We Keep Coding

RegExpr and variable

I have just one question maybe stupid (like every day)
var word = []; (an array with 100 words for example)
var tab = []; // resultat
var root = "test";
var debut = "Anti";
var reg1=new RegExp("^"+debut + "+." + root,"g")
for(var i = 0;i<word.length; i++){
// a word begin with Anti and contain test pls
if (word[i].match(reg1)){´
tab.push(word[i])
}
}
console.log(tab.join(', ');
but it is dont work, i dont know how to use variable with regexpr, thanks, sorry for my english

var r = new RegExp('anti.*esis', 'ig')
document.write('antithesis'.match(r), '<br/>') // ["antithesis"]
document.write('antihero'.match(r), '<br/>') // null

Here is the code, but is used the test() instead of match()
var word=["yea","antiboyahtest","antigssjshbztest"];
var debut="anti";
var root="test";
var reg=new RegExp("^"+debut+".*"+root,"g");
var tabs=[];
for(i in word){
if(reg.test(word[i])){
tabs.push(word[i]);
}
}
alert(tabs);

The solution using RegExp.test and Array.filter functions:
var word = ['Antitest', 'Antidot', 'Anti-next-test', 'testAnti'],
root = "test", debut = "Anti",
reg1 = new RegExp("^"+debut + ".*?" + root, "g");
var result = word.filter(function (w) {
return reg1.test(w);
});
console.log(result); // ["Antitest", "Anti-next-test"]
Also, there's an additional approach using Array.indexOf function without any regex which will give the same result:
...
var result = word.filter(function (w) {
return w.indexOf(debut) === 0 && w.indexOf(root) !== -1;
});

Find smallest substring containing a given set of letters in a larger string

Say you have the following string:
FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNT
LDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFY
FFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQ
XBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR
AMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR
I'm trying to find the smallest substring containing the letters ABCDA.
I tried a regex approach.
console.log(str.match(/[A].*?[B].*?[C].*?[D].*?[A]/gm).sort((a, b) => a.length - b.length)[0]);
This works, but it only find strings where ABCDA appear (in that order). Meaning it won't find substring where the letters appear in a order like this: BCDAA
I'm trying to change my regex to account for this. How would I do that without using | and type out all the different cases?

You can't.
Let's consider a special case: Assume the letters you are looking for are A, A, and B. At some point in your regexp there will certainly be a B. However, the parts to the left and to the right of the B are independent of each other, so you cannot refer from one to the other. How many As are matched in the subexpression to the right of the B depends on the number of As being already matched in the left part. This is not possible with regular expressions, so you will have to unfold all the different orders, which can be many!
Another popular example that illustrates the problem is to match opening brackets with closing brackets. It's not possible to write a regular expression asserting that in a given string a sequence of opening brackets is followed by a sequence of closing brackets of the same length. The reason for this is that to count the brackets you would need a stack machine in contrast to a finite state machine but regular expressions are limited to patterns that can be matched using FSMs.

This algorithm doesn't use a regex, but found both solutions as well.
var haystack = 'FJKAUNOJDCUTCRHBYDLXKEODVBWTYPTSHASQQFCPRMLDXIJMYPVOHBDUGSMBLMVUMMZYHULSUIZIMZTICQORLNTOVKVAMQTKHVRIFMNTSLYGHEHFAHWWATLYAPEXTHEPKJUGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGRAMELUTEPYILBIUOCKKUUBJROQFTXMZRLXBAMHSDTEKRRIKZUFNLGTQAEUINMBPYTWXULQNIIRXHHGQDPENXAJNWXULFBNKBRINUMTRBFWBYVNKNKDFR';
var needle = 'ABCDA'; // the order of letters doesn't matter
var letters = {};
needle.split('').forEach(function(ch) {
letters[ch] = letters[ch] || 0;
letters[ch]++;
});
var shortestSubstringLength = haystack.length;
var shortestSubstrings = []; // storage for found substrings
var startingPos = 0;
var length;
var currentPos;
var notFound;
var letterKeys = Object.keys(letters); // unique leters
do {
lettersLeft = JSON.parse(JSON.stringify(letters)); // copy letters count object
notFound = false;
posStart = haystack.length;
posEnd = 0;
letterKeys.forEach(function(ch) {
currentPos = startingPos;
while (!notFound && lettersLeft[ch] > 0) {
currentPos = haystack.indexOf(ch, currentPos);
if (currentPos >= 0) {
lettersLeft[ch]--;
posStart = Math.min(currentPos, posStart);
posEnd = Math.max(currentPos, posEnd);
currentPos++;
} else {
notFound = true;
}
}
});
if (!notFound) {
length = posEnd - posStart + 1;
startingPos = posStart + 1; // starting position for next iteration
}
if (!notFound && length === shortestSubstringLength) {
shortestSubstrings.push(haystack.substr(posStart, length));
}
if (!notFound && length < shortestSubstringLength) {
shortestSubstrings = [haystack.substr(posStart, length)];
shortestSubstringLength = length;
}
} while (!notFound);
console.log(shortestSubstrings);

Maybe not as clear as using regex could be (well, for me regex are never really clear :D ) you can use brute force (not so brute)
Create an index of "valid" points of your string (those with the letters you want) and iterate with a double loop over it getting substrings containing at least 5 of those points, checking that they are valid solutions. Maybe not the most efficient way, but easy to implement, to understand, and probably to optimize.
var haystack="UGDVWUDDPRQLUZMSZOJPSIKAIHLTONYXAULECXXKWFQOIKELWOHRVRUCXIAASKHMWTMAJEWGEESLWRTQKVHRRCDYXNTLDSUPXMQTQDFAQAPYBGXPOLOCLFQNGNKPKOBHZWHRXAWAWJKMTJSLDLNHMUGVVOPSAMRUJEYUOBPFNEHPZZCLPNZKWMTCXERPZRFKSXVEZTYCXFRHRGEITWHRRYPWSVAYBUHCERJXDCYAVICPTNBGIODLYLMEYLISEYNXNMCDPJJRCTLYNFMJZQNCLAGHUDVLYIGASGXSZYPZKLAWQUDVNTWGFFYFFSMQWUNUPZRJMTHACFELGHDZEJWFDWVPYOZEVEJKQWHQAHOCIYWGVLPSHFESCGEUCJGYLGDWPIWIDWZZXRUFXERABQJOXZALQOCSAYBRHXQQGUDADYSORTYZQPWGMBLNAQOFODSNXSZFURUNPMZGHTAJUJROIGMRKIZHSFUSKIZJJTLGOEEPBMIXISDHOAIFNFEKKSLEXSJLSGLCYYFEQBKIZZTQQXBQZAPXAAIFQEIXELQEZGFEPCKFPGXULLAHXTSRXDEMKFKABUTAABSLNQBNMXNEPODPGAORYJXCHCGKECLJVRBPRLHORREEIZOBSHDSCETTTNFTSMQPQIJBLKNZDMXOTRBNMTKHHCZQQMSLOAXJQKRHDGZVGITHYGVDXRTVBJEAHYBYRYKJAVXPOKHFFMEPHAGFOOPFNKQAUGYLVPWUJUPCUGGIXGR";
var needle="ABCD";
var size=haystack.length;
var candidate_substring="";
var minimal_length=size;
var solutions=new Array();
var points=Array();
for(var i=0;i<size;i++){
if(needle.indexOf(haystack[i])>-1) points.push(i);
}
var limit_i= points.length-4;
var limit_k= points.length;
for (var i=0;i<limit_i;i++){
for(var k=i;k<limit_k;k++){
if(points[k]-points[i]+1<=minimal_length){
candidate_substring=haystack.substr(points[i],points[k]-points[i]+1);
if(is_valid(candidate_substring)){
solutions.push(candidate_substring);
if(candidate_substring.length < minimal_length) minimal_length=candidate_substring.length;
}
}
}
}
document.write('<p>Solution length:'+minimal_length+'<p>');
for(var i=0;i<solutions.length;i++){
if(solutions[i].length<=minimal_length) document.write('<p>Solution:'+solutions[i]+'<p>');
}
function is_valid(candidate_substring){
//verify we've got all characters
for(var j=0;j<candidate_substring.length;j++){
if(candidate_substring.indexOf(needle.charAt(j))<0) return false;
}
//...and verify we have two "A"
if(candidate_substring.indexOf("A")==candidate_substring.lastIndexOf("A")) return false;
return true;
}

Just had this problem in an interview as a coding assignment and came up with another solution, (it's not as optimal as the one above but maybe it's easier to understand).
function MinWindowSubstring(strArr) {
const N = strArr[0];
const K = strArr[1];
const letters = {};
K.split('').forEach( (character) => {
letters[character] = letters[character] ? letters[character] + 1 : 1;
});
let possibleSequencesList = [];
const letterKeys = Object.keys(letters);
for(let i=0; i< N.length; i++) {
const char = N[i];
if (new String(letterKeys).indexOf(char) !== -1) {
// found a character in the string
// update all previus sequences
possibleSequencesList.forEach((seq) => {
if(!seq.sequenceComplete) {
seq[char] = seq[char]-1;
seq.lastIndex = i;
// check if sequence is complete
var sequenceComplete = true;
letterKeys.forEach( (letter) => {
if(seq[letter] > 0) {
sequenceComplete = false;
}
});
seq.sequenceComplete = sequenceComplete
}
})
// create a new sequence starting from it
const newSeq = {
startPoint: i,
lastIndex: i,
sequenceComplete: false,
...letters
}
newSeq[char] = newSeq[char]-1;
possibleSequencesList.push(newSeq);
}
}
// cleanup sequences
let sequencesList = possibleSequencesList.filter(sequence => sequence.sequenceComplete);
let output = [];
let minLength = N.length;
// find the smalles one
sequencesList.forEach( seq => {
if( (seq.lastIndex - seq.startPoint) < minLength) {
minLength = seq.lastIndex - seq.startPoint;
output = N.substring(seq.startPoint, seq.lastIndex + 1);
}
})
return output;
}

How to efficiently check if any substring in an array is contained in another string

I want to have something that checks if any substring in a list of strings is included in string. I have something that works, but I'm hoping there is something cleaner and more efficient, preferably just in one line I can call like if(string.contains(list.any)) or something.
var list = ['aa','bb','cc','dd'];
var string = "nygaard"; // true because "aa" is in "nygaard".
for (var i = 0; i < list.length; i++) {
if( string.indexOf( list[i] ) > -1 ) {
alert("True");
break;
}
}

var list = ['aa','bb','cc','dd'];
var string = "nygaard";
var patt = new RegExp(list.join('|'))
// regEx - /aa|bb|cc|dd/
patt.test(string)
//output true
Demo
Explanation
Dynamically create a regEx using new RegExp
for checking existance of multiple SubString we have | operator in RegEx
Use Array.join('|') to dynamically make a regExp like this aa|bb|cc
use test func to validate String
Edit - For Complex Cases - Problem is strings in list may have to be escaped for RegEx to work, Pointed By - #GabrielRatener
Compairing to my solu, #GabrielRatener's solution is better
var list = ['aa','bb','cc','dd', 'ab?', '(aa)'];
list = list.sort(function(a, b) {
if (a>b) return -1;
else if (a<b) return 1;
else return 0;
});
list = list.join(" ").replace(/[^\w\s]/g, function($1) {
return '\\' + $1
}).split(/\s/);
//["dd", "cc", "bb", "ab\?", "aa", "\(aa\)"]
//sorting needed to match "ab?" before single "a" as "b" will become optional
//due to "?"
//after processing
var string = "nygaard";
var patt = new RegExp(list.join('|')) // RegExp -> /dd|cc|bb|ab\?|aa|\(aa\)/
patt.test(string)
//true

Why not just put your loop in its own function you can call?
function stringContains(string, list) {
for (var i = 0; i < list.length; i++) {
if( string.indexOf( list[i] ) > -1 )
return true;
}
return false;
}
and then call it like this:
var list = ['aa','bb','cc','dd'];
var string = "nygaard";
if(stringContains(string, list))
alert("True");
If you're looking for a javascript library function, I don't believe there is one.

EcmaScript 6 is proposing a contains method for strings such that you could do this:
function stringContainsArray(str, array){
for (var i = 0; i < array.length; i++){
if (str.contains(array[i])){
return true;
}
}
return false;
}
var list = ['aa','bb','cc','dd'];
var string = "nygaard"; // true because "aa" is in "nygaard".
console.log(stringContainsArray(list));
// => true
Since implementation will likely remain spotty in the major browsers and other runtimes for the near future, you should add the following code for compatibility before the stringContainsArray function:
// String.prototype.contains will be implemented in
// the major browsers in the not too distant future
// this will add the same functionality to browsers
// or other runtimes that do not currently support it
if ( !String.prototype.contains ) {
String.prototype.contains = function() {
return String.prototype.indexOf.apply( this, arguments ) !== -1;
};
}
this may help: https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/String/contains
You can also extend the string prototype to have a simple method:
String.prototype.containsArray = function(array){
return stringContainsArray(this, array);
}
Then you can simpy do:
"nygaard".containsArray(list);
// => true

Determine token value by comparing with token replaced string

I have a tokenised string like so;
var route = a/b/{firstId}/c/d/{nextId}
and I am wondering if it is possible with regex to get the value of "firstId" via a second string with tokens already replaced.
Example, if I have a given string;
var partPath = a/b/33
I can do something like;
function getValueFromPath(path, route){
//regex stuff
return tokenValue; //Expected result 33
}
getValueFromPath(partPath, route);
Thank you,
C.

A regex solution would be overly complicated (if you didn't define the route with a regexp right away). I'd just use
function getValueFromPath(path, route){
var actualParts = path.split("/"),
expectedParts = route.split("/"),
result = {};
for (var i=0; i<expectedParts.length; i++) {
if (i >= actualParts.length)
return result;
var actual = actualParts[i],
expected = expectedParts[i];
if (/^\{.+\}$/.test(expected))
result[ expected.slice(1, -1) ] = actual;
else if (actual != expected)
// non-matching literals found, abort
return result;
}
return result;
}
> getValueFromPath("a/b/33", "a/b/{firstId}/c/d/{nextId}")
{firstId: "33"}
> getValueFromPath("a/b/33/c/d/42/x", "a/b/{firstId}/c/d/{nextId}")
{firstId: "33", nextId: "42"}
Here's the same thing with "regex stuff" (notice that regex special characters in the route are not escaped, you have to take care about that yourself):
function getValueFromPath(path, route){
var keys = [];
route = "^"+route.split("/").reduceRight(function(m, part) {
return part + "(?:/" + m + ")?"; // make right parts optional
}).replace(/\{([^\/{}]+)\}/g, function(m, k) {
keys.push(k); // for every "variable"
return "([^/]+)"; // create a capturing group
});
var regex = new RegExp(route); // build an ugly regex:
// regex == /^a(?:\/b(?:\/([^/]+)(?:\/c(?:\/d(?:\/([^/]+))?)?)?)?)?/
var m = path.match(regex),
result = {};
for (var i=0; m && i<keys.length; i++)
result[keys[i]] = m[i+1];
return result;
}

You can create a regexp like this:
function getValueFromPath(path, route){
tokenValue = path.match(route)[1];
return tokenValue; //Expected result 33
}
var route = /\/a\/b\/([^\/]+)(\/c\/d\/([^\/]+))?/;
var partPath = '/a/b/33';
getValueFromPath(partPath, route); // == 33
http://jsfiddle.net/firstclown/YYvvn/2/
This will let you extract the first value at the first match with [1] and you can get the nextId by changing that to [3] (since [2] will give you the whole path after the 33).

Parse query string in JavaScript [duplicate]

This question already has answers here:
How can I get query string values in JavaScript?
(73 answers)
Closed 3 years ago.
The community reviewed whether to reopen this question 1 year ago and left it closed:
Duplicate This question has been answered, is not unique, and doesn’t differentiate itself from another question.
I need to parse the query string www.mysite.com/default.aspx?dest=aboutus.aspx.
How do I get the dest variable in JavaScript?

Here is a fast and easy way of parsing query strings in JavaScript:
function getQueryVariable(variable) {
var query = window.location.search.substring(1);
var vars = query.split('&');
for (var i = 0; i < vars.length; i++) {
var pair = vars[i].split('=');
if (decodeURIComponent(pair[0]) == variable) {
return decodeURIComponent(pair[1]);
}
}
console.log('Query variable %s not found', variable);
}
Now make a request to page.html?x=Hello:
console.log(getQueryVariable('x'));

function parseQuery(queryString) {
var query = {};
var pairs = (queryString[0] === '?' ? queryString.substr(1) : queryString).split('&');
for (var i = 0; i < pairs.length; i++) {
var pair = pairs[i].split('=');
query[decodeURIComponent(pair[0])] = decodeURIComponent(pair[1] || '');
}
return query;
}
Turns query string like hello=1&another=2 into object {hello: 1, another: 2}. From there, it's easy to extract the variable you need.
That said, it does not deal with array cases such as "hello=1&hello=2&hello=3". To work with this, you must check whether a property of the object you make exists before adding to it, and turn the value of it into an array, pushing any additional bits.

You can also use the excellent URI.js library by Rodney Rehm. Here's how:-
var qs = URI('www.mysite.com/default.aspx?dest=aboutus.aspx').query(true); // == { dest : 'aboutus.aspx' }
alert(qs.dest); // == aboutus.aspx
And to parse the query string of current page:-
var $_GET = URI(document.URL).query(true); // ala PHP
alert($_GET['dest']); // == aboutus.aspx

Me too! http://jsfiddle.net/drzaus/8EE8k/
(Note: without fancy nested or duplicate checking)
deparam = (function(d,x,params,p,i,j) {
return function (qs) {
// start bucket; can't cheat by setting it in scope declaration or it overwrites
params = {};
// remove preceding non-querystring, correct spaces, and split
qs = qs.substring(qs.indexOf('?')+1).replace(x,' ').split('&');
// march and parse
for (i = qs.length; i > 0;) {
p = qs[--i];
// allow equals in value
j = p.indexOf('=');
// what if no val?
if(j === -1) params[d(p)] = undefined;
else params[d(p.substring(0,j))] = d(p.substring(j+1));
}
return params;
};//-- fn deparam
})(decodeURIComponent, /\+/g);
And tests:
var tests = {};
tests["simple params"] = "ID=2&first=1&second=b";
tests["full url"] = "http://blah.com/?third=c&fourth=d&fifth=e";
tests['just ?'] = '?animal=bear&fruit=apple&building=Empire State Building&spaces=these+are+pluses';
tests['with equals'] = 'foo=bar&baz=quux&equals=with=extra=equals&grault=garply';
tests['no value'] = 'foo=bar&baz=&qux=quux';
tests['value omit'] = 'foo=bar&baz&qux=quux';
var $output = document.getElementById('output');
function output(msg) {
msg = Array.prototype.slice.call(arguments, 0).join("\n");
if($output) $output.innerHTML += "\n" + msg + "\n";
else console.log(msg);
}
var results = {}; // save results, so we can confirm we're not incorrectly referencing
$.each(tests, function(msg, test) {
var q = deparam(test);
results[msg] = q;
output(msg, test, JSON.stringify(q), $.param(q));
output('-------------------');
});
output('=== confirming results non-overwrite ===');
$.each(results, function(msg, result) {
output(msg, JSON.stringify(result));
output('-------------------');
});
Results in:
simple params
ID=2&first=1&second=b
{"second":"b","first":"1","ID":"2"}
second=b&first=1&ID=2
-------------------
full url
http://blah.com/?third=c&fourth=d&fifth=e
{"fifth":"e","fourth":"d","third":"c"}
fifth=e&fourth=d&third=c
-------------------
just ?
?animal=bear&fruit=apple&building=Empire State Building&spaces=these+are+pluses
{"spaces":"these are pluses","building":"Empire State Building","fruit":"apple","animal":"bear"}
spaces=these%20are%20pluses&building=Empire%20State%20Building&fruit=apple&animal=bear
-------------------
with equals
foo=bar&baz=quux&equals=with=extra=equals&grault=garply
{"grault":"garply","equals":"with=extra=equals","baz":"quux","foo":"bar"}
grault=garply&equals=with%3Dextra%3Dequals&baz=quux&foo=bar
-------------------
no value
foo=bar&baz=&qux=quux
{"qux":"quux","baz":"","foo":"bar"}
qux=quux&baz=&foo=bar
-------------------
value omit
foo=bar&baz&qux=quux
{"qux":"quux","foo":"bar"} <-- it's there, i swear!
qux=quux&baz=&foo=bar <-- ...see, jQuery found it
-------------------

Here's my version based loosely on Braceyard's version above but parsing into a 'dictionary' and support for search args without '='. In use it in my JQuery $(document).ready() function. The arguments are stored as key/value pairs in argsParsed, which you might want to save somewhere...
'use strict';
function parseQuery(search) {
var args = search.substring(1).split('&');
var argsParsed = {};
var i, arg, kvp, key, value;
for (i=0; i < args.length; i++) {
arg = args[i];
if (-1 === arg.indexOf('=')) {
argsParsed[decodeURIComponent(arg).trim()] = true;
}
else {
kvp = arg.split('=');
key = decodeURIComponent(kvp[0]).trim();
value = decodeURIComponent(kvp[1]).trim();
argsParsed[key] = value;
}
}
return argsParsed;
}
parseQuery(document.location.search);

Following on from my comment to the answer #bobby posted, here is the code I would use:
function parseQuery(str)
{
if(typeof str != "string" || str.length == 0) return {};
var s = str.split("&");
var s_length = s.length;
var bit, query = {}, first, second;
for(var i = 0; i < s_length; i++)
{
bit = s[i].split("=");
first = decodeURIComponent(bit[0]);
if(first.length == 0) continue;
second = decodeURIComponent(bit[1]);
if(typeof query[first] == "undefined") query[first] = second;
else if(query[first] instanceof Array) query[first].push(second);
else query[first] = [query[first], second];
}
return query;
}
This code takes in the querystring provided (as 'str') and returns an object. The string is split on all occurances of &, resulting in an array. the array is then travsersed and each item in it is split by "=". This results in sub arrays wherein the 0th element is the parameter and the 1st element is the value (or undefined if no = sign). These are mapped to object properties, so for example the string "hello=1&another=2&something" is turned into:
{
hello: "1",
another: "2",
something: undefined
}
In addition, this code notices repeating reoccurances such as "hello=1&hello=2" and converts the result into an array, eg:
{
hello: ["1", "2"]
}
You'll also notice it deals with cases in whih the = sign is not used. It also ignores if there is an equal sign straight after an & symbol.
A bit overkill for the original question, but a reusable solution if you ever need to work with querystrings in javascript :)

If you know that you will only have that one querystring variable you can simply do:
var dest = location.search.replace(/^.*?\=/, '');

The following function will parse the search string with a regular expression, cache the result and return the value of the requested variable:
window.getSearch = function(variable) {
var parsedSearch;
parsedSearch = window.parsedSearch || (function() {
var match, re, ret;
re = /\??(.*?)=([^\&]*)&?/gi;
ret = {};
while (match = re.exec(document.location.search)) {
ret[match[1]] = match[2];
}
return window.parsedSearch = ret;
})();
return parsedSearch[variable];
};
You can either call it once without any parameters and work with the window.parsedSearch object, or call getSearch subsequently.
I haven't fully tested this, the regular expression might still need some tweaking...

How about this?
function getQueryVar(varName){
// Grab and unescape the query string - appending an '&' keeps the RegExp simple
// for the sake of this example.
var queryStr = unescape(window.location.search) + '&';
// Dynamic replacement RegExp
var regex = new RegExp('.*?[&\\?]' + varName + '=(.*?)&.*');
// Apply RegExp to the query string
var val = queryStr.replace(regex, "$1");
// If the string is the same, we didn't find a match - return false
return val == queryStr ? false : val;
}
..then just call it with:
alert('Var "dest" = ' + getQueryVar('dest'));
Cheers

I wanted a simple function that took a URL as an input and returned a map of the query params.
If I were to improve this function, I would support the standard for array data in the URL, and or nested variables.
This should work back and for with the jQuery.param( qparams ) function.
function getQueryParams(url){
var qparams = {},
parts = (url||'').split('?'),
qparts, qpart,
i=0;
if(parts.length <= 1 ){
return qparams;
}else{
qparts = parts[1].split('&');
for(i in qparts){
qpart = qparts[i].split('=');
qparams[decodeURIComponent(qpart[0])] =
decodeURIComponent(qpart[1] || '');
}
}
return qparams;
};

I wanted to pick up specific links within a DOM element on a page, send those users to a redirect page on a timer and then pass them onto the original clicked URL. This is how I did it using regular javascript incorporating one of the methods above.
Page with links: Head
function replaceLinks() {
var content = document.getElementById('mainContent');
var nodes = content.getElementsByTagName('a');
for (var i = 0; i < document.getElementsByTagName('a').length; i++) {
{
href = nodes[i].href;
if (href.indexOf("thisurl.com") != -1) {
nodes[i].href="http://www.thisurl.com/redirect.aspx" + "?url=" + nodes[i];
nodes[i].target="_blank";
}
}
}
}
Body
<body onload="replaceLinks()">
Redirect page
Head
function getQueryVariable(variable) {
var query = window.location.search.substring(1);
var vars = query.split('&');
for (var i = 0; i < vars.length; i++) {
var pair = vars[i].split('=');
if (decodeURIComponent(pair[0]) == variable) {
return decodeURIComponent(pair[1]);
}
}
console.log('Query variable %s not found', variable);
}
function delayer(){
window.location = getQueryVariable('url')
}
Body
<body onload="setTimeout('delayer()', 1000)">