How to get raw html passed as jQuery selector?

Question

Note: Answer given by SLacks requires the least code to get what I want compared to answers in the duplicate questions. Accepting this.

I'm trying to do get the raw html that's passed in as the selector:

$('<div></div>')

I need just the html so '<div></div>'

Thanks!

score 3 · Accepted Answer · edited Oct 10 '16 at 16:20

3

You're looking for the native outerHTML property: el[0].outerHTML

Note: this is not quite "raw"; this is the HTML as parsed by the browser.

example of parsed output:

$("<div></div>")[0].outerHTML = "<div></div>";    
$("<div/>")[0].outerHTML = "<div></div>";

edited Oct 10 '16 at 16:20

freedomn-m

answered Oct 10 '16 at 16:17

SLaks

1

Doing this in the console gives an error $('
').outerHtml(); – User123 Oct 10 '16 at 16:19
Note the `[0]` in the example: `$('
')[0].outerHtml();` – Rory McCrossan Oct 10 '16 at 16:19
@User123 : `$('
')[0].outerHtml();` – Pranav C Balan Oct 10 '16 at 16:19
property is not function... `$('
')[0].outerHTML` – musefan Oct 10 '16 at 16:21
1

$('
')[0].outerHtml(); – User123 Oct 10 '16 at 16:22
Here's a working example: https://jsfiddle.net/huyod8sz/ – Rory McCrossan Oct 10 '16 at 16:24
Sorry i didnt captialize the HTML ->, $("
")[0].outerHTML.. this works. Thanks SLaks! – User123 Oct 10 '16 at 16:26
on version 3.3.1 not aperar – Rubén Ruíz Mar 30 '18 at 13:24

1 Answers1