jquery id array
you can use $("div[id^='x['").click(function() // write code here }). So this will execute on click of those ids of div which start from x. So as per my ..., You have to escape the brackets with -- $("#info--[text1--]"). see http://api.jquery.com/category/selectors/ (second paragraph).,For id selectors, jQuery uses the JavaScript function document.getElementById() , which is extremely efficient. When another selector is attached to the id ... ,Join the array elements into a single string: var id = ["#id1", "#id2", "#id3"]; $(id.join(',')).css('color', '#000');. Or use a loop: for (var i=0; i<id.length; i++) ... , First off, an id attribute should not contain square brackets. It's just not valid. It can contain letters, numbers, underscores, hyphens, colons and ..., this would solve your problem: <div id="variable_1">3</div> <div id="jinja_1" class="jinja">I'm Jinja 1</div> <div id="jinja_2" class="jinja">I'm ...,jQuery selectors always return an array, even if that array is only one actual HTML element. Well, not really an array, it's a jQuery wrapper object, but it acts like ... , $j("#divID").val() will work just fine. Per the jQuery documentation, .val() will return the value of the first element in the set of matched elements., Yes, but #id will get you a single object (at most) anyway. Also, note that [0] will get you the DOM element, not a jQuery object, so you can't use jQuery's function without wrapping it again. Yes it is valid. It retrieves the first element of th, This piece of code: some.push([$('ul#jdLists li').attr("id")]); will push id of first li found by ul#jdLists li selector, what you need to do is to get id of ...
| 相關軟體 WinMerge 資訊 | |
|---|---|
 jquery id array 相關參考資料
How to pass an array id into jquery selector? - Stack Overflow
you can use $("div[id^='x['").click(function() // write code here }). So this will execute on click of those ids of div which start from x. So as per my ... https://stackoverflow.com How to selector ID textbox array using jquery - Stack Overflow
You have to escape the brackets with -- $("#info--[text1--]"). see http://api.jquery.com/category/selectors/ (second paragraph). https://stackoverflow.com ID Selector (“#id”) | jQuery API Documentation
For id selectors, jQuery uses the JavaScript function document.getElementById() , which is extremely efficient. When another selector is attached to the id ... https://api.jquery.com input an array of id's into a jquery selector - Stack Overflow
Join the array elements into a single string: var id = ["#id1", "#id2", "#id3"]; $(id.join(',')).css('color', '#000');. Or use a loop: for (var i=... https://stackoverflow.com jQuery - help needed on ID selector when id is an array type ...
First off, an id attribute should not contain square brackets. It's just not valid. It can contain letters, numbers, underscores, hyphens, colons and ... https://stackoverflow.com jQuery create id's in array - Stack Overflow
this would solve your problem: <div id="variable_1">3</div> <div id="jinja_1" class="jinja">I'm Jinja 1</div> <div id="jinja_2" ... https://stackoverflow.com JQuery find by id returning an array - Stack Overflow
jQuery selectors always return an array, even if that array is only one actual HTML element. Well, not really an array, it's a jQuery wrapper object, but it acts like ... https://stackoverflow.com jQuery ID Selector ("#id") Returns Array - Stack Overflow
$j("#divID").val() will work just fine. Per the jQuery documentation, .val() will return the value of the first element in the set of matched elements. https://stackoverflow.com jQuery object as a array: $('#id') - Stack Overflow
Yes, but #id will get you a single object (at most) anyway. Also, note that [0] will get you the DOM element, not a jQuery object, so you can't use jQuery's function without wrapping it again... https://stackoverflow.com Jquery push all li's ID's into array - Stack Overflow
This piece of code: some.push([$('ul#jdLists li').attr("id")]); will push id of first li found by ul#jdLists li selector, what you need to do is to get id of ... https://stackoverflow.com |