whats wrong here !!!!!!!

[webtuto]

hi

whats the error here =>

 

$web="SELECT SUM(`rate`) / COUNT(`rate`) from `rate`";
$tuto=mysql_query($web) or die('Database error: ' . mysql_error());
$mas=mysql_fetch_array($tuto);
echo"The Rating of this tutorial is : ".$mas['rate']."<br><br>";

 

it douesnt show the result :s

[nikefido]

what error are you getting when you run the script?

 

edit: im also bad with sql, but im guessing how you are doing math in the sql query is the problem

[rlindauer]

You aren't returing "rate" from the sql query.

 

$web="SELECT SUM(`rate`) / COUNT(`rate`) as rate from `rate`";

[Donovan]

hi

whats the error here =>

 

$web="SELECT SUM(`rate`) / COUNT(`rate`) from `rate`";
$tuto=mysql_query($web) or die('Database error: ' . mysql_error());
$mas=mysql_fetch_array($tuto);
echo"The Rating of this tutorial is : ".$mas['rate']."<br><br>";

 

it douesnt show the result :s

 

 

I usually use and alias such as SELECT SUM(`rate`) AS rate_avg

[webtuto]

so whats the solution

it douesnt give any error

[rlindauer]

so whats the solution

it douesnt give any error

$web="SELECT SUM(`rate`) / COUNT(`rate`) as rate from `rate`";

[webtuto]

thanks dude it works