Question

I’m trying to select data from a certain row in a my sql db. This is the code im using:

mysql_connect($hostname,$username,$password) or die('connection error '.mysql_error()); mysql_select_db($dbname) or die('selection error '.mysql_error());

$select = mysql_query("SELECT FROM WordPress_theme_customers WHERE ". "GDI_Username = '$sponsor_GDI_id' AND Unique_id = '$sponsor_refference'");


while($check = mysql_fetch_array($select)) {
    $sponsor_email = $check["Email"];
    $sponsor = $check["GDI_Username"];
    $sponsor_first_nme = $check["First_Name"];
    $sponsor_second_nme = $check["Last_Name"];
    $sponsor_domain = $check["GDI_Domain"];
    $unq_id = $check["Unique_id"];
 } 

I keep getting this error:
“Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /SC/sponsor.php on line 10”

line 10 is: while($check = mysql_fetch_array($select)) {

I’ve made sure the db info is correct including the table name, still nothing.

thanks

Answer 1

Your query has a syntax error, causing msyql_query() to return a boolean FALSE, not a result handle. Try doing:

$sql = "SELECT FROM WordPress_theme_customers WHERE ". "GDI_Username = '$sponsor_GDI_id' AND Unique_id = '$sponsor_refference'";
$select = mysql_query($sql) or die($sql . "<br>" . mysql_error());

Putting the query into its own variable lets you examine it separately, and the or die(...) will tell you exactly why the query failed.

The syntax error itself is here:

SELECT FROM WordPress_theme_customers
      ^--- here

you are not specifying any fields to retrieve. At minimum, it should be:

SELECT * FROM ...