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I'm trying to solve a problem similar to the popular fizzbuzz or pingpong question. I already figured out how to write this portion of the code, however I also need to make a counter in which the user inputs their number and the page gives back 1 to the number they input, as well as the fizzbuzz/popcorn numbers. I think the code is something along the lines of
for (i = 1; i <= 20; i++) {...}
But I don't know how to combine this with my other code
var pingpong = function(number) {
if (number % 15 === 0) {
return 'pingpong';
} else if (number % 5 === 0) {
return 'pong';
} else if (number % 3 === 0) {
return 'ping';
} else {
return false;
}
};
Sorry, I know this is a super beginner question, but I'm just starting out and am having a hard time figuring out how everything works together.
You are close. To read more about JavaScripts loops.
var pingpong = function(number) {
if (number % 15 === 0) {
return 'pingpong';
} else if (number % 5 === 0) {
return 'pong';
} else if (number % 3 === 0) {
return 'ping';
} else {
return false;
}
};
for (var i = 1; i <= 20; i++) {
var num = pingpong(i);
// Just show it on the screen
document.body.innerHTML += i + ': ' + num + '<br />';
}
This question already has answers here:
How do you convert numbers between different bases in JavaScript?
(21 answers)
Closed 1 year ago.
In JavaScript, is there any built-in function for converting an integer from one given base to another given base? I've noticed that it's already possible to convert a decimal number to another base using toString(numberToConvertTo), but I haven't yet found a general-purpose function that can convert from any base to any other base, as shown:
convertFrom(toConvert, baseToConvertFrom, baseToConvertTo){
//convert the number from baseToConvertFrom to BaseToConvertTo
}
Call parseInt(str, fromBase) to convert to base 10 (or rather, to an actual number), then call num.toString(toBase).
You might want to customise your input or output. You might want more than 36 bases. So I made this
function baseCon(number,fromBase,toBase,fromChars,toChars) {
if (!fromChars) {
fromChars = ['0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
number = number.toString().toUpperCase();
}
if (!toChars) {
toChars = ['0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z'];
}
if (typeof number != 'object') {
number = number.toString();
}
if (fromBase > fromChars.length) {
console.log('fromBase was too high');
} else if (toBase > toChars.length) {
console.log('toBase was too high');
} else {
if (fromBase == 10) {
var base10Num = Math.min(number);
} else {
var base10Num = 0;
var place = 0;
for (var index = number.length - 1; index > -1; index--) {
base10Num = base10Num + (fromChars.indexOf(number[index]) * Math.pow(fromBase,place));
place++;
}
}
var string = '';
var looping = true;
var stringLen = 0;
var stringVal = 0;
while (looping) {
stringLen++;
if (Math.pow(toBase,stringLen) == base10Num) {
stringLen++;
break;
} else if (Math.pow(toBase,stringLen) >= base10Num) {
break;
}
}
for (var placePos = stringLen; placePos > 0; placePos--) {
for (var placeVal = 0; placeVal < (toBase + 1); placeVal++) {
if (((placeVal * Math.pow(toBase,placePos - 1)) > (base10Num - stringVal)) || (placeVal == 0) && ((base10Num - stringVal) == 0)) {
if (!((placeVal == 0) && ((base10Num - stringVal) == 0))) {
stringVal = stringVal + ((placeVal - 1) * Math.pow(toBase,placePos - 1));
string = string + toChars[placeVal - 1];
} else {
string = string + toChars[0];
}
break;
}
}
}
if (stringVal == base10Num) {
return string;
} else {
console.log('baseCon failed');
return string;
}
}
};
You can customise the input chars so you can do things like this
> baseCon([true,true,false,true,false],2,10,[false,true]);
< "26"
and of coarse the output chars
> baseCon('3942',10,8,false,['!','#','#','$','%','^','&','*']);
< "*^%&"
It took me all day and then I remembered parseInt(number,fromBase).toString(toBase); haha
I'm looking for a way to construct regular expressions to match numeric inputs specified by a given integer range, ie. if I pass in a range of 1,3-4 then a regex would be returned matching just 1, 3 and 4.
I wrote the following method to try and do this:
function generateRegex(values) {
if (values == "*") {
return new RegExp("^[0-9]+$");
} else {
return new RegExp("^[" + values + "]+$");
}
}
I'm having issues however as sometimes I need to match double digits, such as "8-16", and I also need to ensure that if I am passed a single digit value, such as "1", that the generated regex matches only 1, and not say 11.
I really would like this to remain a pretty small snippet of code, but am not sure enough about regexs to know how to do this. Would be massively grateful for any help!
EDIT: I realise I wasn't clear, with my original paragraph, so have edited it. I realise the regex's that I originally generated do not work at all
Regexes don't know anything about numbers, only digits. So [8-16] is invalid because you say match between 8 and 1 (instead of 1 and 8 e.g.) plus the digit 6.
If you want to match numbers, you have to consider them lexically. For example, to match numbers between 1 and 30, you have to write something like (other regexes exist):
/^(30|[1-2]\d|[1-9])$/
I was sure it was 4-8 hours :-) In the end (and in its uselessness) it was a good exercise in composing Regexes. You are free to try it. If we exclude one use of continue and the use of the Array constructor, it's fully jsLint ok.
var BuildRegex = function(matches) {
"use strict";
var splits = matches.split(','),
res = '^(',
i, subSplit, min, max, temp, tempMin;
if (splits.length === 0) {
return new RegExp('^()$');
}
for (i = 0; i < splits.length; i += 1) {
if (splits[i] === '*') {
return new RegExp('^([0-9]+)$');
}
subSplit = splits[i].split('-');
min = BuildRegex.Trim(subSplit[0], '0');
if (min === '') {
return null;
}
if (subSplit.length === 1) {
res += min;
res += '|';
continue;
} else if (subSplit.length > 2) {
return null;
}
max = BuildRegex.Trim(subSplit[1], '0');
if (max === '') {
return null;
}
if (min.length > max.length) {
return null;
}
// For 2-998 we first produce 2-9, then 10-99
temp = BuildRegex.DifferentLength(res, min, max);
tempMin = temp.min;
if (tempMin === null) {
return null;
}
res = temp.res;
// Then here 100-998
res = BuildRegex.SameLength(res, tempMin, max);
}
res = res.substr(0, res.length - 1);
res += ')$';
return new RegExp(res);
};
BuildRegex.Repeat = function(ch, n) {
"use strict";
return new Array(n + 1).join(ch);
};
BuildRegex.Trim = function(str, ch) {
"use strict";
var i = 0;
while (i < str.length && str[i] === ch) {
i += 1;
}
return str.substr(i);
};
BuildRegex.IsOnlyDigit = function(str, start, digit) {
"use strict";
var i;
for (i = start; i < str.length; i += 1) {
if (str[i] !== digit) {
return false;
}
}
return true;
};
BuildRegex.RangeDigit = function(min, max) {
"use strict";
if (min === max) {
return min;
}
return '[' + min + '-' + max + ']';
};
BuildRegex.DifferentLength = function(res, min, max) {
"use strict";
var tempMin = min,
i, tempMax;
for (i = min.length; i < max.length; i += 1) {
tempMax = BuildRegex.Repeat('9', i);
res = BuildRegex.SameLength(res, tempMin, tempMax);
tempMin = '1' + BuildRegex.Repeat('0', i);
}
if (tempMin > tempMax) {
return null;
}
return {
min: tempMin,
res: res
};
};
BuildRegex.SameLength = function(res, min, max) {
"use strict";
var commonPart;
// 100-100
if (min === max) {
res += min;
res += '|';
return res;
}
for (commonPart = 0; commonPart < min.length; commonPart += 1) {
if (min[commonPart] !== max[commonPart]) {
break;
}
}
res = BuildRegex.RecursivelyAddRange(res, min.substr(0, commonPart), min.substr(commonPart), max.substr(commonPart));
return res;
};
BuildRegex.RecursivelyAddRange = function(res, prefix, min, max) {
"use strict";
var only0Min, only9Max, i, middleMin, middleMax;
if (min.length === 1) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
res += '|';
return res;
}
// Check if
only0Min = BuildRegex.IsOnlyDigit(min, 1, '0');
only9Max = BuildRegex.IsOnlyDigit(max, 1, '9');
if (only0Min && only9Max) {
res += prefix;
res += BuildRegex.RangeDigit(min[0], max[0]);
for (i = 1; i < min.length; i += 1) {
res += '[0-9]';
}
res += '|';
return res;
}
middleMin = min;
if (!only0Min) {
res = BuildRegex.RecursivelyAddRange(res, prefix + min[0], min.substr(1), BuildRegex.Repeat('9', min.length - 1));
if (min[0] !== '9') {
middleMin = String.fromCharCode(min.charCodeAt(0) + 1) + BuildRegex.Repeat('0', min.length - 1);
} else {
middleMin = null;
}
}
middleMax = max;
if (!only9Max) {
if (max[0] !== '0') {
middleMax = String.fromCharCode(max.charCodeAt(0) - 1) + BuildRegex.Repeat('9', max.length - 1);
} else {
middleMax = null;
}
}
if (middleMin !== null && middleMax !== null && middleMin[0] <= middleMax[0]) {
res = BuildRegex.RecursivelyAddRange(res, prefix + BuildRegex.RangeDigit(middleMin[0], middleMax[0]), middleMin.substr(1), middleMax.substr(1));
}
if (!only9Max) {
res = BuildRegex.RecursivelyAddRange(res, prefix + max[0], BuildRegex.Repeat('0', max.length - 1), max.substr(1));
}
return res;
};
// ----------------------------------------------------------
var printRegex = function(p) {
"use strict";
document.write(p + ': ' + BuildRegex(p) + '<br>');
};
printRegex('*');
printRegex('1');
printRegex('1,*');
printRegex('1,2,3,4');
printRegex('1,11-88');
printRegex('1,11-88,90-101');
printRegex('1-11111');
printRegex('75-11119');
Test here http://jsfiddle.net/dnqYV/
The C# version is here http://ideone.com/3aEt3E
I'm not sure there is a (sane) way to test integer ranges with RegExp. I believe you're fixated on RegExp, where there are much simpler (more flexible) approaches. Take a look at IntRangeTest().
var range = new IntRangeTest('0,10-20');
console.log(
"0,10-20",
range.test("") == false,
range.test("-5") == false,
range.test("0") == true,
range.test("5") == false,
range.test("11") == true,
range.test("123.23") == false
);
If you feel like it, you can easily add this to Number.prototype. You could also quite easily make this an extension to RegExp, if that's what you're worried about.
Ok so it seems that there are 4 main cases that I need to address:
Single digits, ie 1, would simply generate the regex /^1$/
Multiple digits, ie 12, would require the regex /^12&/
Single digit ranges, ie 3-6, would generate the regex /^[3-6]$/
And finally, multiple digit ranges work in a similar method to multiple digits but with a range, ie 11-14 would become /^1[1-4]$/. These would need to be split into multiple regexes if they span over multiple start digits, Ie 23-31 would become /^2[3-9]|3[0-1]$/
Therefore, all I need to do is identify each of these cases and create a compound regex using | like xanatos suggested. Ie, to match all of the above criteria would generate a regex like:
/^( 1 | 12 | [3-6] | 1[1-4] | 2[3-9]|3[0-1] )$/
Do other agree this seems like a decent way to progress?
a script returns either a number like 0.0580 so in x.xxxx format or a (x) for X units left.
I want to format the number 0.0580 and return 5.8 cent or return x units left.
Any ideas how to do that in javascript? Especially how do I format the x.xxxx?
In case the first x is not 0 I want to return e.g. 1.75$.
MS has written a nice plugin for jquery. it's especially useful if you're localizing. Give it a go:
http://weblogs.asp.net/scottgu/archive/2010/06/10/jquery-globalization-plugin-from-microsoft.aspx
I'm not sure if this can be used outside of jquery...
I may be spoiling you here, but whatever. Here's a function that I found somewhere at some point and have been recycling since. I haven't actually bothered to look much into it to figure out what it does exactly, but it has been rather useful:
function FormatMoneyAmount(starting_string, ending_string) {
//check validity of input (true = invalid, false = valid)
var valid_exp = new RegExp ('[^0-9,.$]', 'gi');
input_invalid = (typeof(ending_string) == 'undefined' && valid_exp.test(starting_string));
//check if more than 2 digits follow decimal or no decimal
decimal_invalid = typeof(ending_string) == 'undefined' && (starting_string.indexOf('.') > -1) && ((starting_string.length - starting_string.indexOf('.')) > 3);
if (input_invalid || decimal_invalid) {
ending_string = starting_string;
} else {
//remove commas, dollar signs
var replace_exp = new RegExp ('[,$]', 'gi');
starting_string = starting_string.replace(replace_exp, '');
//remove decimal if ending string not set, save for adding on later
var decimal_substring = '';
if (typeof(ending_string) == 'undefined' && starting_string.indexOf('.') > -1) {
decimal_substring = starting_string.substring(starting_string.indexOf('.'), starting_string.length);
remaining_string = starting_string.substring(0,starting_string.indexOf('.'));
} else {
remaining_string = starting_string;
}
//if string is already 3 characters or less, do nothing
if (remaining_string.length > 3) {
//separate last 3 characters of string from rest of string
var final_three = remaining_string.substring(remaining_string.length - 3, remaining_string.length);
remaining_string = remaining_string.substring(0, remaining_string.length - 3);
//if not first group of 3, add new group before old group with comma, else set to new group
ending_string = (typeof(ending_string) == 'undefined') ? final_three + ((typeof(decimal_substring) == 'undefined') ? '' : decimal_substring) : final_three + ',' + ending_string;
//call function again if more than 3 digits remaining to process, else add to end string
if (remaining_string.length > 3) {
ending_string = FormatMoneyAmount(remaining_string, ending_string);
} else {
ending_string = remaining_string + ',' + ending_string;
}
} else {
ending_string = (typeof(ending_string) == 'undefined') ? remaining_string : remaining_string + ',' + ending_string + ((typeof(decimal_substring) == 'undefined') ? '' : decimal_substring);
}
}
return ending_string;
}
The first thing to do is check the format of the string, since you will have two code paths depending on the result:
if (typeof num = "string" && num.slice(0,1) == "(" && num.slice(-1) == ")") {
// String is in the format (x), so we just need to return that number
return num.slice(1,-1) + " units left";
}
The next part is to check if the number is less than 1, indicating that it is cents and not whole dollars. If it is less than 1, multiplying it by 100 will give you the number of cents you're after:
if (+num < 1)
// 0.0580 * 100 = 5.8
return (num * 100) + " cents";
else
return +num + "$";
I want to be able to test whether a value is within a number range. This is my current code:
if ((year < 2099) && (year > 1990)){
return 'good stuff';
}
Is there a simpler way to do this? For example, is there something like this?
if (1990 < year < 2099){
return 'good stuff';
}
In many languages, the second way will be evaluated from left to right incorrectly with regard to what you want.
In C, for instance, 1990 < year will evaluate to 0 or 1, which then becomes 1 < 2099, which is always true, of course.
Javascript is a quite similar to C: 1990 < year returns true or false, and those boolean expressions seem to numerically compare equal to 0 and 1 respectively.
But in C#, it won't even compile, giving you the error:
error CS0019: Operator '<' cannot be applied to operands of type 'bool' and 'int'
You get a similar error from Ruby, while Haskell tells you that you cannot use < twice in the same infix expression.
Off the top of my head, Python is the only language that I'm sure handles the "between" setup that way:
>>> year = 5
>>> 1990 < year < 2099
False
>>> year = 2000
>>> 1990 < year < 2099
True
The bottom line is that the first way (x < y && y < z) is always your safest bet.
You could make your own method:
// jquery
$(function() {
var myNumber = 100;
try {
if (myNumber.isBetween(50, 150))
alert(myNumber + " is between 50 and 100.");
else
alert(myNumber + " is not between 50 and 100.");
} catch (e) {
alert(e.message());
}
});
// js prototype
if (typeof(Number.prototype.isBetween) === "undefined") {
Number.prototype.isBetween = function(min, max, notBoundaries) {
var between = false;
if (notBoundaries) {
if ((this < max) && (this > min)) between = true;
alert('notBoundaries');
} else {
if ((this <= max) && (this >= min)) between = true;
alert('Boundaries');
}
alert('here');
return between;
}
}
hope this helps.
Max
The fast and simple way to make this is to create a function like this:
function inRange(n, nStart, nEnd)
{
if(n>=nStart && n<=nEnd) return true;
else return false;
}
Then use that as follows:
inRange(500, 200, 1000) => this return true;
Or like this:
inRange(199, 200, 1000) => this return false;
If you don't like the boolean operator, you could always use nested if statements:
if (1990 < year)
{
if( year < 2099)
return 'good stuff';
}
From a similar solution here: http://indisnip.wordpress.com/2010/08/26/quicktip-check-if-a-number-is-between-two-numbers/
$.fn.between = function(a,b){
return (a < b ? this[0] >= a && this[0] <= b : this[0] >= b && this[0] <= a);
}