You are here: Re: Link parameter problem « PHP Programming Language « IT news, forums, messages
Re: Link parameter problem

Posted by Rami Elomaa on 03/21/07 19:01

Lennart Anderson kirjoitti:
> I want to present a table with main data. Each revord will have a field
> acting like a link to a new page with detailed data on the selected record.
> My problem is that I can't get the record-ID parsed into the link parameter.
> Whatever I do will just let my $_GET['id'] give me what is after the
> equal-sign in the link prameter.
> The code is:
> while($row = mysql_fetch_object($result))
> {
> $mid = ($row->catid);
> $name = ($row->catname);
> echo '<tr>';
> echo '<td >' . $mid . '</td>';
> echo '<td>' . '<a href="advertinfo.php?id=$mid">' . $name . '</a></td>';
> echo '</tr>';
> }
> echo '</table>';
>
> In this case the $_GET on advertinfor.php will only give me $mid.
> I think the problem might be in the quotes but I also think I have tested
> every possible combinaion without success.
> Any solution or hint is very much appreciated.
>
>

The difference between ' and " is that php variables inside "" are
parsed but inside '' they are not. So "$foo" will be parsed as $foo the
variable, but '$foo' is seen as a literal string, a dollar sign followed
by the string foo.

--
Rami.Elomaa@gmail.com
"Olemme apinoiden planeetalla."

 

Navigation:

[Reply to this message]


Удаленная работа для программистов  •  Как заработать на Google AdSense  •  England, UK  •  статьи на английском  •  PHP MySQL CMS Apache Oscommerce  •  Online Business Knowledge Base  •  DVD MP3 AVI MP4 players codecs conversion help
Home  •  Search  •  Site Map  •  Set as Homepage  •  Add to Favourites

Copyright © 2005-2006 Powered by Custom PHP Programming

Сайт изготовлен в Студии Валентина Петручека
изготовление и поддержка веб-сайтов, разработка программного обеспечения, поисковая оптимизация