[SOLVED] error with simple mysql query

[daveh33]

if ($free) {
$query = mysql_query("SELECT * FROM vids WHERE number='$free'") or die("cant get free vid info");
$row = mysql_fetch_array($query) or die("error with this row");
$videourlfree = $row['videourl'];
$imgurlfree = $row['imgurl'];
}

 

The code stops and displays the message: error with this row  -any ideas???

[chiprivers]

What is error?

[daveh33]

The page shows: error with this row which is the or die message for $row line

[rajivgonsalves]

try and tell me if you see any errors

 

if ($free) {
$query = mysql_query("SELECT * FROM vids WHERE number='$free'") or die("cant get free vid info");
$row = mysql_fetch_array($query) or die("error with this row");
if (mysql_error()) { echo mysql_error();}
$videourlfree = $row['videourl'];
$imgurlfree = $row['imgurl'];
}

[daveh33]

Exact same error, the screen shows nothing but:  error with this row

[kenrbnsn]

Change you code to:

<?php
if ($free) {
     $query = "SELECT * FROM vids WHERE number='$free'";
     $rs = mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error());
     $row = mysql_fetch_array($rs) or die("error with this row<br>" . mysql_error());
     $videourlfree = $row['videourl'];
     $imgurlfree = $row['imgurl'];
}?>

 

This will tell you what the problem is.

 

Ken

 

[daveh33]

I changed that code - it just shows error with this row

 

html source of the page is:

error with this row<br>

[rajivgonsalves]

why don't you just try something like

 

$row = mysql_fetch_array($rs);

[kenrbnsn]

Please repost the code you're using.

 

Ken

[daveh33]

if ($free) {
     $query = "SELECT * FROM vids WHERE number='$free'";
     $rs = mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error());
     $row = mysql_fetch_array($rs) or die("error with this row<br>" . mysql_error());
     $videourlfree = $row['videourl'];
     $imgurlfree = $row['imgurl'];
}

[daveh33]

why don't you just try something like

 

$row = mysql_fetch_array($rs);

 

I tried this but it doesn't give any values to the below variables

    $videourlfree = $row['videourl'];

    $imgurlfree = $row['imgurl'];

 

 

[kenrbnsn]

Are you sure rows are being returned by the query?

 

Try this:

<?php
if ($free) {
     $query = "SELECT * FROM vids WHERE number='$free'";
     $rs = mysql_query($query) or die("Problem with the query: $query<br>" . mysql_error());
     if (mysql_num_rows($rs) > 0) {
         $row = mysql_fetch_array($rs);
         $videourlfree = $row['videourl'];
         $imgurlfree = $row['imgurl']; }
     else
         echo 'Problem with the query: ' . $query . '<br>No information returned from the database<br>';
}?>

 

Ken

[daveh33]

Problem with the query: SELECT * FROM vids WHERE number='2147483647'

No information returned from the database

 

I do not know where this number has come from - it was inserted into the db as 121333789826 ... strange

 

 

[daveh33]

Ah .......... its because the number field on the database is set to INT - when I change to VARCHAR it works - thanks for your help

[toplay]

Do not double post same topic again. I answered and closed your other one found at:

 

http://www.phpfreaks.com/forums/index.php/topic,167516.0.html

 

 

[daveh33]

Sorry the page crashed when I submitted and it double posted - theres no way to delete - sorry!

[revraz]

The forums have been erroring out the last few days and people think their post didn't go through, so they post again.

 

Do not double post same topic again. I answered and closed your other one found at:

 

http://www.phpfreaks.com/forums/index.php/topic,167516.0.html