okay so after spending sometime i figured it out options = [‘color’, ‘size’]; optionsValues= [[‘Red’,’Blue’], [‘L’, ‘XS’]]; function combination() { var r = [], arg = arguments[0], max = arg.length-1; function helper(arr, i) { for (var j=0, l=arg[i].length; j<l; j++) { var a = arr.slice(0); // clone arr var obj = {}; obj[options[i]] = arg[i][j]; … Read more
JavaScript 101, check your console for errors, there are a few: extra ; inside json undefined _pcost which should be _cost unfinished/incorrect for loop: ; instead of , inside for loops in JavaScript, + the iteration counter missing var data = { “products”: [{ “p_id”: 111, “p_name”: “p_one”, “p_cost”: 100 }] }; var results = … Read more
You where comparing the wrong array in your code, you can fix it like following: for (var currentPokemon in FullGame.pokemon) { // HERE added .pokemon at the end if (FullGame.pokemon[currentPokemon].name == name) { var Detail = FullGame.pokemon[currentPokemon]; alert(Detail); } else { alert(“Type Again”); } } 3 solved Iterate over JS [closed]
You havent closed ready function properly: $(document).ready(function(){ $(‘#venture’).change(function(){ alert($(this).val()); }); }); //^—- typo here Demo 2 solved jquery .change() not triggering on select [closed]
var i = parseInt(‘6250’), res = (i / 100).toFixed(2); 3 solved Integer value to decimal only last 2 digit in Javascript
You can use binary XOR with 3 to do this function returnTwo(input) { return input ^ 3; } Or as Phil suggested function returnTwo(input) { return ~input + 4; } Or function returnTwo(input) { return (input * 2) % 3; } Or function returnTwo(input) { return Math.abs(input – 3); } Or function returnTwo(input) { return … Read more
Make sure you run the javascript code only when the document is ready- $( document ).ready(function() { $(‘.hamburger’).on(‘click’, function(e) { e.preventDefault(); $(‘.menu’).toggleClass(‘slide-down’); }); }); 1 solved How to make that responsive hamburger menu work? It just doesnt
Here’s the order in which that code runs: The variables index and myProduct are created and given the initial value undefined. $(‘.product img’) is used to look up elements and then click is used to assign an event handler to them. ‘product’+ (index + 1)+’.php’ is assigned to myProduct; note that index is still undefined. … Read more
As mentioned in the comments, you haven’t included jQuery or jQuery UI, and your HTML closing tags are incorrect. If you view the source of the demo on the jqueryui site, you will find the code is a little different you yours. <!DOCTYPE html> <html> <head> <link rel=”stylesheet” href=”https://code.jquery.com/ui/1.12.1/themes/base/jquery-ui.css”> <script src=”https://code.jquery.com/jquery-1.12.4.js”></script> <script src=”https://code.jquery.com/ui/1.12.1/jquery-ui.js”></script> <script> $(function() … Read more
Without your code we will not be able to identify your problem. But Different browsers render and act differently in some situation. if you use jQuery like library it will be easy to get done what you want. $(document).ready(function() { $(‘input[type=radio][name=type1]’).change(function() { if (this.value == ‘Response’) {//if the value of radio button is male //your … Read more
assuming that your JSON is object named data var data = { “destination_addresses” : [ “Chennai, Tamil Nadu, India” ], “origin_addresses” : [ “Kolkata, West Bengal, India” ], “rows” : [ { “elements” : [ { “distance” : { “text” : “1,671 km”, “value” : 1671269 }, “duration” : { “text” : “1 day 5 … Read more
Do it like this: var array = [{t_id:”1″,val1:”1″,title:”cash to purchase”,unit:”bag”},{t_id:”1″,val1:”1″,title:”cash to purchase”,unit:”bag”},{t_id:”1″,val1:”1″,title:”cash to purchase”,unit:”bag”},{t_id:”2″,val1:”4″,title:”offload”,unit:”bag”},{t_id:”2″,val1:”5″,title:”onroad”,unit:”bag”},{t_id:”3″,val1:”5″,title:”Onroad”,unit:”bag”},{t_id:”3″,val1:”6″,title:”Onroad”,unit:”bag”}]; var grouped = []; array.forEach(function(o) { var count = 0; if (!this[o.t_id]) { this[o.t_id] = { t_id: o.t_id, val1: 0, title: o.title, counter: count }; grouped.push(this[o.t_id]); } this[o.t_id].val1 += Number(o.val1); this[o.t_id].counter += Number(++count); }, Object.create(null)); console.log(grouped); Now in HTML show like … Read more
Use this code document.getElementById(“aaa:j_idt89:”+ i +”:asd”).innerHTML Note the change I made inside. “+ i +” . You actually needed a String Concatenation. So explaining the code. when i = 1 “aaa:j_idt89:”+ i +”:asd” = “aaa:j_idt89:”+ 1 +”:asd” = “aaa:j_idt89:1:asd” = thats what you need 0 solved How to use a variable inside a string
In order to ensure both browser compatibility AND a gradient/image background, you may find you will have to use multiple elements, as well as a pseudo element on each of the nested divs. A quick mock up can be seen below. html { background: radial-gradient(red, black); } div.wrap { height: 300px; width: 300px; position: relative; … Read more
Something like this might work for you: var i; var k; for(i = 0; i < existingArray.length; i++){ var count = 2; for(k = 0; k < existingArray.length; k ++){ if(i != k && existingArray[i].content.host.name === existingArray[k].content.host.name){ existingArray[k].content.host.name += ‘_’ + count; count++; } } } 3 solved How to modify an array of objects … Read more