Why is this query giving me a value of resource 10?

[ghurty]

I am trying to have a sql command pull up the "name" value for a "group" number of the same record. I have a table has both fields.

 

$queryid = "SELECT 'name'  FROM `users` WHERE `group` = " . $GROUP;
$resultid = mysql_query($queryid)
    or die("Web site query failed");

if ($debug) :
fputs($stdlog, "Name has been set to=". $resultid. "\n" ); 
endif ;

 

For some reason the debug is showing that it has been set to "resource #10".

 

What should I change?

 

Thank You

[trq]

The code is giving the expected result. mysql_query returns a result resource. You'll likely want to pass it to mysql_fetch_assoc (or similar) if you actually want to get to the data itself.

[ghurty]

I just need the value that is in the "name" field.

 

How would I rewrite it to get that value?

 

Thanks

[oni-kun]

I just need the value that is in the "name" field.

 

How would I rewrite it to get that value?

 

Thanks

 

while ($row = mysql_fetch_assoc($resultid)) {
    echo $row['name'];
}

[trq]

oni-kun's example is good, but you really don't want / need the loop.

[ghurty]

I will try to use oni-kun's example, but is there a better way?

 

Thanks

[oni-kun]

I will try to use oni-kun's example, but is there a better way?

 

Thanks

 

Oh yes, I believe you can just do this:

 fputs($stdlog, "Name has been set to=". mysql_fetch_row($resultid) . "\n" );

[trq]

$queryid = "SELECT 'name'  FROM `users` WHERE `group` = " . $GROUP;
$resultid = mysql_query($queryid) or die("Web site query failed");
$row = mysql_fetch_assoc($resultid);
if ($debug) :
fputs($stdlog, "Name has been set to={$rwo['name']}\n" ); 
}

[Buddski]

You will need to change the query for it to work as expected..

$queryid = "SELECT 'name'  FROM `users` WHERE `group` = " . $GROUP;

to

$queryid = "SELECT `name`  FROM `users` WHERE `group` = " . $GROUP;

[ghurty]

When I have the code like:

$queryid = "SELECT `name` FROM `users` WHERE `group` = " . $GROUP;
$resultid = mysql_query($queryid) or die("Web site query failed");
$row = mysql_fetch_assoc($resultid);
if ($debug) :
fputs($stdlog, "Name has been set to={$row[`name`]}\n" ); 
}

It locks up and doesnt process it. It appears the problem is this line:

fputs($stdlog, "Name has been set to={$row[`name`]}\n" ); 

Because when I remove it, it does complete the script.

 

When I have:

fputs($stdlog, "Name has been set to= $row\n" ); 

 

It displays:

Name has been set to=Array

 

 

Thank You

[trq]

You using backticks in place of single quotes.

 

fputs($stdlog, "Name has been set to={$row['name']}\n" );