#10224 closed bug (duplicate)
wrong value for selector property on chained finds with comma in selector
Reported by: | anonymous | Owned by: | anonymous |
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Priority: | undecided | Milestone: | None |
Component: | selector | Version: | 1.6.2 |
Keywords: | Cc: | ||
Blocked by: | Blocking: |
Description
If I do: var a = jQuery("table:first").find("tr").find('img,a')
I correctly get all the images and links in the rows of the first table.
However, the returned jQuery object has this as the value for its selector property:
console.log(a.selector); "table:first tr img,a"
which is just a concatenation and would not return the same thing at all:
jQuery("table:first tr img,a")
Change History (8)
comment:1 Changed 11 years ago by
Component: | unfiled → selector |
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Owner: | set to anonymous |
Status: | new → pending |
comment:2 Changed 11 years ago by
Status: | pending → new |
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Here is the test: http://jsfiddle.net/sf2Fn/3/
comment:3 Changed 11 years ago by
So, which part is not doing what is expected? http://gyazo.com/3048146e8db6b69e65a0fddd7e980c8e.png
comment:4 Changed 11 years ago by
Status: | new → pending |
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comment:5 Changed 11 years ago by
Status: | pending → new |
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The second link should not be red, and the first link should read: "table img, table a" assuming the selector property has any meaning.
comment:6 Changed 11 years ago by
Resolution: | → duplicate |
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Status: | new → closed |
Don't look at the .selector
property, it's not documented to survive traversal methods and is only to be used internally.
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