I'm trying to select the 3rd element of a JQuery object, by using eq() method. But for some reason the 2nd and 3rd selections pop out in changed order:
var selection = $("[name=input0], [name=input1], [name=input2], [name=input3]");
selection.eq(1); //turns out to be input2!!
What could be the reasons for this behavior? Can I trust acessing it by index in my script?
According to:
https://api.jquery.com/multiple-selector/
The order of the DOM elements in the returned jQuery object may not be identical, as they will be in document order.
With the help of the answers and comments and a bit of reading on the documentation.
Turns out that jQuery selects elements in the order they appear in the DOM (HTML), so:
Using a direct array access to a jQuery selection will work if you use that order, but you can only trust it if you are also responsible for the HTML, and you would need to remember this if you are ever going to change the layout.
Bottom line: not the best way to select a specific element.
Related
I have a page with duplicate ID's for a form element. The catch is I the elements show up separately based on a toggle. So both ID's never show up simultaneously.
However when I do form validation on that element, it always selects the element displayed last in the code (even if its hidden).
Is there a selector to select the visible duplicate ID?
I've tried the following but to no avail:
$('#my_element:visible').val();
As the myriad of other questions about this premise will tell you, you cannot use the ID selector # in this case; you have to use something like $('div[id=foo]') to find it.
Duplicate IDs are invalid HTML and will nearly always cause issues with scripting. Avoid if at all possible.
The reason this is occurring is because of Duplicate IDs. IDs must be unique for the HTML to be considered valid. Whenever you aren't working against valid HTML, your results are often erratic.
In this case, even though you are only showing one of the forms at a time, they're both still present in the mark up which is why the last one in the code is always the one that's getting run.
Since you're using jQuery, I'd highly recommend using classes for this instead.
Avoid duplicates ids on the page. It is not a valid HTML.
as Rwwl said, duplicate IDs are invalid. Assign classes instead of ids to them.
Then you can do
alert($('.my_element:visible').val());
try :hidden
$("#my_element").find(":hidden").val();
Elements can be considered hidden for several reasons:
They have a CSS display value of none.
They are form elements with type="hidden".
Their width and height are explicitly set to 0.
An ancestor element is hidden, so the element is not shown on the page.
NOTE: Elements with visibility: hidden or opacity: 0 are considered to be visible,
Do not use same id for multiple elements, classes are better!
You can not specify using the # id selector only, you need to be more specific. One way is choose the type of element first then id:
For an input element:
$('input#my_element:visible').val();
or for a div element:
$('div#my_element:visible').val();
An alternative solution to select the element with jQuery and then get value from from the element directly:
$('#my_element:visible')[0].value
I see on jquery documentation that I can use .parent() to filter my matched elements based on the parent. But in the process, the final result I get is the set of parent elements, not the original set of elements. So I see that I can use filter to achieve what I want. But I found so few documentation about how to use filter to filter based on the parent.
For example, my html is:
<div id="social">
Facebook<br/>
Twitter<br/>
</div>
<div id="topsites">
Facebook<br/>
Stack Overflow<br/>
</div>
I want to get a set of elements, which consist of <a> tag that has facebook in it's href attribute, but within the social div parents.
I suspect the code will be something like this:
$('a[href*="facebook"]').filter( ... ).click(function() {
});
But I have absolutely no idea what to put on the filter. "parent#social" ?
Another way to put it is to use filter function.
$('a[href*="facebook"]').filter(function(index) {
return ...
}
.click(function() {
});
I also don't know what code to put on the ... . Is it something like this.parent.id == "social" ? If possible, I prefer the first form, but if the solution can only be achieved by using the second form (filter function) then it's okay. Thank you very much.
.parent() doesn't filter, it traverses. It takes a jQuery object, that lists an array (of size [0, n) ), and generates a new jQuery object with each element's parent.
Getting a jQuery object with the list you're looking for is much simpler though...
CSS Selectors, which jQuery is based upon (with various extensions) are heirarchical by nature. That means selecting specific children of some parent(s) is quite trivial. a CSS selector to pick the element you want is:
#social a[href*="facebook"]
and if you use this inside a jQuery constructor, you'll get you object:
$('#social a[href*="facebook"]')
I want to get a set of elements, which consist of <a> tag that has facebook in it's href attribute, but within the social div parents.
No need of filter here,
$('#social a[href*="facebook"]')
will do.
I have a page with duplicate ID's for a form element. The catch is I the elements show up separately based on a toggle. So both ID's never show up simultaneously.
However when I do form validation on that element, it always selects the element displayed last in the code (even if its hidden).
Is there a selector to select the visible duplicate ID?
I've tried the following but to no avail:
$('#my_element:visible').val();
As the myriad of other questions about this premise will tell you, you cannot use the ID selector # in this case; you have to use something like $('div[id=foo]') to find it.
Duplicate IDs are invalid HTML and will nearly always cause issues with scripting. Avoid if at all possible.
The reason this is occurring is because of Duplicate IDs. IDs must be unique for the HTML to be considered valid. Whenever you aren't working against valid HTML, your results are often erratic.
In this case, even though you are only showing one of the forms at a time, they're both still present in the mark up which is why the last one in the code is always the one that's getting run.
Since you're using jQuery, I'd highly recommend using classes for this instead.
Avoid duplicates ids on the page. It is not a valid HTML.
as Rwwl said, duplicate IDs are invalid. Assign classes instead of ids to them.
Then you can do
alert($('.my_element:visible').val());
try :hidden
$("#my_element").find(":hidden").val();
Elements can be considered hidden for several reasons:
They have a CSS display value of none.
They are form elements with type="hidden".
Their width and height are explicitly set to 0.
An ancestor element is hidden, so the element is not shown on the page.
NOTE: Elements with visibility: hidden or opacity: 0 are considered to be visible,
Do not use same id for multiple elements, classes are better!
You can not specify using the # id selector only, you need to be more specific. One way is choose the type of element first then id:
For an input element:
$('input#my_element:visible').val();
or for a div element:
$('div#my_element:visible').val();
An alternative solution to select the element with jQuery and then get value from from the element directly:
$('#my_element:visible')[0].value
I'm curious if anyone knows why this piece of jQuery code doesn't remove the images?
var a = $('#tblMain').clone().remove('img');
The table is being selected. This is trying to take the table on the webpage and export to excel but I do not want the images to export.
Thank you,
Do it like this:
$("#tblMain").clone().find("img").remove();
EDIT: Okay, here's the problem:
selector: A selector expression that
filters the set of matched elements to
be removed.
http://api.jquery.com/remove/
The img in .remove('img') is to filter the set of items in the jquery object, NOT to find elements within the items themselves. In this case, the jquery object contains only one item, the cloned table. Therefore, .remove('img') removes nothing, since the jquery object does not contain any images (only images within items it contains).
I don't know what's happening behind the scenes, but you're referring to some variable called img whilst you most probably just want to select all img elements. In that case, you ought to use a selector as a string:
var a = $('#tblMain').clone().remove('img');
EDIT: .clone.remove does not seem to work indeed. I used this workaround which actually works:
.find('img').each(function() {$(this).remove()});
I want to write a select something like...
#window_4 > content > p:eq(0)
I think I have this correct, but I have a few selectors that are all similar but I can't test them all at once.
Am I right in saying this is selecting an element, who is the fist p tag child of a content tag that is a child of a tag with id 'window_4'
If I have gotten this wrong, can you give me some pointers. Would love to be able to simplify this code, I have more code selecting the tag I am after then actually doing stuff with them.
Looks good to me, although you can make it a bit more readable by substituting p:eq(0) for p:first.
Edit for comment:
jQuery always returns an array of elements, no matter whether 0, 1 or many elements were found. On these elements, yes, you can perform JS functions, such as innerHTML. You can access each element returned by jQuery just as if you would any other array:
$(".red")[0].innerHTML = "Glen Crawford";
More info: http://groups.google.com/group/jquery-ui/browse_thread/thread/34551a757f139ae1/20111f82c2596426