CODEWITHSUNDEEP
phpjquerydebuggingfirebug
Klausos Klausos
Klausos Klausos

Reputation: 16060

Firebug for debugging jQuery elements

I'm debugging my jQuery DataTable using Firebug. So, in the Firebub panel I see:

<div id="newspaper-b_length" class="dataTables_length">
<label>
Show
<select name="newspaper-b_length" size="1" aria-controls="newspaper-b">
entries
</label>
</div>

but I don't have this piece of code in my php files. How actually I can find these lines of code in my php files?

Upvotes: 0

Views: 250

Answers (2)

Ricardo Lohmann
Ricardo Lohmann

Reputation: 26320

"Firebug's HTML tab shows you what the HTML looks like right now." as you can see here.
Probably it's being generated by other code php, js or html.

Upvotes: 1

Selvakumar Arumugam
Selvakumar Arumugam

Reputation: 79850

Those are additional wrappers created by datatables plugin to show entries drop down.

Most widget create wrappers/additional div to show their widget controls. Those div's are created by datatables plugin dynamically and so you will not find it in your php code.

Upvotes: 2

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