You are here: PHP/MySql noob falls at first hurdle « PHP « IT news, forums, messages
PHP/MySql noob falls at first hurdle

Posted by Paul Jinks on 12/09/05 13:50

Hi all

I've been asked to put simple database interactivity on an academic
site. They want users to enter a few details of their projects so other
researchers can search and compare funding etc. How difficult can that
be, I thought....

I've built the database in MySQL and entered some dummy data, and I'm
now trying in the first place to get the data to display with a simple
select query to display the variable "projTitle" from the table
"project" thus:

<head>
<snip>
<?

$SQLquery = "SELECT projTitle FROM project";
$result = mysql_query($SQLquery)
or die ("couldn't execute query");
mysql_close($connect)

?>

<body>
<p>Result of <b><?=$SQLquery ?></b></p>

<p>
<?
while($ouput_row = mysql_fetch_array($result)) {
?>
<?=$output_row["projTitle"]?><br />
<?
}
?>

</p>
</body>

When I view the page I get this:

<p>Result of <b>SELECT projTitle FROM project</b></p>

<p>
<br />
<br />
<br />
<br />


</p>

There are indeed 4 entries in the database, but I can't figure out why
it's not displaying the data. It worked fine on my "PHP/Mysql-in-a-box"
course. No, we didn't study the syntax :(

Any help gratefully received.

Thanks

Paul

 

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