mysql_fetch_array not working inside function

Question

I was trying to make a weighted avg. value for my 5 star rating script:

function raten($setvl)
{
    global $book_id, $resulten, $dbhandle;
    $resulten =  mysql_query("SELECT SUM(rate) 
                                FROM review 
                               WHERE rate = '$setvl' AND 
                                     book_id = '$book_id",
                            $dbhandle);
    $rowen = mysql_fetch_array($resulten);
    return $rowen['SUM(rate)'];
}

Can anyone explain what I'm doing wrong?

how about making SELECT SUM(rate) as tot_rate in the query and then return $rowen['tot_rate']; — Abhik Chakraborty
– Abhik Chakraborty, Commented Mar 9, 2014 at 15:48
It's impossible to say where the issue is if you don't provide more details. What's $resulten? Is the query actually returning any row (if you run it in the console)? Is the connection actually working? What does $rowen contain (is $rowen['SUM(rate)'] actually returned by the query)? — ItalyPaleAle
– ItalyPaleAle, Commented Mar 9, 2014 at 15:49
Try if(!mysql_query(...)) /*error*/ else{ if(!mysql_fetch_array(...)) /*error*/ { /*handle result*/} and see what happens — rath
– rath, Commented Mar 9, 2014 at 15:49
PS: your coding style is really PHP 4-ish. If you're following a book, I'd recommend you to update it. The use of mysql_* functions is not recommended, and using global parameters is not really the best practice... — ItalyPaleAle
– ItalyPaleAle, Commented Mar 9, 2014 at 15:49

Ne Ma · Accepted Answer · 2014-03-09 16:31:24Z

1

You have missed a ' in your query. Change...

"SELECT SUM(rate) FROM review WHERE rate = '$setvl' AND book_id = '$book_id"

to...

"SELECT SUM(rate) FROM review WHERE rate = '$setvl' AND book_id = '$book_id'"

answered Mar 9, 2014 at 16:31

Ne Ma

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