Re: mysql_query and SELECT syntax (Using php4, mysql, apache on WIndows XP home)
| From: | DL Neil | Date: | Wed, 07 Aug 2002 15:03:04 +0000 |
| Subject: | Re: mysql_query and SELECT syntax (Using php4, mysql, apache on WIndows XP home) | ||
| References: | 1 | Groups: | php.windows |
| Request: | Send a blank email to php-windows+get-15154@lists.php.net to get a copy of this message | ||
Marlene,
I want to SELECT a field from a table but only if it meets TWO criteria and
then assign it to $result. I can get this code (only requiring one
criteria) to work:
After connecting to my database...
Partial code in a function:
$query = "SELECT username FROM $user_tablename
WHERE userid = '$userid'";
$result = mysql_query($query);
if (!mysql_num_rows($result)) return 0;
else {
$query_data = mysql_fetch_row($result);
return $query_data[0];
}
But when I try to add another crtiteria...
WHERE userid = "$userid'
AND userpassword =
password('$userpassword')";
I get an error about mysql_num_rows() supplied argument is not a valid MySQL
result resource. Can anyone help?
=If the final query is:
$query = "SELECT username FROM $user_tablename
WHERE userid = "$userid'
AND userpassword =
password('$userpassword')";
then the quotation marks don't balance correctly.
Standard advice: print the SQL before calling the RDBMS so that you can see
what is actually transferred.
=dn