After examining the jQuery source, I see that the problem I am having is because replaceWith
calls html
which does not exist for XML documents. Is replaceWith
not supposed to work on XML documents?
I have found this admittedly simple workaround, in case anybody needs it in the future, that will accomplish what I'm trying to do:
xml.find('b').each(function() {
$(this).replaceWith($('<c>yo</c>')) // this way you can custom taylor the XML based on each node's att开发者_如何学Goributes and such
});
But I would still like to know why the easy way doesn't work.
I don't know much about jQuery, but shouldn't this work?
xml = $.parseXML('<a><b>hey</b></a>')
$(xml).find('b').replaceWith('<c>yo</c>')
Instead of xml
representing <a><c>yo</c></a>
it fails and represents <a></a>
. Did I do something wrong? I am using jQuery 1.6.2.
Edit:
As a side note, if I try to use the function version of replaceWith
, like so:
$(xml).find('b').replaceWith(function() {
return '<c>yo</c>' // doesn't matter what I return here
})
I get this error:
TypeError: Cannot call method 'replace' of undefined
Edit 2:
replaceAll
works however, but I need to use the function version so I can't settle for this:
$('<c>yo</c>').replaceAll($(xml).find('b')) // works
Edit 3:
This also works:
xml.find('b').replaceWith($('<c>yo</c>')) // but not with the $() around the argument
This looks like either a design limitation with replaceWith()
or a bug.
When I run:
$(xml).find('b').replaceWith(function() {
return '<c>yo</c>';
})
I get a "this[0].innerHTML is undefined"
exception. See this jsFiddle.
Drilling into xml
, the b
node doesn't have an innerHTML member -- which makes a little sense, since it's not HTML. ;)
So, it's looking like replaceWith()
may not always play nice with XML. Consider reporting a bug.
yes. this is old bug and it still exists. you can workaround it:
$.ajax
dataType: "xml"
...
success: (data) ->
$(data).find("section").each ->
ugly_but_working_clone = $($(".existing_dom_element").append(this).html())
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