ID: 14273
User updated by: [EMAIL PROTECTED]
Reported By: [EMAIL PROTECTED]
Status: Open
Bug Type: URL related
Operating System: Linux
PHP Version: 4.0.6
New Comment:

upgrading to http://www.php.net/~zeev/php-4.1.0.tar.gz worked!

hmm??   now should I should I run beta code on a production site?

Bill

Previous Comments:
------------------------------------------------------------------------

[2001-11-29 09:53:41] [EMAIL PROTECTED]

var_dump($url) returns 'string(206) 
"http://maps.yahoo.com/py/pmap.py?Pyt=Tmap&addr=6133 A Backlick 
Road&city=Springfield&state=VA&zipcode=22150&BFKey=&mag=9&cs=9&name=Jhoon Rhee 
Institute of Tae Kwon Do&name=Concord Centre&desc=(703) 923-0052"'

I tried 
 urlencode($url);
This worked...

I then tried
 $url = urlencode($url);
And this also cored!

------------------------------------------------------------------------

[2001-11-29 09:49:31] [EMAIL PROTECTED]

Its impossible to track this down with actual data which crashes this, please provide 
such data.

Feedback.

------------------------------------------------------------------------

[2001-11-29 09:47:47] [EMAIL PROTECTED]

I forgot the comma in the list when I rewrote the example.  I have the comma in my 
'real' code.

------------------------------------------------------------------------

[2001-11-28 17:51:47] [EMAIL PROTECTED]

Um ... list($name$url) looks really weird to me. You should write list($name, $url) 
instead.

Can you reproduce this with latest (not official) release?

http://www.php.net/~zeev/php-4.1.0.tar.gz

Also, the actual content of '$url' would be interesting anyway (since coredump is 
never ok), something like 'var_dump($url);' before passing it to urlencode().

Feedback.

------------------------------------------------------------------------

[2001-11-28 17:23:08] [EMAIL PROTECTED]

This works!!!

$cnt = count($arrayA);
for ( $ix=0 ; $ix < $cnt; $ix++ ) 
{
       $aurl = mysql_query("select name, url from sites where 
code='".$arrayA[$ix][1]."'");
       list($name$url) = mysql_fetch_row($aurl);
       mysql_free_result($aurl);

       echo "$url $name";
}

This Fails!!
$cnt = count($arrayA);
for ( $ix=0 ; $ix < $cnt; $ix++ ) 
{
       $aurl = mysql_query("select name, url from sites where 
code='".$arrayA[$ix][1]."'");
       list($name$url) = mysql_fetch_row($aurl);
       mysql_free_result($aurl);

       echo urlencode($url)," $name";
}

Only change is addition of urlencode,  What gives?

------------------------------------------------------------------------



Edit this bug report at http://bugs.php.net/?id=14273&edit=1


-- 
PHP Development Mailing List <http://www.php.net/>
To unsubscribe, e-mail: [EMAIL PROTECTED]
For additional commands, e-mail: [EMAIL PROTECTED]
To contact the list administrators, e-mail: [EMAIL PROTECTED]

Reply via email to