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Get substings between Specific Characters

I am trying to solve some JS problem. I want to check if an IP address is a valid one.
So the numbers must be between 0-255.
So what I want to do at this point, is to get an IP ex 192.168.1.1 and get substrings and load them to an array, so I want to create an array that looks like that:
array = ['192' , '168' , '1' , '1'];
I've tried various approaches in my algorithm but can't manage to target dynamically the numbers and split them between every dot.
I've done several tries, and thats the closest I could get.
let str = '192.168.1.1';
isValidIp(str);
function isValidIP(str) {
let array = [];
let substringArray = [];
for (let i=0; i<str.length; i++){
if (str[i] == '.') array.push(i);
}
let counter = 0;
for (let i in array){
substringArray.push(str.substring(counter, array[i]));
counter = array[i];
}
console.log(substringArray);
}
Which returns:
[ '192', '.168', '.1' ]

You can use the split() function of JavaScript which returns an array of every element separated by the digit specified. Or, which I wouldn't recommend, you could use RegEx. Here is an example of both:
function isValidIPwRegEx(str){
if (/^(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)\.(25[0-5]|2[0-4][0-9]|[01]?[0-9][0-9]?)$/.test(str))
{
return true;
}
return false;
}
function isValidIP(str) {
let array = str.split("."),
isIP = true;
array = array.filter( block => !block.includes("+") && !block.includes("e") );
if(array.length!=4) return false;
array.forEach((number) => {
if ( !(+number >=0 && +number <= 255) ) { //As #p.s.w.g kindly suggested
isIP = false;
}
});
return isIP;
}
//With RegEx
console.log("With RegEx");
console.log(isValidIPwRegEx("192.168.1.1"));
console.log(isValidIPwRegEx("blah.blah.blah.blah")); //As #georg suggested
console.log(isValidIPwRegEx("1e1.2e1.+3e1.+5e1")); //As #georg again suggested to #Nina Scholz
console.log("");
//Without RegEx
console.log("Without RegEx");
console.log(isValidIP("192.168.1.1"));
console.log(isValidIP("blah.blah.blah.blah")); //As #georg suggested
console.log(isValidIP("1e1.2e1.+3e1.+5e1")); //As #georg again suggested to #Nina Scholz
console.log(isValidIP("1e1.2e1.3e1.5e1"));

Use String's split function.
So, something like "192.168.1.1".split(".")

You could split the string and check if the length is four and all values are integers and smaller than 256.
var ip = '192.168.1.1',
values = ip.split('.'),
valid = values.length === 4 && values.every(v => +v >= 0 && +v < 256);
console.log(values);
console.log(valid);

function isValidIP(str) {
let re = /^(\d{1,3})\.(\d{1,3})\.(\d{1,3})\.(\d{1,3})$/;
let m = str.match(re);
return m &&
m[1] >= 0 && m[1] <= 255 &&
m[2] >= 0 && m[2] <= 255 &&
m[3] >= 0 && m[3] <= 255 &&
m[4] >= 0 && m[4] <= 255
;
}
If you wish to be more precise, each digit check can be:
(0|[1-9]\d{0:2})
This prevents extraneous leading 0's.

Check how many times a char appears in a string

Simply trying to find how many times a given character appears in a string but I can't solve it any other way then this simple for-loop. Is there a method that would solve this quicker or more eloquently other than using Regex?
function countCharacter(str, char) {
var count = 0;
for(var i = 0; i < str.length; i++){
if(str.charAt(i) === char)
count++;
}
return count;
}

There are many possible ways are available in the market.
I am adding a few of them.
Method 1:
str = "The man is as good as his word"
str.split('a')
output: (4) ["The m", "n is ", "s good ", "s his word"]
str.split('a').length - 1
output: 3
Method 2:
str = "The man is as good as his word"
str.split('').map( function(char,i)
{ if(char === 'a')
return i;
}
).filter(Boolean)
Output: (3) [5, 11, 19]
str.split('').map( function(char,i)
{ if(char === 'a')
return i;
}
).filter(Boolean).length
ouput: 3
Edit: As per comment we can also make use of filter().
str.split('').filter(function(char, i){
if(char == 'a'){
return i;
}
})
output: (3) ["a", "a", "a"]
str.split('').filter(function(char, i){
if(char == 'a'){
return i;
}
}).length
output: 3

----edited by adding more cases from answers----
there are several ways, you can use split/for/regex/reduce/indexOf like this:
function countCharacter_reduce(str, ch) {
return Array.prototype.reduce.call(str, (prev, cur) => cur === ch && ++prev && prev, 0);
}
function countCharacter_split(str, ch) {
return str.split(ch).length - 1;
}
function countCharacter_for(str, ch) {
for (var count = 0, ii = 0; ii < str.length; ii++) {
if (str[ii] === ch)
count++;
}
return count;
}
function countCharacter_regex(str, ch) {
return str.length - str.replace(new RegExp(ch, 'g'), '').length;
}
function countCharacter_indexOf(str, char) {
var start = 0;
var count = 0;
while ((start = str.indexOf(char, start) + 1) !== 0) {
count++;
}
return count;
}
performance of them by running 1,000,000 times on counting '/' in a string.
-- case1: running 1000000 times on ( 'this/is/a/path/with/extension', '/' )
countCharacter_reduce: 2389.880ms
countCharacter_regex: 496.251ms
countCharacter_split: 114.709ms
countCharacter_for: 152.012ms
countCharacter_indexOf: 90.330ms
-- case2: running 1000000 times on ( '////////////', '/' )
countCharacter_reduce: 1138.051ms
countCharacter_regex: 619.886ms
countCharacter_split: 121.945ms
countCharacter_for: 74.542ms
countCharacter_indexOf: 204.242ms
Conclusion ('>' means 'has better performance'):
for|split|indexOf > regex > reduce.
furthermore,
if the string contains more searching characters (like case2),
for>split>indexOf,
otherwise (like case1)
indexOf > split > for.
BTW: you can change the for indexOf method to fit multi-characters search (example is single character)

using reduce:
function countCharacter(str, char) {
return str.split('').reduce((a, x) => x === char ? ++a : a, 0);
}

I guess this involves regex which you wanted to avoid but it's pretty quick:
function countCharacter(str, char) {
return str.length - str.replace(new RegExp(char,"g"),"").length;
}
You can also try the str.split(char).length-1 approach suggested by Jaromanda.
Or, go all out with some fun recursion (pass 0 to startingFrom):
function countCharacter(str, char, startingFrom) {
var idx = str.indexOf(char, startingFrom);
return idx == -1 ? 0 : 1 + countCharacter(str, char, idx + 1);
}
You can get rid of the annoying extra argument at the cost of some efficiency:
function countCharacter(str, char) {
var idx = str.indexOf(char);
return idx == -1 ? 0 : 1 + countCharacter(str.substr(idx+1), char);
}
And here's a version optimized for speed (this is about 3 times faster on my browser than your original and much faster than the regex versions, according to jsperf):
function countCharacter(str, char) {
var start = 0;
var count = 0;
while((start = str.indexOf(char, start)+1) !== 0) {
count++;
}
return count;
}
Note that the indexOf approaches will generally be substantially faster than manually iterating through the string. See jsperf

Here you go. One line code
"hello".match(new RegExp('l','g')).length
replace 'l' with any char here, new RegExp('l','g').
that is
str.match(new RegExp(char,'g')).length

Have you thought of using the split() method? How about this -
function myFunction(str, char) {
return string.split(char).length - 1
}
Let me know what you think of this solution.

Inserting into a number string

Have the function DashInsert(num) insert dashes ('-') between each two odd numbers in num. For example: if num is 454793 the output should be 4547-9-3. Don't count zero as an odd number.
Here is my code (not working). When I run it, I get the same response as an infinite loop where I have to kill the page but I can't see why. I know there are ways to do this by keeping it as a string but now I'm wondering why my way isn't working. Thanks...
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
}
}
num = num.join("");
return num;
}

Using num.splice you are inserting new entries into the array, therefor increasing its length – and that makes the value of i “running behind” the increasing length of the array, so the break condition is never met.
And apart from that, on the next iteration after inserting a -, num[i-1] will be that - character, and therefor you are practically trying to check if '-' % 2 != 0 … that makes little sense as well.
So, when you insert a - into the array, you have to increase i by one as well – that will a) account for the length of the array having increased by one, and also it will check the next digit after the - on the next iteration:
function DashInsert(num) {
num = num.split("");
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
num.splice(i, 0, "-");
i++; // <- this is the IMPORTANT part!
}
}
num = num.join("");
return num;
}
alert(DashInsert("454793"));
http://jsfiddle.net/37wA9/

Once you insert a dash -, the if statement is checking this '-'%2 != 0 which is always true and thus inserts another dash, ad infinitum.
Here's one way to do it with replace using a regex and function:
function DashInsert(n) {
var f = function(m,i,s) { return m&s[i+1]&1 ? m+'-' : m; };
return String(n).replace(/\d/g,f);
}
DashInsert(454793) // "4547-9-3"

When you are adding a dash, this dash will be processed as a number on the next iteration. You need to forward one step.
function DashInsert(num) {
var num = num.split("");
for (var i = 1; i < num.length; i++) {
if ((num[i - 1] % 2 != 0) && (num[i] % 2 != 0)) {
num.splice(i, 0, "-");
i++; // This is the only thing that needs changing
}
}
num = num.join("");
return num;
}

It's because there are cases when you use the % operator on dash '-' itself, e.g. right after you splice a dash into the array.
You can correct this behavior by using a clone array.
function DashInsert(num) {
num = num.split("");
var clone = num.slice(0);
var offset = 0;
for (i = 1; i < num.length; i++) {
if (num[i - 1] % 2 != 0 && num[i] % 2 != 0) {
clone.splice(i + offset, 0, "-");
offset++;
}
}
return clone.join("");
}
alert(DashInsert("45739"));
Output: 45-7-3-9
Demo: http://jsfiddle.net/262Bf/

To complement the great answers already given, I would like to share an alternative implementation, that doesn't modify arrays in-place:
function DashInsert(num) {
var characters = num.split("");
var numbers = characters.map(function(chr) {
return parseInt(chr, 10);
});
var withDashes = numbers.reduce(function(result, current) {
var lastNumber = result[result.length - 1];
if(lastNumber == null || current % 2 === 0 || lastNumber % 2 === 0) {
return result.concat(current);
} else {
return result.concat("-", current);
}
}, []);
return withDashes.join("");
}
It's longer, but IMHO reveals the intention better, and avoids the original issue.

How to sort strings in JavaScript numerically

I would like to sort an array of strings (in JavaScript) such that groups of digits within the strings are compared as integers not strings. I am not worried about signed or floating point numbers.
For example, the result should be ["a1b3","a9b2","a10b2","a10b11"] not ["a1b3","a10b11","a10b2","a9b2"]
The easiest way to do this seems to be splitting each string on boundaries around groups of digits. Is there a pattern I can pass to String.split to split on character boundaries without removing any characters?
"abc11def22ghi".split(/?/) = ["abc","11","def","22","ghi"];
Or is there another way to compare strings that does not involve splitting them up, perhaps by padding all groups of digits with leading zeros so they are the same length?
"aa1bb" => "aa00000001bb", "aa10bb" => "aa00000010bb"
I am working with arbitrary strings, not strings that have a specific arrangement of digit groups.
I like the /(\d+)/ one liner from Gaby to split the array. How backwards compatible is that?
The solutions that parse the strings once in a way that can be used to rebuild the originals are much more efficient that this compare function. None of the answers handle some strings starting with digits and others not, but that would be easy enough to remedy and was not explicit in the original question.
["a100", "a20", "a3", "a3b", "a3b100", "a3b20", "a3b3", "!!", "~~", "9", "10", "9.5"].sort(function (inA, inB) {
var result = 0;
var a, b, pattern = /(\d+)/;
var as = inA.split(pattern);
var bs = inB.split(pattern);
var index, count = as.length;
if (('' === as[0]) === ('' === bs[0])) {
if (count > bs.length)
count = bs.length;
for (index = 0; index < count && 0 === result; ++index) {
a = as[index]; b = bs[index];
if (index & 1) {
result = a - b;
} else {
result = !(a < b) ? (a > b) ? 1 : 0 : -1;
}
}
if (0 === result)
result = as.length - bs.length;
} else {
result = !(inA < inB) ? (inA > inB) ? 1 : 0 : -1;
}
return result;
}).toString();
Result: "!!,9,9.5,10,a3,a3b,a3b3,a3b20,a3b100,a20,a100,~~"

Another variant is to use an instance of Intl.Collator with the numeric option:
var array = ["a100", "a20", "a3", "a3b", "a3b100", "a3b20", "a3b3", "!!", "~~", "9", "10", "9.5"];
var collator = new Intl.Collator([], {numeric: true});
array.sort((a, b) => collator.compare(a, b));
console.log(array);

I think this does what you want
function sortArray(arr) {
var tempArr = [], n;
for (var i in arr) {
tempArr[i] = arr[i].match(/([^0-9]+)|([0-9]+)/g);
for (var j in tempArr[i]) {
if( ! isNaN(n = parseInt(tempArr[i][j])) ){
tempArr[i][j] = n;
}
}
}
tempArr.sort(function (x, y) {
for (var i in x) {
if (y.length < i || x[i] < y[i]) {
return -1; // x is longer
}
if (x[i] > y[i]) {
return 1;
}
}
return 0;
});
for (var i in tempArr) {
arr[i] = tempArr[i].join('');
}
return arr;
}
alert(
sortArray(["a1b3", "a10b11", "a10b2", "a9b2"]).join(",")
);

Assuming you want to just do a numeric sort by the digits in each array entry (ignoring the non-digits), you can use this:
function sortByDigits(array) {
var re = /\D/g;
array.sort(function(a, b) {
return(parseInt(a.replace(re, ""), 10) - parseInt(b.replace(re, ""), 10));
});
return(array);
}
It uses a custom sort function that removes the digits and converts to a number each time it's asked to do a comparison. You can see it work here: http://jsfiddle.net/jfriend00/t87m2/.

Use this compare function for sorting...
function compareLists(a, b) {
var alist = a.split(/(\d+)/), // Split text on change from anything
// to digit and digit to anything
blist = b.split(/(\d+)/); // Split text on change from anything
// to digit and digit to anything
alist.slice(-1) == '' ? alist.pop() : null; // Remove the last element if empty
blist.slice(-1) == '' ? blist.pop() : null; // Remove the last element if empty
for (var i = 0, len = alist.length; i < len; i++) {
if (alist[i] != blist[i]){ // Find the first non-equal part
if (alist[i].match(/\d/)) // If numeric
{
return +alist[i] - +blist[i]; // Compare as number
} else {
return alist[i].localeCompare(blist[i]); // Compare as string
}
}
}
return true;
}
Syntax
var data = ["a1b3", "a10b11", "b10b2", "a9b2", "a1b20", "a1c4"];
data.sort(compareLists);
alert(data);
There is a demo at http://jsfiddle.net/h9Rqr/7/.

Here's a more complete solution that sorts according to both letters and numbers in the strings
function sort(list) {
var i, l, mi, ml, x;
// copy the original array
list = list.slice(0);
// split the strings, converting numeric (integer) parts to integers
// and leaving letters as strings
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].match(/(\d+|[a-z]+)/g);
for( mi = 0, ml = list[i].length; mi < ml ; mi++ ) {
x = parseInt(list[i][mi], 10);
list[i][mi] = !!x || x === 0 ? x : list[i][mi];
}
}
// sort deeply, without comparing integers as strings
list = list.sort(function(a, b) {
var i = 0, l = a.length, res = 0;
while( res === 0 && i < l) {
if( a[i] !== b[i] ) {
res = a[i] < b[i] ? -1 : 1;
break;
}
// If you want to ignore the letters, and only sort by numbers
// use this instead:
//
// if( typeof a[i] === "number" && a[i] !== b[i] ) {
// res = a[i] < b[i] ? -1 : 1;
// break;
// }
i++;
}
return res;
});
// glue it together again
for( i = 0, l = list.length; i < l; i++ ) {
list[i] = list[i].join("");
}
return list;
}

I needed a way to take a mixed string and create a string that could be sorted elsewhere, so that numbers sorted numerically and letters alphabetically. Based on answers above I created the following, which pads out all numbers in a way I can understand, wherever they appear in the string.
function padAllNumbers(strIn) {
// Used to create mixed strings that sort numerically as well as non-numerically
var patternDigits = /(\d+)/g; // This recognises digit/non-digit boundaries
var astrIn = strIn.split( patternDigits ); // we create an array of alternating digit/non-digit groups
var result = "";
for (var i=0;i<astrIn.length; i++) {
if (astrIn[i] != "") { // first and last elements can be "" and we don't want these padded out
if (isNaN(astrIn[i])) {
result += astrIn[i];
} else {
result += padOneNumberString("000000000",astrIn[i]);
}
}
}
return result;
}
function padOneNumberString(pad,strNum,left) {
// Pad out a string at left (or right)
if (typeof strNum === "undefined") return pad;
if (typeof left === "undefined") left = true;
var padLen = pad.length - (""+ strNum).length;
var padding = pad.substr(0,padLen);
return left? padding + strNum : strNum + padding;
}

Sorting occurs from left to right unless you create a custom algorithm. Letters or digits are compared digits first and then letters.
However, what you want to accomplish as per your own example (a1, a9, a10) won’t ever happen. That would require you knowing the data beforehand and splitting the string in every possible way before applying the sorting.
One final alternative would be:
a) break each and every string from left to right whenever there is a change from letter to digit and vice versa; &
b) then start the sorting on those groups from right-to-left. That will be a very demanding algorithm. Can be done!
Finally, if you are the generator of the original "text", you should consider NORMALIZING the output where a1 a9 a10 could be outputted as a01 a09 a10. This way you could have full control of the final version of the algorithm.

How to get the next letter of the alphabet in Javascript?

I am build an autocomplete that searches off of a CouchDB View.
I need to be able to take the final character of the input string, and replace the last character with the next letter of the english alphabet. (No need for i18n here)
For Example:
Input String = "b"
startkey = "b"
endkey = "c"
OR
Input String = "foo"
startkey = "foo"
endkey = "fop"
(in case you're wondering, I'm making sure to include the option inclusive_end=false so that this extra character doesn't taint my resultset)
The Question
Is there a function natively in Javascript that can just get the next letter of the alphabet?
Or will I just need to suck it up and do my own fancy function with a base string like "abc...xyz" and indexOf()?

my_string.substring(0, my_string.length - 1)
+ String.fromCharCode(my_string.charCodeAt(my_string.length - 1) + 1)

// This will return A for Z and a for z.
function nextLetter(s){
return s.replace(/([a-zA-Z])[^a-zA-Z]*$/, function(a){
var c= a.charCodeAt(0);
switch(c){
case 90: return 'A';
case 122: return 'a';
default: return String.fromCharCode(++c);
}
});
}

A more comprehensive solution, which gets the next letter according to how MS Excel numbers it's columns... A B C ... Y Z AA AB ... AZ BA ... ZZ AAA
This works with small letters, but you can easily extend it for caps too.
getNextKey = function(key) {
if (key === 'Z' || key === 'z') {
return String.fromCharCode(key.charCodeAt() - 25) + String.fromCharCode(key.charCodeAt() - 25); // AA or aa
} else {
var lastChar = key.slice(-1);
var sub = key.slice(0, -1);
if (lastChar === 'Z' || lastChar === 'z') {
// If a string of length > 1 ends in Z/z,
// increment the string (excluding the last Z/z) recursively,
// and append A/a (depending on casing) to it
return getNextKey(sub) + String.fromCharCode(lastChar.charCodeAt() - 25);
} else {
// (take till last char) append with (increment last char)
return sub + String.fromCharCode(lastChar.charCodeAt() + 1);
}
}
return key;
};

Here is a function that does the same thing (except for upper case only, but that's easy to change) but uses slice only once and is iterative rather than recursive. In a quick benchmark, it's about 4 times faster (which is only relevant if you make really heavy use of it!).
function nextString(str) {
if (! str)
return 'A' // return 'A' if str is empty or null
let tail = ''
let i = str.length -1
let char = str[i]
// find the index of the first character from the right that is not a 'Z'
while (char === 'Z' && i > 0) {
i--
char = str[i]
tail = 'A' + tail // tail contains a string of 'A'
}
if (char === 'Z') // the string was made only of 'Z'
return 'AA' + tail
// increment the character that was not a 'Z'
return str.slice(0, i) + String.fromCharCode(char.charCodeAt(0) + 1) + tail
}

Just to explain the main part of the code that Bipul Yadav wrote (can't comment yet due to lack of reps). Without considering the loop, and just taking the char "a" as an example:
"a".charCodeAt(0) = 97...hence "a".charCodeAt(0) + 1 = 98 and String.fromCharCode(98) = "b"...so the following function for any letter will return the next letter in the alphabet:
function nextLetterInAlphabet(letter) {
if (letter == "z") {
return "a";
} else if (letter == "Z") {
return "A";
} else {
return String.fromCharCode(letter.charCodeAt(0) + 1);
}
}

var input = "Hello";
var result = ""
for(var i=0;i<input.length;i++)
{
var curr = String.fromCharCode(input.charCodeAt(i)+1);
result = result +curr;
}
console.log(result);

I understand the original question was about moving the last letter of the string forward to the next letter. But I came to this question more interested personally in changing all the letters in the string, then being able to undo that. So I took the code written by Bipul Yadav and I added some more code. The below code takes a series of letters, increments each of them to the next letter maintaining case (and enables Zz to become Aa), then rolls them back to the previous letter (and allows Aa to go back to Zz).
var inputValue = "AaZzHello";
console.log( "starting value=[" + inputValue + "]" );
var resultFromIncrementing = ""
for( var i = 0; i < inputValue.length; i++ ) {
var curr = String.fromCharCode( inputValue.charCodeAt(i) + 1 );
if( curr == "[" ) curr = "A";
if( curr == "{" ) curr = "a";
resultFromIncrementing = resultFromIncrementing + curr;
}
console.log( "resultFromIncrementing=[" + resultFromIncrementing + "]" );
inputValue = resultFromIncrementing;
var resultFromDecrementing = "";
for( var i2 = 0; i2 < inputValue.length; i2++ ) {
var curr2 = String.fromCharCode( inputValue.charCodeAt(i2) - 1 );
if( curr2 == "#" ) curr2 = "Z";
if( curr2 == "`" ) curr2 = "z";
resultFromDecrementing = resultFromDecrementing + curr2;
}
console.log( "resultFromDecrementing=[" + resultFromDecrementing + "]" );
The output of this is:
starting value=[AaZzHello]
resultFromIncrementing=[BbAaIfmmp]
resultFromDecrementing=[AaZzHello]