Hi
[snip]
<?
while($ouput_row = mysql_fetch_array($result)) {
?>
[/snip]
Try:
while($ouput_row = mysql_fetch_array($result, MYSQL_ASSOC))
[snip]
<?=$output_row["projTitle"]?><br />
[/snip]
You have missed the ; (semicolon) after ["projTitle"] also
Dan
-----Original Message-----
From: Paul Jinks [mailto:[EMAIL PROTECTED]
Sent: 09 December 2005 11:51
To: php-general@lists.php.net
Subject: [PHP] PHP/MySql noob falls at first hurdle
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
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