Tek-Tips is the largest IT community on the Internet today!

Members share and learn making Tek-Tips Forums the best source of peer-reviewed technical information on the Internet!

  • Congratulations Chriss Miller on being selected by the Tek-Tips community for having the most helpful posts in the forums last week. Way to Go!

Resource ID problem???

Status
Not open for further replies.

k3bap

Programmer
Joined
Jul 9, 2003
Messages
8
Location
GB
I have the following problem:

$sql_car = "SELECT Car FROM cars c INNER JOIN numberplate n ON n.Model = c.Model WHERE NumberPlate='RVVG12'";

$carModel = mysql_query($sql_car) or die("MySQL Error: " . mysql_error());

The query itself works fine in something like PhpMyAdmin however when I use it in PHP just as shown above instead of the the actual car model I get something like "Resource id #3"

Does anyone have an idea of what I am doing wrong here?

 
$carModel is the resultset that is holding the value of the column car.

try this:
$carModel = mysql_query($sql_car) or die("MySQL Error: " . mysql_error());
$row=mysql_fetch_row($carModel);
echo $row[0];



Known is handfull, Unknown is worldfull
 
thats done it. Thanks!
 
Status
Not open for further replies.

Part and Inventory Search

Sponsor

Back
Top