[SOLVED] SQL Query not working?

[scottjcampbell]

Hi, I have been trying to get my sql query working for some time now. I would greatly appreiciate it if anyone could find a solution to this:

 

      <?
$query = "SELECT * FROM messages WHERE trigger = '1'";
$result = mysql_query($query);

while($row = mysql_fetch_array($result)){

$type =  $row["message_type"];
$text =  $row["message_text"];

echo "<div id=\"$type\"><div class=\"bl\"><div class=\"br\"><div class=\"tl\"><div class=\"tr\">$text</div></div></div></div></div>";
}
?>

and i get this error:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /public_html/index.php on line 40

I have tried removing: WHERE trigger = '1'" and it works, but i dont know why?

 

Cheers Scott!

 

[PFMaBiSmAd]

I have tried removing: WHERE trigger = '1'" and it works

 

That would imply that trigger is not a column in your table. What does echoing mysql_error(); show?

[scottjcampbell]

Hi,

The echo mysql_error(); returns nothing (blank) and the "trigger" is a column in the table. Is there another method i need to use to make it so only trigger=1 displays.

 

cheers, scott.

[PFMaBiSmAd]

Since trigger is a reserved mysql keyword, it would need to be enclosed in back-ticks ` for it to be treated as an identifier.

[scottjcampbell]

thank you - SO MUCH!