How to test if an id is present in an associative array

How to test if an id is present in an associative array - javascript

I'm am starting in javascript. I'm trying to do a little program that make a statistic upon the number of answer found in a text document.
The situation is this: each question has one id, e.g 8000001 and W if answer is good or R if answer is not good, e.g for an user answer is 8000001W. I have many user so many question of the same id. I want to get number of good answers per questions. E.g id: 800001 have W: 24 and "R": 5.
I have split the answer into id for 8000001 and ans for W or R. I wanted to create an associative table to get question[id]=["W": 0, "R": 0]. But I'm blocking on this. I've tried this code:
var tab = [];
tab[0] = [];
tab[0] = ['8000001W', '8000002W', '8000003W', '8000004R', '8000005W', '8000006R'];
tab[1] = [];
tab[1] = ['8000001R', '8000002W', '8000003R', '8000004W', '8000005R', '8000006W'];
var question = [];
var id;
for (var i=0;i<tab.length;i++) {
document.write("<dl><dt>tableau n° "+i+"<\/dt>");
for (var propriete in tab[i]) {
id = tab[i][propriete].slice(0,7);
var ans = tab[i][propriete].slice(7,8);
question[id] = [];
if(question[id]){
incrementResp.call(rep, ans);
}else{
var rep = initResp(ans);
question[id] = rep;
}
}
document.write("<\/dl>");
}
function incrementResp(type){
this.this++;
}
function initResp(t){
rep = [];
rep.W = (t=='W'?1:0);
rep.R = (t=='R'?1:0);
}

Based on what your want finally, the 'question' should be used as an object literal, defined as question = {} (similar to association array), what you defined here is an array literal. You can check this for more information about different types of literals in JavaScript:
https://developer.mozilla.org/en/docs/Web/JavaScript/Guide/Values,_variables,_and_literals#Literals
In terms of your code, you can simple do like this:
if (question[id]) {
question[id][ans] += 1;
}
else {
var rep = initResp(ans);
question[id] = rep;
}
Also your 'initResp' function better to return an object literal 'rep', not as an array literal:
function initResp(t){
var rep = {};
rep.W = (t=='W'?1:0);
rep.R = (t=='R'?1:0);
return rep;
}

For an "associative array" in JavaScript, use a regular object. In the code below, "results" is one of these objects. It has two keys, "W" and "R" that point to numbers starting at 0. Just iterate through your answer arrays and continuously increment the correct key.
There are two ways to access a key in an object: 1) using brackets, 2) using "dot" notation. In the loop I use brackets because 'key' is a variable--it will resolve to "W" or "R" and therefore access the "W" or "R" key in that object. In the final two lines I use dot notation because "W" and "R" are literally the keys I want to access. It would also work if I did this instead: results['W']++ and results['R']++.
var tab = [];
tab[0] = ['8000001W', '8000002W', '8000003W', '8000004R', '8000005W', '8000006R'];
tab[1] = ['8000001R', '8000002W', '8000003R', '8000004W', '8000005R', '8000006W'];
var results = {
W: 0,
R: 0
};
// go through each tab
for (var tabIdx = 0; tabIdx < tab.length; tabIdx++) {
// go through each answer and apppend to an object that keeps the results
for (var i = 0; i < tab[tabIdx].length; i++) {
var answer = tab[tabIdx][i];
// the last character in the string is the "key", (W or R)
var key = answer.slice(-1);
// append to the results object
results[key]++;
}
}
console.log(results);
console.log(results.W); // 7
console.log(results.R); // 5
Open up your development console (on Chrome it's F12) to see the output.

This is how i resolved my problem for associative array.
var tab = [];
tab[0] = ['8000001W', '8000002W', '8000003W', '8000004R', '8000005W', '8000006R'];
tab[1] = ['8000001R', '8000002W', '8000003R', '8000004W', '8000005R', '8000006W'];
tab[2] = ['8000001R', '8000002W', '8000003R', '8000004W', '8000005R', '8000006W'];
var question = {};
for (var tabIndex = 0; tabIndex < tab.length; tabIndex++) {
for (var i = 0; i < tab[tabIndex].length; i++) {
var answer = tab[tabIndex][i];
var id = answer.slice(0,7);
var ans = answer.slice(-1);
if (question[id]) {
question[id][ans] += 1;
}else {
var results = initResp(ans);
question[id] = results;
}
}
}
console.log(question);
function initResp(t) {
var results = [];
results.W = (t === 'W' ? 1 : 0);
results.R = (t === 'R' ? 1 : 0);
//console.log(results);
return results;
}

Related

Using RegExp With Replace and Arrays

I am running a RegExp on a user input to test for capitals of 3 or more in a row. I have a loop that finds the RegExp and then adds it to an array. Another loop that creates a new array. When I run the .replace it seems that that array isn't being looped over.
var abbrBracket = /\([A-Z]{3,30}\)/g;
var abbrBracketArr = [];
var a;
while (a = abbrBracket.exec(nonCode)) {
abbrBracketArr.push(a[0]);
}
var capFoundInArr = /[A-Z]{3,30}/g;
var abbrBracketArrRemove = [];
var b;
while (b = capFoundInArr.exec(abbrBracketArr)) {
abbrBracketArrRemove.push('(<abbr>' + b[0] + '</abbr>)');
}
for(var c = 0; c < abbrBracketArrRemove.length; c++){
nonCode = document.getElementById('cleanse').innerHTML;
//nonCode = nonCode.replace(new RegExp(/\([A-Z]{3,30}\)/), abbrBracketArrRemove[c]);
nonCode = nonCode.replace(new RegExp(abbrBracketArr[c]), abbrBracketArrRemove[c]);
document.getElementById('cleanse').innerHTML = nonCode;
}
The results show if there are two (or more) of the same abbreviations, the first is executed multiple times the next is skipped.
Saying that, I am using the exact same code to run a second query for replace and I am not getting this error.
var abbrNoBracket = /\s[A-Z]{3,30}/g;
var abbrNoBracketArr = [];
var d;
while (d = abbrNoBracket.exec(nonCode)) {
abbrNoBracketArr.push(d[0]);
}
var abbrNoBracketArrRemove = [];
var e;
while (e = capFoundInArr.exec(abbrNoBracketArr)) {
abbrNoBracketArrRemove.push(' <abbr title="">' + e[0] + '</abbr>');
}
for(var f = 0; f < abbrNoBracketArrRemove.length; f++){
nonCode = document.getElementById('cleanse').innerHTML;
nonCode = nonCode.replace(new RegExp(abbrNoBracketArr[f]), abbrNoBracketArrRemove[f]);
document.getElementById('cleanse').innerHTML = nonCode;
}
In the first block, you can see I commented out a line, if I use the RegExp instead of the array. It works. Curious, why this would work for one, but not the other.

Found my error, which was very obvious after a good nights sleep.
I called a new RegExp on the array. Once I removed that, everything worked as it should.
for(var c = 0; c < abbrBracketArrRemove.length; c++){
nonCode = document.getElementById('cleanse').innerHTML;
//mistake
//nonCode = nonCode.replace(new RegExp(abbrNoBracketArr[f]), abbrNoBracketArrRemove[f]);
//corrected
nonCode = nonCode.replace(abbrBracketArr[c], abbrBracketArrRemove[c]);
document.getElementById('cleanse').innerHTML = nonCode;
}
Hopefully that helps someone else.

Manipulate more javascript array based on another array

I've a strange thing to do but I don't know how to start
I start with this vars
var base = [1,1,1,2,3,5,7,9,14,19,28,40,56,114,232,330];
var sky = [0,0,0,3,4,5,6,7,8,9,10,11,12,14,16,17];
var ite = [64,52,23,38,13,15,6,4,6,3,2,1,2,1,1,1];
So to start all the 3 array have the same length and the very first operation is to see if there is a duplicate value in sky array, in this case the 0 is duplicated and only in this case is at the end, but all of time the sky array is sorted. So I've to remove all the duplicate (in this case 0) from sky and remove the corresponding items from base and sum the corresponding items on ite. So if there's duplicate on position 4,5 I've to manipulate this conditions. But let see the new 3 array:
var new_base = [1,2,3,5,7,9,14,19,28,40,56,114,232,330];
var new_sky = [0,3,4,5,6,7,8,9,10,11,12,14,16,17];
var new_ite = [139,38,13,15,6,4,6,3,2,1,2,1,1,1];
If you see the new_ite have 139 instead the 64,52,23, that is the sum of 64+52+23, because the first 3 items on sky are the same (0) so I remove two corresponding value from base and sky too and I sum the corresponding value into the new_ite array.
There's a fast way to do that? I thought a for loops but I stuck at the very first for (i = 0; i < sky.length; i++) lol, cuz I've no idea on how to manipulate those 3 array in that way
J

When removing elements from an array during a loop, the trick is to start at the end and move to the front. It makes many things easier.
for( var i = sky.length-1; i>=0; i--) {
if (sky[i] == prev) {
// Remove previous index from base, sky
// See http://stackoverflow.com/questions/5767325/how-to-remove-a-particular-element-from-an-array-in-javascript
base.splice(i+1, 1);
sky.splice(i+1, 1);
// Do sum, then remove
ite[i] += ite[i+1];
ite.splice(i+1, 1);
}
prev = sky[i];
}
I won't speak to whether this is the "fastest", but it does work, and it's "fast" in terms of requiring little programmer time to write and understand. (Which is often the most important kind of fast.)

I would suggest this solution where j is used as index for the new arrays, and i for the original arrays:
var base = [1,1,1,2,3,5,7,9,14,19,28,40,56,114,232,330];
var sky = [0,0,0,3,4,5,6,7,8,9,10,11,12,14,16,17];
var ite = [64,52,23,38,13,15,6,4,6,3,2,1,2,1,1,1];
var new_base = [], new_sky = [], new_ite = [];
var j = -1;
sky.forEach(function (sk, i) {
if (!i || sk !== sky[i-1]) {
new_ite[++j] = 0;
new_base[j] = base[i];
new_sky[j] = sk;
}
new_ite[j] += ite[i];
});
console.log('new_base = ' + new_base);
console.log('new_sky = ' + new_sky);
console.log('new_ite = ' + new_ite);

You can use Array#reduce to create new arrays from the originals according to the rules:
var base = [1,1,1,2,3,5,7,9,14,19,28,40,56,114,232,330];
var sky = [0,0,0,3,4,5,6,7,8,9,10,11,12,14,16,17];
var ite = [64,52,23,38,13,15,6,4,6,3,2,1,2,1,1,1];
var result = sky.reduce(function(r, n, i) {
var last = r.sky.length - 1;
if(n === r.sky[last]) {
r.ite[last] += ite[i];
} else {
r.base.push(base[i]);
r.sky.push(n);
r.ite.push(ite[i]);
}
return r;
}, { base: [], sky: [], ite: [] });
console.log('new base:', result.base.join(','));
console.log('new sky:', result.sky.join(','));
console.log('new ite:', result.ite.join(','));

atltag's answer is fastest. Please see:
https://repl.it/FBpo/5

Just with a single .reduce() in O(n) time you can do as follows; (I have used array destructuring at the assignment part. One might choose to use three .push()s though)
var base = [1,1,1,2,3,5,7,9,14,19,28,40,56,114,232,330],
sky = [0,0,0,3,4,5,6,7,8,9,10,11,12,14,16,17],
ite = [64,52,23,38,13,15,6,4,6,3,2,1,2,1,1,1],
results = sky.reduce((r,c,i) => c === r[1][r[1].length-1] ? (r[2][r[2].length-1] += ite[i],r)
: ([r[0][r[0].length],r[1][r[1].length],r[2][r[2].length]] = [base[i],c,ite[i]],r),[[],[],[]]);
console.log(JSON.stringify(results));

Javascript sort and order

So i have this array
[ 'vendor/angular/angular.min.js',
'vendor/angular-nice-bar/dist/js/angular-nice-bar.min.js',
'vendor/angular-material/modules/js/core/core.min.js',
'vendor/angular-material/modules/js/backdrop/backdrop.min.js',
'vendor/angular-material/modules/js/dialog/dialog.min.js',
'vendor/angular-material/modules/js/button/button.min.js',
'vendor/angular-material/modules/js/icon/icon.min.js',
'vendor/angular-material/modules/js/tabs/tabs.min.js',
'vendor/angular-material/modules/js/content/content.min.js',
'vendor/angular-material/modules/js/toolbar/toolbar.min.js',
'vendor/angular-material/modules/js/input/input.min.js',
'vendor/angular-material/modules/js/divider/divider.min.js',
'vendor/angular-material/modules/js/menu/menu.min.js',
'vendor/angular-material/modules/js/select/select.min.js',
'vendor/angular-material/modules/js/radioButton/radioButton.min.js',
'vendor/angular-material/modules/js/checkbox/checkbox.min.js',
'vendor/angular-material/modules/js/switch/switch.min.js',
'vendor/angular-material/modules/js/tooltip/tooltip.min.js',
'vendor/angular-material/modules/js/toast/toast.min.js',
'vendor/angular-clipboard/angular-clipboard.js',
'vendor/angular-animate/angular-animate.min.js',
'vendor/angular-aria/angular-aria.min.js',
'vendor/angular-messages/angular-messages.min.js',
'vendor/angular-ui-router/release/angular-ui-router.js',
'src/app/about/about.js',
'src/app/hekate.cfg.js',
'src/app/hekate.ctrl.js',
'src/app/hekate.module.js',
'src/app/home/home.js',
'src/app/user/dialog/user.signIn.ctrl.js',
'src/app/user/dialog/user.signIn.module.js',
'src/app/user/user.cfg.js',
'src/app/user/user.ctrl.js',
'src/app/user/user.module.js',
'src/common/services/toast.service.js',
'templates-common.js',
'templates-app.js'
]
And taking the following part from the above array as example:
[
'src/app/hekate.cfg.js',
'src/app/hekate.ctrl.js',
'src/app/hekate.module.js',
]
I want to sort it like
[
'src/app/hekate.module.js',
'src/app/hekate.cfg.js',
'src/app/hekate.ctrl.js',
]
So more specific of what i want is to find in that array where string is duplicated and after check if has at the end [.cfg.js, .ctrl.js, .module.js] and automatic order them to [.module.js, .cfg.js, .ctrl.js]
Can anyone please help me with that?

A single sort proposal.
var array = ['src/app/about/about.js', 'src/app/hekate.cfg.js', 'src/app/hekate.ctrl.js', 'src/app/hekate.module.js', 'src/app/home/home.js', 'src/app/user/dialog/user.signIn.ctrl.js', 'src/app/user/dialog/user.signIn.module.js', 'src/app/user/user.cfg.js', 'src/app/user/user.ctrl.js', 'src/app/user/user.module.js'];
array.sort(function (a, b) {
function replaceCB(r, a, i) { return r.replace(a, i); }
var replace = ['.module.js', '.cfg.js', '.ctrl.js'];
return replace.reduce(replaceCB, a).localeCompare(replace.reduce(replaceCB, b));
});
document.write('<pre>' + JSON.stringify(array, 0, 4) + '</pre>');
To prevent so much replaces, i suggest to have a look to sorting with map.

You can try something like this:
Algo:
Group based on path and store file names as value.
Check for existence of one of special file ".cfg.js"
Sort following list based on custom sort.
Loop over object's property and join key with values to form full path again.
If you wish to sort full array, you can sort keys itself and then merge path with names. I have done this. If you do not wish to do this, just remove sort function from final loop.
Sample
var data=["vendor/angular/angular.min.js","vendor/angular-nice-bar/dist/js/angular-nice-bar.min.js","vendor/angular-material/modules/js/core/core.min.js","vendor/angular-material/modules/js/backdrop/backdrop.min.js","vendor/angular-material/modules/js/dialog/dialog.min.js","vendor/angular-material/modules/js/button/button.min.js","vendor/angular-material/modules/js/icon/icon.min.js","vendor/angular-material/modules/js/tabs/tabs.min.js","vendor/angular-material/modules/js/content/content.min.js","vendor/angular-material/modules/js/toolbar/toolbar.min.js","vendor/angular-material/modules/js/input/input.min.js","vendor/angular-material/modules/js/divider/divider.min.js","vendor/angular-material/modules/js/menu/menu.min.js","vendor/angular-material/modules/js/select/select.min.js","vendor/angular-material/modules/js/radioButton/radioButton.min.js","vendor/angular-material/modules/js/checkbox/checkbox.min.js","vendor/angular-material/modules/js/switch/switch.min.js","vendor/angular-material/modules/js/tooltip/tooltip.min.js","vendor/angular-material/modules/js/toast/toast.min.js","vendor/angular-clipboard/angular-clipboard.js","vendor/angular-animate/angular-animate.min.js","vendor/angular-aria/angular-aria.min.js","vendor/angular-messages/angular-messages.min.js","vendor/angular-ui-router/release/angular-ui-router.js","src/app/about/about.js","src/app/hekate.cfg.js","src/app/hekate.ctrl.js","src/app/hekate.module.js","src/app/home/home.js","src/app/user/dialog/user.signIn.ctrl.js","src/app/user/dialog/user.signIn.module.js","src/app/user/user.cfg.js","src/app/user/user.ctrl.js","src/app/user/user.module.js","src/common/services/toast.service.js","templates-common.js","templates-app.js"];
// Create groups based on path
var o = {};
data.forEach(function(item) {
var lastIndex = item.lastIndexOf('/') + 1;
var path = item.substring(0, lastIndex);
var fname = item.substring(lastIndex);
if (!o[path]) o[path] = [];
o[path].push(fname);
});
var manualOrder= [".module.js", ".cfg.js", ".ctrl.js"];
Array.prototype.fuzzyMatch = function(search){
return this.some(function(item){
return item.indexOf(search)>-1;
});
}
Array.prototype.fuzzySearchIndex = function(search){
var pos = -1;
this.forEach(function(item, index){
if(search.indexOf(item)>-1){
pos = index;
}
});
return pos;
}
function myCustomSort(a,b){
var a_pos = manualOrder.fuzzySearchIndex(a);
var b_pos = manualOrder.fuzzySearchIndex(b);
return a_pos > b_pos ? 1 : a_pos < b_pos ? -1 : 0;
}
// Check for ".cfg.js" and apply custom sort
for (var k in o) {
if (o[k].fuzzyMatch(".cfg.js")) {
o[k].sort(myCustomSort);
}
}
// Merge Path and names to create final value
var final = [];
Object.keys(o).sort().forEach(function(item) {
if (Array.isArray(o[item])) {
final = final.concat(o[item].map(function(fn) {
return item + fn
}));
} else
final = final.concat(o[item]);
});
console.log(final);

First make an array for names like 'hekate'.
Then make an array for final results.
We need 3 searching loops for ctrls, cfgs and modules.
If string contains arrayWithNames[0] + '.module' push the whole record to new array that you created. Same with ctrls and cfgs.
var allItems = []; //your array with all elements
var namesArray = [];
var finalResultsArray = [];
//fill name array here:
for(var i=0; i<=allItems.length; i++){
//you have to split string and find the module name (like 'hekate'). i hope you know how to split strings
}
//sort by modules, cfgs, ctrls:
for(var i=0; i<=namesArray.length; i++){
if(allItems[i].indexOf(namesArray[i] + '.module') > -1) {
finalResultsArray.push(allItems[i]);
}
}
for(var i=0; i<=namesArray.length; i++){
if(allItems[i].indexOf(namesArray[i] + '.cfg') > -1) {
finalResultsArray.push(allItems[i]);
}
}
for(var i=0; i<=namesArray.length; i++){
if(allItems[i].indexOf(namesArray[i] + '.ctrl') > -1) {
finalResultsArray.push(allItems[i]);
}
}
//now finalResultsArray have what you wanted

You can provide your own compare function to array.sort (see https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort)
Write one that returns the correct order for modules, ctrls and cfgs:
It should first remove the suffixes, and if the rest is the same, use the correct logic to return the order according to the suffix. Otherwise return a value according to the alphabetical order.
Update
I didn't test this code (not is it finished), but it should look something like that:
arr.sort(function(a, b) {
if ((a.endsWith(".cfg.js") || a.endsWith(".ctrl.js") || a.endsWith(".module.js")) &&
(b.endsWith(".cfg.js") || b.endsWith(".ctrl.js") || b.endsWith(".module.js"))) {
var sortedSuffixes = {
".module.js": 0,
".cfg.js": 1,
".ctrl.js": 2
};
var suffixAIdx = a.lastIndexOf(".cfg.js");
if (suffixAIdx < 0) suffixAIdx = a.lastIndexOf(".ctrl.js");
if (suffixAIdx < 0) suffixAIdx = a.lastIndexOf(".module.js");
var suffixBIdx = b.lastIndexOf(".cfg.js");
if (suffixBIdx < 0) suffixBIdx = b.lastIndexOf(".ctrl.js");
if (suffixBIdx < 0) suffixBIdx = b.lastIndexOf(".module.js");
var prefixA = a.substring(0, suffixAIdx);
var prefixB = b.substring(0, suffixAIdx);
if (prefixA != prefixB)
{
return a.localeCompare(b);
}
var suffixA = a.substring(suffixAIdx);
var suffixB = b.substring(suffixBIdx);
return sortedSuffixes[suffixA] - sortedSuffixes[suffixB];
} else {
return a.localeCompare(b);
}
});
Update 2
Here is a fiddle (https://jsfiddle.net/d4fmc7ue/) that works.

Counting the frequency of elements in an array in JavaScript

how do I count the frequency of the elements in the array, I'm new to Javascript and completely lost, I have looked at other answers here but can't get them to work for me. Any help is much appreciated.
function getText() {
var userText;
userText = document.InputForm.MyTextBox.value; //get text as string
alphaOnly(userText);
}
function alphaOnly(userText) {
var nuText = userText;
//result = nuText.split("");
var alphaCheck = /[a-zA-Z]/g; //using RegExp create variable to have only alphabetic characters
var alphaResult = nuText.match(alphaCheck); //get object with only alphabetic matches from original string
alphaResult.sort();
var result = freqLet(alphaResult);
document.write(countlist);
}
function freqLet(alphaResult) {
count = 0;
countlist = {
alphaResult: count
};
for (i = 0; i < alphaResult.length; i++) {
if (alphaResult[i] in alphaResult)
count[i] ++;
}
return countlist;
}

To count frequencies you should use an object which properties correspond to the letters occurring in your input string.
Also before incrementing the value of the property you should previously check whether this property exists or not.
function freqLet (alphaResult) {
var count = {};
countlist = {alphaResult:count};
for (i = 0; i < alphaResult.length; i++) {
var character = alphaResult.charAt(i);
if (count[character]) {
count[character]++;
} else {
count[character] = 1;
}
}
return countlist;
}

If you can use a third party library, underscore.js provides a function "countBy" that does pretty much exactly what you want.
_.countBy(userText, function(character) {
return character;
});
This should return an associative array of characters in the collection mapped to a count.
Then you could filter the keys of that object to the limited character set you need, again, using underscore or whatever method you like.

Do as below:
var __arr = [6,7,1,2,3,3,4,5,5,5]
function __freq(__arr){
var a = [], b = [], prev
__arr.sort((a,b)=>{return a- b} )
for(let i = 0; i<__arr.length; i++){
if(__arr[i] !== prev){
a.push(__arr[i])
b.push(1)
}else{
b[b.length - 1]++
}
prev = __arr[i]
}
return [a , b]
}

how can I refactor this?

As for now, when I prepare my data to be sent by Ajax request to my web app, I just concat my JS arrays (with placing -1 between them as separator - values can be positive only, so -1 means start of new array). This seems a bit ugly for me, so I'm wondering what would be best practice to refator this.
var online1 = [];
var online2 = [];
var online3 = [];
var online4 = [];
for(i = 0 ; i < listOfPlayers.length ; i++) {
var player = listOfPlayers[i].href;
var uid = player.substring(player.lastIndexOf('=') + 1);
if(onlineStatus[i].className == "online1"){
online1.push(uid);
}
if(onlineStatus[i].className == "online2"){
online2.push(uid);
}
if(onlineStatus[i].className == "online3"){
online3.push(uid);
}
if(onlineStatus[i].className == "online4"){
online4.push(uid);
}
}
online1.push(-1);
online2.push(-1);
online3.push(-1);
online4.push(-1);
var result = online1.concat(online2, online3, online4);
//...
ajaxRequest.send("result="+result);

You could do two things:
Use an object, stringify it using JSON.stringify. You can parse it using JSON.parse, even server-side solutions exist. JSON is available in recent browsers and as library.
Make the if generic.
E.g.:
var online = {1: [],
2: [],
3: [],
4: []};
for(i = 0 ; i < listOfPlayers.length ; i++) {
var player = listOfPlayers[i].href;
var uid = player.substring(player.lastIndexOf('=') + 1);
var number = onlineStatus[i].className.substring(6);
online[number].push(uid);
}
var result = JSON.stringify(online);
//...
ajaxRequest.send("result="+result);

Develop Reference

JavaScript is the programming language of the Web.

How to test if an id is present in an associative array - javascript

Related

Using RegExp With Replace and Arrays

Manipulate more javascript array based on another array

Javascript sort and order

Counting the frequency of elements in an array in JavaScript

how can I refactor this?

Categories

Resources