Solved

SQL question....

Posted on 2002-05-06
5
277 Views
Last Modified: 2006-11-17
Hi,

Why does this sample of code just produce a blank page?

<?php

$cat=test;

$conn=mysql_connect("host","user","pass");
if(!$conn){
       echo"cant connect";
       die;
}

mysql_select_db("testdb");

$qid=mysql_query("SELECT * FROM testtable WHERE cat=$cat");

while($row=mysql_fetch_array($qid)){
?>
<table>
<tr><td><?=$row[heading]?></td></tr>
<tr><td><?=$row[message]?></td></tr>
<tr><td><a href="/folder/subfolder/script.php?id=<?=$row[id]?>">LINK</a></td></tr>
</table>

<br><br>
<?}?>

I am a SQL newbie, and I think there's something wrong with the "SELECT * FROM testtable WHERE cat=$cat" part, but have been unable to fiqure it out.

Thanks,
Nick!
0
Comment
Question by:Nick50000
  • 2
  • 2
5 Comments
 
LVL 32

Accepted Solution

by:
Batalf earned 95 total points
ID: 6991722
Something you could test:

1) Change $cat=test;

to $cat="test";

2) $qid=mysql_query("SELECT * FROM testtable WHERE cat=$cat");

I't always a good thing to put $cat into single-quotes, like this:

$qid=mysql_query("SELECT * FROM testtable WHERE cat='$cat'");

Batalf
0
 
LVL 4

Expert Comment

by:lokeshv
ID: 6991734
put $cat in '',

then echo ur query and check...is that alrite ?

is till doesnt work ...


put this few lines after

$qid=....

if(!$qid){

echo 'Error :'.mysql_error();
}

and check the error...


Hope this helps..

LK

0
 

Author Comment

by:Nick50000
ID: 6991760
O.K thanks, i'm trying that now.
0
 

Author Comment

by:Nick50000
ID: 6991815
Great, your advice worked first time....

Problem solved!

Thanks,
Nick!!!
0
 
LVL 32

Expert Comment

by:Batalf
ID: 6991822
Glad I could help.

Batalf
0

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