
if (headerLength > lastHeaderLength || headerWords.length === 0) {
//If we stopped getting shorter, undo
headerWords.push(lastLine.shift());
break;
}
lastHeaderLength = headerLength;
}
return getLongestLine(lines).join(" ");
};
debugger;
var header = "an apple a day keeps the doctor away";
var longestHeaderLine = getLongestHeaderLine(header);
debugger;
EDIT: I tagged javascript, because ultimately I would like a solution I can implement in that language. It's not super critical to the problem though, and I would take any solution that works.
EDIT#2: While performance is not what I'm most concerned about here, I do need to be able to perform whatever solution I come up with ~100-200 times, on strings that can be up to ~250 characters long. This would be done during a page load, so it needs to not take forever. For example, I've found that trying to offload this problem to the rendering engine by putting each string into a DIV and playing with the dimensions doesn't work, since it (seems to be) incredibly expensive to measure rendered elements.
Try this. For any reasonable N, it should do the job:
function format(srcString, lines) {
var target = "";
var arr = srcString.split(" ");
var c = 0;
var MAX = Math.ceil(srcString.length / lines);
for (var i = 0, len = arr.length; i < len; i++) {
var cur = arr[i];
if(c + cur.length > MAX) {
target += '\n' + cur;
c = cur.length;
}
else {
if(target.length > 0)
target += " ";
target += cur;
c += cur.length;
}
}
return target;
}
alert(format("this is a very very very very " +
"long and convoluted way of creating " +
"a very very very long string",7));
You may want to give this solution a try, using canvas. It will need optimization and is only a quick shot, but I think canvas might be a good idea as you can calculate real widths. You can also adjust the font to the really used one, and so on. Important to note: This won't be the most performant way of doing things. It will create a lot of canvases.
DEMO
var t = `However, I don't care that much about optimization. As long as the lines are (in most cases) roughly even, I'm fine if the solution doesn't work in every single edge case, or can't be proven to be the least time complexity. I just need a real world solution that can take a string, and a number of lines (greater than 2), and give me back an array of strings that will usually look pretty even.`;
function getTextTotalWidth(text) {
var canvas = document.createElement("canvas");
var ctx = canvas.getContext("2d");
ctx.font = "12px Arial";
ctx.fillText(text,0,12);
return ctx.measureText(text).width;
}
function getLineWidth(lines, totalWidth) {
return totalWidth / lines ;
}
function getAverageLetterSize(text) {
var t = text.replace(/\s/g, "").split("");
var sum = t.map(function(d) {
return getTextTotalWidth(d);
}).reduce(function(a, b) { return a + b; });
return sum / t.length;
}
function getLines(text, numberOfLines) {
var lineWidth = getLineWidth(numberOfLines, getTextTotalWidth(text));
var letterWidth = getAverageLetterSize(text);
var t = text.split("");
return createLines(t, letterWidth, lineWidth);
}
function createLines(t, letterWidth, lineWidth) {
var i = 0;
var res = t.map(function(d) {
if (i < lineWidth || d != " ") {
i+=letterWidth;
return d;
}
i = 0;
return "<br />";
})
return res.join("");
}
var div = document.createElement("div");
div.innerHTML = getLines(t, 7);
document.body.appendChild(div);
I'm sorry this is C#. I had created my project already when you updated your post with the Javascript tag.
Since you said all you care about is roughly the same line length... I came up with this. Sorry for the simplistic approach.
private void DoIt() {
List<string> listofwords = txtbx_Input.Text.Split(' ').ToList();
int totalcharcount = 0;
int neededLineCount = int.Parse(txtbx_LineCount.Text);
foreach (string word in listofwords)
{
totalcharcount = totalcharcount + word.Count(char.IsLetter);
}
int averagecharcountneededperline = totalcharcount / neededLineCount;
List<string> output = new List<string>();
int positionsneeded = 0;
while (output.Count < neededLineCount)
{
string tempstr = string.Empty;
while (positionsneeded < listofwords.Count)
{
tempstr += " " + listofwords[positionsneeded];
if ((positionsneeded != listofwords.Count - 1) && (tempstr.Count(char.IsLetter) + listofwords[positionsneeded + 1].Count(char.IsLetter) > averagecharcountneededperline))//if (this is not the last word) and (we are going to bust the average)
{
if (output.Count + 1 == neededLineCount)//if we are writting the last line
{
//who cares about exceeding.
}
else
{
//we're going to exceed the allowed average, gotta force this loop to stop
positionsneeded++;//dont forget!
break;
}
}
positionsneeded++;//increment the needed position by one
}
output.Add(tempstr);//store the string in our list of string to output
}
//display the line on the screen
foreach (string lineoftext in output)
{
txtbx_Output.AppendText(lineoftext + Environment.NewLine);
}
}
(Adapted from here, How to partition an array of integers in a way that minimizes the maximum of the sum of each partition?)
If we consider the word lengths as a list of numbers, we can binary search the partition.
Our max length ranges from 0 to sum (word-length list) + (num words - 1), meaning the spaces. mid = (range / 2). We check if mid can be achieved by partitioning into N sets in O(m) time: traverse the list, adding (word_length + 1) to the current part while the current sum is less than or equal to mid. When the sum passes mid, start a new part. If the result includes N or less parts, mid is achievable.
If mid can be achieved, try a lower range; otherwise, a higher range. The time complexity is O(m log num_chars). (You'll also have to consider how deleting a space per part, meaning where the line break would go, features into the calculation.)
JavaScript code (adapted from http://articles.leetcode.com/the-painters-partition-problem-part-ii):
function getK(arr,maxLength) {
var total = 0,
k = 1;
for (var i=0; i<arr.length; i++) {
total += arr[i] + 1;
if (total > maxLength) {
total = arr[i];
k++;
}
}
return k;
}
function partition(arr,n) {
var lo = Math.max(...arr),
hi = arr.reduce((a,b) => a + b);
while (lo < hi) {
var mid = lo + ((hi - lo) >> 1);
var k = getK(arr,mid);
if (k <= n){
hi = mid;
} else{
lo = mid + 1;
}
}
return lo;
}
var s = "this is a very very very very "
+ "long and convoluted way of creating "
+ "a very very very long string",
n = 7;
var words = s.split(/\s+/),
maxLength = partition(words.map(x => x.length),7);
console.log('max sentence length: ' + maxLength);
console.log(words.length + ' words');
console.log(n + ' lines')
console.log('')
var i = 0;
for (var j=0; j<n; j++){
var str = '';
while (true){
if (!words[i] || str.length + words[i].length > maxLength){
break
}
str += words[i++] + ' ';
}
console.log(str);
}
Using the Java String Split() Method to split a string we will discover How and Where to Apply This String Manipulation Technique:
We'll examine the Java Split() method's explanation and discover how to apply it. The principles are explained simply and with enough programming examples, either as a separate explanation or in the comment part of the programs.
The Java String Split() method is used to divide or split the calling Java String into pieces and return the Array, as the name implies. The delimiters("", " ", ) or regular expressions that we have supplied separately for each component or item of an array.
Syntax
String[ ] split(String regExp)
First Case: It involves initializing a Java String variable with a variety of words separated by spaces, using the Java String Split() method, and evaluating the results. We can effectively print each word without the space using the Java Split() function.
Second Case: In this case, we initialize a Java String variable and attempt to split or deconstruct the main String variable to use the String Split() method utilizing a substring of the initialized String variable.
Third Case: In this case, we will attempt to split a String using its character by taking a String variable (a single word).
You can check out other approaches to this problem on YouTube and even coding websites on google such as Coding Ninjas
This old question was revived by a recent answer, and I think I have a simpler technique than the answers so far:
const evenSplit = (text = '', lines = 1) => {
if (lines < 2) {return [text]}
const baseIndex = Math .round (text .length / lines)
const before = text .slice (0, baseIndex) .lastIndexOf (' ')
const after = text .slice (baseIndex) .indexOf (' ') + baseIndex
const index = after - baseIndex < baseIndex - before ? after : before
return [
text .slice (0, index),
... evenSplit (text .slice (index + (before > -1 ? 1 : 0)), lines - 1)
]
}
const text = `However, I don't care that much about optimization. As long as the lines are (in most cases) roughly even, I'm fine if the solution doesn't work in every single edge case, or can't be proven to be the least time complexity. I just need a real world solution that can take a string, and a number of lines (greater than 2), and give me back an array of strings that will usually look pretty even.`
const display = (lines) => console .log (lines .join ('\n'))
display (evenSplit (text, 7))
display (evenSplit (text, 5))
display (evenSplit (text, 12))
display (evenSplit (`this should be three lines, but it has a loooooooooooooooooooooooooooooooong word`, 3))
.as-console-wrapper {max-height: 100% !important; top: 0}
It works by finding the first line then recurring on the remaining text with one fewer lines. The recursion bottoms out when we have a single line. To calculate the first line, we take an initial target index which is just an equal share of the string based on its length and the number of lines. We then check to find the closest space to that index, and split the string there.
It does no optimization, and could certainly be occasionally misled by long words, but mostly it just seems to work.
I'm sorry for the dumb question. I've been trying to do this for hours now, and i really can't get it to work. So i have a for-loop that loops though some numbers.
But it doesn't take the first value(71990000).
How can this be achieved?
This is what i've got so far:
var minNr = 0000;
var maxNr = 10000;
var prefix = 7199;
function Nummer(min,max)
{
var regex = /^(\d{2})\1$/;
var guld_nr;
for(guld_nr = minNr; guld_nr < maxNr;)
{
if(regex.test(guld_nr))
{
$(".resultat").append(prefix + "" + guld_nr + "<br>");
}
guld_nr++;
}
}
The output is this:
71991010
71991111
71991212
71991313
But i also need the number: 71990000
How can i do that ?
It's because your regex is rejecting the number 0; the first time through the loop, minNr has the numeric value 0 (setting it to 0000 doesn't help; it's just a fancy way of saying 0). The regex expects two digits followed by the same pattern, but what you're giving it is the string '0'.
You could set minNr to be a string instead on the first pass through ('0000'), and this will solve the problem for '0000', but you will miss '0101', '0202', etc. (which will convert to the strings '101', '202', and so on.)
One solution would be to zero pad the string representation of your number. The following function will take any number and left zero pad it to fit a given width:
function zeropad(n, w) {
n = String(n);
while(n.length < w) n = '0' + n;
return n;
}
You can use it to convert minNr for the regex:
regex.test(zeropad(guld_nr, 4))
Also note that Number is a built-in object wrapper for literals in JavaScript (all of the primitives have object wrappers: Number, Boolean, String), and by creating a function called Number, you are occluding this built-in object, which is inadvisable (code that needs to use it will invoke your function instead, which is incompatible and has a different purpose).
Use string:
var minNr = '0000';
It's the start value for the regex test, and you need the four zeroes for that. If it would be a number, then you get only one zero for testing. it would help, if you pad it with leading zeroes.
var minNr = '0000',
maxNr = 10000,
prefix = 7199;
function Nummer(min,max) {
var regex = /^(\d{2})\1$/;
var guld_nr;
for(guld_nr = minNr; guld_nr < maxNr;guld_nr++) {
if(regex.test(guld_nr)) {
document.write(prefix + "" + guld_nr + "<br>");
}
}
}
Nummer(minNr, maxNr);
Numbers don't zero-pad themselves; 0000; // 0
Make a custom zero-pad method for it so you can do zpad(0, 4); // "0000"
function zpad(x, digits) {
var pad = '0';
x = x.toString();
digits -= x.length;
while (digits > 0) {
if (digits & 1) x = pad + x;
pad += pad;
digits >>>= 1;
}
return x;
}
Now adjust Nummer accordingly
function Nummer(min, max, prefix) {
var regex = /^(\d{2})\1$/,
i, str;
prefix = prefix || '';
for(i = min; i < max; ++i) {
str = zpad(i, 4);
if(regex.test(str)) console.log(prefix + str);
}
}
and use
Nummer(minNr, maxNr, '7199');
Side note
Nummer is not constructing an Object, consider camel casing it
You could use arithmetic to do the digit pattern check, and keep the result numerical:
var minNr = 0; // it does not help to put 4 zeroes here.
var maxNr = 10000;
var prefix = 7199;
function Nummer(min,max) {
for (var guld_nr = min; guld_nr < max; guld_nr++) {
if (Math.floor(guld_nr/100) === guld_nr % 100 ) {
$(".resultat").append((prefix * 10000 + guld_nr) + "<br>");
}
}
}
Nummer(minNr, maxNr);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="resultat"></div>
The problem with your code is when the lower numbers are tested against the regular expression, they are implicitly converted to string, and do not get prefixed zeroes, so they fail on the regular expression.
Anyway, the code will be more efficient when sticking to numbers instead of strings, so I would suggest working with numbers all the way up to the point of outputting them in the browser.
Even more efficient is this code:
var minNr = 0; // it does not help to put 4 zeroes here.
var maxNr = 10000;
var prefix = 7199;
function Nummer(min,max) {
var test = Math.floor(min/100)*100 + Math.floor(min/100)%100;
var guld_nr = test < min ? test + 101 : test;
for (; guld_nr < max; guld_nr+=101) {
$(".resultat").append((prefix * 10000 + guld_nr) + "<br>");
}
}
Nummer(minNr, maxNr);
<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<div class="resultat"></div>
For an ObjectId in MongoDB, I work with a 24 digit hexadecimal number. Because I need to keep track of a second collection, I need to add 1 to this hexadecimal number.
In my case, here's my value
var value = "55a98f19b27585d81922ba0b"
What I'm looking for is
var newValue = "55a98f19b25785d81922ba0c"
I tried to create a function for this
function hexPlusOne(hex) {
var num = (("0x" + hex) / 1) + 1;
return num.toString(16);
}
This works with smaller hex numbers
hexPlusOne("eeefab")
=> "eeefac"
but it fails miserably for my hash
hexPlusOne(value)
=> "55a98f19b275840000000000"
Is there a better way to solve this?
This version will return a string as long as the input string, so the overflow is ignored in case the input is something like "ffffffff".
function hexIncrement(str) {
var hex = str.match(/[0-9a-f]/gi);
var digit = hex.length;
var carry = 1;
while (digit-- && carry) {
var dec = parseInt(hex[digit], 16) + carry;
carry = Math.floor(dec / 16);
dec %= 16;
hex[digit] = dec.toString(16);
}
return(hex.join(""));
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("ffffffffffffffffffffffff"));
This version may return a string which is 1 character longer than the input string, because input like "ffffffff" carries over to become "100000000".
function hexIncrement(str) {
var hex = str.match(/[0-9a-f]/gi);
var digit = hex.length;
var carry = 1;
while (digit-- && carry) {
var dec = parseInt(hex[digit], 16) + carry;
carry = Math.floor(dec / 16);
dec %= 16;
hex[digit] = dec.toString(16);
}
if (carry) hex.unshift("1");
return(hex.join(""));
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("ffffffffffffffffffffffff"));
I was curious te see whether user2864740's suggestion of working with 12-digit chunks would offer any advantage. To my surprise, even though the code looks more complicated, it's actually around twice as fast. But the first version runs 500,000 times per second too, so it's not like you're going to notice in the real world.
function hexIncrement(str) {
var result = "";
var carry = 1;
while (str.length && carry) {
var hex = str.slice(-12);
if (/^f*$/i.test(hex)) {
result = hex.replace(/f/gi, "0") + result;
carry = 1;
} else {
result = ("00000000000" + (parseInt(hex, 16) + carry).toString(16)).slice(-hex.length) + result;
carry = 0;
}
str = str.slice(0,-12);
}
return(str.toLowerCase() + (carry ? "1" : "") + result);
}
document.write(hexIncrement("55a98f19b27585d81922ba0b") + "<BR>");
document.write(hexIncrement("000000000000ffffffffffff") + "<BR>");
document.write(hexIncrement("0123456789abcdef000000000000ffffffffffff"));
The error comes from attempting to covert the entire 24-digit hex value to a number first because it won't fit in the range of integers JavaScript can represent distinctly2. In doing such a conversion to a JavaScript number some accuracy is lost.
However, it can be processed as multiple (eg. two) parts: do the math on the right part and then the left part, if needed due to overflow1. (It could also be processed one digit at a time with the entire addition done manually.)
Each chunk can be 12 hex digits in size, which makes it an easy split-in-half.
1 That is, if the final num for the right part is larger than 0xffffffffffff, simply carry over (adding) one to the left part. If there is no overflow then the left part remains untouched.
2 See What is JavaScript's highest integer value that a Number can go to without losing precision?
The range is 2^53, but the incoming value is 16^24 ~ (2^4)^24 ~ 2^(4*24) ~ 2^96; still a valid number, but outside the range of integers that can be distinctly represented.
Also, use parseInt(str, 16) instead of using "0x" + str in a numeric context to force the conversion, as it makes the intent arguably more clear.
I have a string, 15.Prototypal-Inheritance-and-Refactoring-the-Slider.txt, I'd like to make it looks like 15.Prototypal...-Slider.txt
The length of the text is 56, how can I keep the first 12 letters and 10 last letters (incuding punctuation marks) and replace the others to ...
I don't really know how to commence the code, I made something like
var str="15.Prototypal-Inheritance-and-Refactoring-the-Slider.txt";
str.split("// ",1);
although this gives me what I need, how do I have the results base on letters not words.
You can use str.slice().
function middleEllipsis(str, a, b) {
if (str.length > a + b)
return str.slice(0, a) + '...' + str.slice(-b);
else
return str;
}
middleEllipsis("15.Prototypal-Inheritance-and-Refactoring-the-Slider.txt", 12, 10);
// "15.Prototypa...Slider.txt"
middleEllipsis("mpchc64.mov", 12, 10);
// "mpchc64.mov"
This function will do what you ask for:
function fixString(str) {
var LEN_PREFIX = 12;
var LEN_SUFFIX = 10;
if (str.length < LEN_PREFIX + LEN_SUFFIX) { return str; }
return str.substr(0, LEN_PREFIX) + '...' + str.substr(str.length - LEN_SUFFIX - 1);
}
You can adjust the LEN_PREFIX and LEN_SUFFIX as needed, but I've the values you specified in your post. You could also make the function more generic by making the prefix and suffix length input arguments to your function:
function fixString(str, prefixLength, suffixLength) {
if (str.length < prefixLength + suffixLength) { return str; }
return str.substr(0, prefixLength) + '...' + str.substr(str.length - suffixLength - 1);
}
I'd like to make it looks like 15.Prototypal...-Slider.txt
LIVE DEMO
No matter how long are the suffixed and prefixed texts, this will get the desired:
var str = "15.Prototypal-Inheritance-and-Refactoring-the-Slider.txt",
sp = str.split('-'),
newStr = str;
if(sp.length>1) newStr = sp[0]+'...-'+ sp.pop() ;
alert( newStr ); //15.Prototypal...-Slider.txt
Splitting the string at - and using .pop() method to retrieve the last Array value from the splitted String.
Instead of splitting the string at some defined positions it'll also handle strings like:
11.jQuery-infinite-loop-a-Gallery.txt returning: 11.jQuery...-Gallery.txt
Here's another option. Note that this keeps the first 13 characters and last 11 because that's what you gave in your example.:
var shortenedStr = str.substr(0, 13) + '...' + str.substring(str.length - 11);
You could use the javascript substring command to find out what you want.
If you string is always 56 characters you could do something like this:
var str="15.Prototypal-Inheritance-and-Refactoring-the-Slider.txt";
var newstr = str.substring(0,11) + "..." + str.substring(45,55)
if your string varies in length I would highly recommend finding the length of the string first, and then doing the substring.
have a look at: http://www.w3schools.com/jsref/jsref_substring.asp
Example: We have the number 1122. I would like to check that if given number contains the digit 1 more than once. In this case, it should return true.
I need the code to be flexible, it has to work with any number, like 3340, 5660, 4177 etc.
You can easily "force" JS to coerce any numeric value to a string, either by calling the toString method, or concatenating:
var someNum = 1122;
var oneCount = (someNum + '').split('1').length;
by concatenating a number to an empty string, the variable is coerced to a string, so you can use all the string methods you like (.match, .substring, .indexOf, ...).
In this example, I've chosen to split the string on each '1' char, count and use the length of the resulting array. If the the length > 2, than you know what you need to know.
var multipleOnes = ((someNum + '').split('1').length > 2);//returns a bool, true in this case
In response to your comment, to make it flexible - writing a simple function will do:
function multipleDigit(number, digit, moreThan)
{
moreThan = (moreThan || 1) + 1;//default more than 1 time, +1 for the length at the end
digit = (digit !== undefined ? digit : 1).toString();
return ((someNum + '').split(digit).length > moreThan);
}
multipleDigit(1123, 1);//returns true
multipleDigit(1123, 1, 2);//returns false
multipleDigit(223344,3);//returns 3 -> more than 1 3 in number.
Use javascript's match() method. Essentially, what you'd need to do is first convert the number to a string. Numbers don't have the RegExp methods. After that, match for the number 1 globally and count the results (match returns an array with all matched results).
var number = 1100;
console.log(number.toString().match(/1/g).length);
function find(num, tofind) {
var b = parseInt(num, 10);
var c = parseInt(tofind, 10);
var a = c.split("");
var times = 0;
for (var i = 0; i < a.length; i++) {
if (a[i] == b) {
times++;
}
}
alert(times);
}
find('2', '1122');
Convert the number to a string and iterate over it. Return true once a second digit has been found, for efficiency.
function checkDigitRepeat(number, digit) {
var i, count = 0;
i = Math.abs(number);
if(isNaN(i)) {
throw(TypeError('expected Number for number, got: ' + number));
}
number = i.toString();
i = Math.abs(digit);
if(isNaN(i)) {
throw(TypeError('expected Number for digit, got: ' + digit));
}
digit = i.toString();
if(digit > 9) {
throw(SyntaxError('expected a digit for digit, got a sequence of digits: ' + digit));
}
for(i = 0; i < number.length; i += 1) {
if(number[i] === digit) {
count += 1;
if(count >= 2) { return true; }
}
}
return false;
}
In the event that you want to check for a sequence of digits, your solution may lie in using regular expressions.
var myNum = '0011';
var isMultipleTimes = function(num) {
return !!num.toString().match(/(\d)\1/g);
}
console.log(isMultipleTimes(myNum));
JavaScript Match
Using #Aspiring Aqib's answer, I made a function that actually works properly and in the way I want.
The way it works is:
Example execution: multDig('221','2')
Split the number (first argument) to an array where each element is one digit.Output: ['2','2','1']
Run a for loop, which checks each of the array elements if they match with the digit (second argument), and increment the times variable if there is a match.Output: 2
Check inside the for loop if the match was detected already to improve performance on longer numbers like 2211111111111111
Return true if the number was found more than once, otherwise, return false.
And finally the code itself:
function multDig(number, digit){
var finalSplit = number.toString().split(''), times = 0;
for (i = 0; i < finalSplit.length; i++){
if (finalSplit[i] == digit){
times++
}
if (times > 1){
return true;
}
}
return false;
}