simple question

[Dimava]

$result = "SELECT catgroup FROM tutorials WHERE id=32";
mysql_query($result);
print $catgroup;


how can I define the variable $catrgoup to equal the thing that it pulled out from that database, because with the code above $catgroup is equal to nothing, but i just check with phpmyadmin, and there is something in that cell

thanks

Dimava

[Mark Hensler]

$result = "SELECT catgroup FROM tutorials WHERE id=32";
mysql_query($result);
extract(mysql_fetch_array($result, MYSQL_ASSOC));
print $catgroup;

[Dimava]

Warning: Undefined variable: catgroup in /home/virtual/site3/fst/var/www/html/admins/tutorials_remove.php on line 35
cat is


line 35 contains:
extract(mysql_fetch_array($result, MYSQL_ASSOC));


thanks

-Dimava

[megahard]

Undefined variable: catgroup

i think we can safely assume:

print $catgroup;

is line 35 seeming as you havent defined it and it matches up with what the error is complaing about.

[megahard]

this is how I do it, of course you can seperate each function if you want, but i usually just use a class to do all the work:


$result = mysql_fetch_array(mysql_query("SELECT catgroup FROM tutorials WHERE id=32"));
echo $result[0] . " This is the same as " . $result[catgroup];

[scoutt]

I didn't know $result[0] worked in that way, cool :)

or simply put


$search = mysql_query("SELECT catgroup FROM tutorials WHERE id=32");

while($row = mysql_fetch_array($search)){
echo $result[catgroup];

}

[Dan615]

$result = mysql_query("SELECT catgroup FROM tutorials WHERE id = 32");

$tutorial = mysql_fetch_array($result);

echo "$tutorial[catgroup]";



Echos one row, or


$result = mysql_query("SELECT catgroup FROM tutorials WHERE id = 32");

while ($tutorial = mysql_fetch_array($result)) {
echo "$tutorial[catgroup]";
}


Echos all of them :)

[Dimava]

ok, thanks for your help guys