asked on

Query Selection Help

I am trying to display the data in my database but I get the error below:

Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/anudaysp/public_html/displayServices.php on line 19

Below is my code:

#CONNECT TO MYSQL
$dbcnx = @mysql_connect('localhost',$username,$password) or die("Unable to connect to MySQL");

#CONNECT TO DATABASE
@mysql_select_db($database) or die( "Unable to select database");

#CREATE THE QUERY
$sql = "SELECT S_NAME,SERVICE FROM SERVICE";

#EXECUTE QUERY
$rs = mysql_query($sql,$dbcnx);

#Write the data
while($row = mysql_fetch_array($rs)) <- Line 19
{
echo("Services: " .$row["s_name"]);
}

Also, What's the best way to debug in PHP? Believe it or not, today is my second day coding in PHP.

SOLUTION

Zontar

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ASKER CERTIFIED SOLUTION

shivsa

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shivsa

or for better debugging.

$rs = mysql_query($sql,$dbcnx) or die ("Query '$sql' failed with error message: \"" . mysql_error () . '"');

Michael701

i think your sql is the problem. you MUST qualify the field name when it is the same as the table name.

$sql = "SELECT S_NAME,SERVICE.SERVICE FROM SERVICE";
echo mysql_errno() . ": " . mysql_error(). "\n";

p.s. to resolve this i normally use plural tables

$sql = "SELECT S_NAME,SERVICE FROM SERVICES";

Zontar

> you MUST qualify the field name when it is the same as the table name.

This is incorrect.

While it's true that it is not a good practice to name a column with the same name as the table it's in, MySQL has no problem with it whatsiever if you choose to do so -- and no special syntax is required to access the column:

CREATE TABLE sometable (
id INT(8) AUTO_INCREMENT PRIMARY KEY,
sometable VARCHAR(25)
);

INSERT INTO sometable VALUES (('', 'foo'), ('', 'bar'), ('', 'baz'));

SELECT id,sometable FROM sometable;

id sometable
--- -----
1 foo
2 bar
3 baz

3 rows in set (0.00 sec)