Hi Paul,
Why you are closing your connection before finishing your work on
the database?
Zareef Ahmed
----- Original Message -----
From: "Paul Jinks" <[EMAIL PROTECTED]>
To: <php-general@lists.php.net>
Sent: Friday, December 09, 2005 6:50 AM
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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