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javascript to add id attribute that is the same as name attribute

Posted on 2008-06-11
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Last Modified: 2012-06-27
I need a simple PHP or javascript script that parses an html page returned as a string, when it finds a name tage that doesn't have an id tag between the < > carrots.it creates an id attribute with the same name as name="objectname".  Anybody have any idea how to do this?
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Question by:kcg-witchdoctor
3 Comments
 
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Accepted Solution

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KhoiNqq earned 250 total points
ID: 21766240
Here the PHP code you needs
<?php

$str = "<html><body><div name=\"something\"><div name=\"something2\" id=\"test\"></div></div></body></html>";
 

$re  = "/<([^\s>]+) ([^>]+)>/";

if (preg_match_all($re, $str, $match)){

	foreach ($match[2] as $k => $attr_str){

		$attrs = explode(" ", $attr_str);

		$id_found = false;

		$name     = "";

		foreach ($attrs as $attr){

			if (substr($attr, 0, 3) == "id="){

				$id_found = true;

				break;

			}

			if (substr($attr, 0, 5) == "name="){

				$name = substr($attr, 5);

				echo "Found name:" . $name;

			}

		}

		if (!$id_found && $name != ""){

			$current_tag = $match[0][$k];

			$new_tag = substr($current_tag, 0, -1) . " id=" . $name . ">";

			$str = str_replace($current_tag, $new_tag, $str);

		}

 	}

}

echo "Result:\n" . $str;

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LVL 6

Assisted Solution

by:ysfx
ysfx earned 250 total points
ID: 21766309
Here's a js solution:
	<script>
 
 

		if(document.body.all)

		{

			for(i in document.body.all)

			{

				if(document.body.all[i].name != "" && document.body.all[i].id == "")

					document.body.all[i].id = document.body.all[i].name;

			}

		}

		else

		{

			for(f in document.forms)

			{
 

				for(e in document.forms[f].elements)

				{

					if(document.forms[f].elements[e].name != "" && document.forms[f].elements[e].id == "")

						document.forms[f].elements[e].id = document.forms[f].elements[e].name;
 

				}

			}

		}
 

	</script>

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0
 
LVL 2

Author Closing Comment

by:kcg-witchdoctor
ID: 31466434
Thank you both, i wish i knew how to give 100000 points to yuou both.
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