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If i have elements such as this
<img src='0.jpg' id='images' />
<img src='...' id='myEle' />
<img src='...' id='myEle' />
in jQuery can i do something like this
$(document).ready(function() {
$('#myEle').mouseup(function () {
$('#images').attr("src", myEle.getNumber() + ".jpg");
}
}
Obviously each element is sorted in the correct number format that corresponds to the myEle array number
276
The HTML id attribute is used to specify a unique id for an HTML element. You cannot have more than one element with the same id in an HTML document.
In simple way if you use id to select all div having same ids then it will select first id only but if you want to select ,want go through each div or any other element with same id then I will explain you how you can do that. $( '#red' ). css( 'background-color' , 'red' ); $( '#green' ).
Given an HTML document and the task is to select the elements with different ID's at the same time using JQuery. Approach: Select the ID's of different element and then use each() method to apply the CSS property on all selected ID's element.
Use the querySelectorAll() method to select elements by multiple ids, e.g. document. querySelectorAll('#box1, #box2, #box3') . The method takes a string containing one or more selectors as a parameter and returns a collection of the matching elements. Here is the HTML for the examples in this article.
Do not create markup that contains elements with duplicate IDs. This will break things, and you will be mauled by a velociraptor faster than you can say "goto".
Use classes instead:
<img src='0.jpg' id='images' />
<img src='...' class='myEle' />
<img src='...' class='myEle' />
then...
$(document).ready(function() {
$('.myEle').live('mouseup', function () {
$('#images').attr("src", myEle.getNumber() + ".jpg");
});
});
"How do i know which image is pressed?"
Use the this keyword:
$(document).ready(function() {
$('.myEle').live('mouseup', function () {
$('#images').attr("src", $(this).attr('src'));
});
});
...I think that's what you're looking for.
153
If you've inherited code so horrible that you can only work with the faulty output, use jQuery("[id=theidthatshouldbeaclassinstead]") or jQuery("[id*=theidthatshouldbeaclassinstead]") if there are multiple ids for some reason.
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