Why am I getting this error ?

[baffled_in_UK]

I must be "snow-blind" because I have been looking for hours and I just cannot see where this SQL syntax error is...

"You have an error in your SQL syntax near 'Resource id #2' at line 1"

the code is :

$db = @mysql_connect("localhost","user","pass") or header("Location: maint.php");
mysql_select_db("database",$db) or header("Location: maint.php");

$result = mysql_query("SELECT field1 FROM table WHERE field2 =  '$field3'",$db);
if ($myrow = mysql_fetch_array($result)) {

// do stuff

}

Any help is much appreciated.

[bernouli]

try this

$result = mysql_query("SELECT field1 FROM table WHERE field2 =  '".$field3."'",$db);

[volleytotal_ch]

i guess that it's not an SQL statement error - it's rather a question of finding a corresponding entry ...
try to check if the query was executed successfully AND has found an entry before using the mysql_fetch_array-command, for example:

if($res)
{
  // Query executed successfully, check for entries found:
  if(mysql_num_rows($result)>0)
  {
    // entry/entries found - do something
  }
  else
  {
    // no entries found ...
  }
}
else
{
  // Query failed (no db connection or wrong sql-statement)
}

[baffled_in_UK]

Thanks guys but neither of these ideas work.

By trial and error it seems not to like

WHERE field2 ='$field3'

but I've used this before (in fact I used cut 'n paste from a working page)...... it's strange.

[cgraz]

Try changing

$result = mysql_query("SELECT field1 FROM table WHERE field2 = '$field3'",$db);

to

$result = mysql_query("SELECT field1 FROM table WHERE field2 = '$field3'");

I'm pretty sure the ,$db is causing the problem. You don't need to use that since you already established a connection to your mysql server.

Cgraz

[ereptur]

can you show where $field3 is defined?