Ticket #10224 (closed bug: duplicate)
wrong value for selector property on chained finds with comma in selector
| Reported by: | anonymous | Owned by: | anonymous |
|---|---|---|---|
| Priority: | undecided | Milestone: | None |
| Component: | selector | Version: | 1.6.2 |
| Keywords: | Cc: | ||
| Blocking: | Blocked by: |
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
comment:1 Changed 21 months ago by rwaldron
- Owner set to anonymous
- Status changed from new to pending
- Component changed from unfiled to selector
comment:2 Changed 21 months ago by anonymous
- Status changed from pending to new
Here is the test: http://jsfiddle.net/sf2Fn/3/
comment:3 Changed 21 months ago by rwaldron
So, which part is not doing what is expected? http://gyazo.com/3048146e8db6b69e65a0fddd7e980c8e.png
comment:5 Changed 21 months ago by anonymous
- Status changed from pending to new
The second link should not be red, and the first link should read: "table img, table a" assuming the selector property has any meaning.
Please follow the bug reporting guidlines and use jsFiddle when providing test cases and demonstrations instead of pasting the code in the ticket.

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