CODEWITHSUNDEEP
phpmysqlfunctionmysqliecho
bonny
bonny

Reputation: 3247

query result wont be echoed

I have a function. in that function in have a select query. another query should return the number of results that will be in the first query. I have this code:

    function name ($f){  
     global $db;
     ...
     $results = "SELECT `a` FROM `b` WHERE $where"; 
     $results_num = ($query = mysqli_query($db, $results)) ? mysqli_num_rows($results) : 0; 

     echo $results;
     echo $results_num;
    }

will echo:

SELECT `a` FROM `b` WHERE `keywords` LIKE '%abc%'0

what is just $results but not $results_num? I do not understand why echo $results_num wont be shown and why there is a 0 at the end of $results so if there is someone who could give me advise to solve this I really would appreciate. thanks a lot.

Upvotes: 0

Views: 92

Answers (3)

Cal
Cal

Reputation: 7157

Firstly, $results_num is zero and is being output (that's why there's a 0 at the end of your output). That's because this code is wrong:

$results_num = ($query = mysqli_query($db, $results)) ? mysqli_num_rows($results) : 0;

Breaking it out a little:

$query = mysqli_query($db, $results);
$results_num = $query ? mysqli_num_rows($results) : 0;

And the second line should actually be:

$results_num = $query ? mysqli_num_rows($query) : 0;

You need to pass the query handle to mysqli_num_rows(), not the SQL.

Upvotes: 1

lafor
lafor

Reputation: 12776

$results is your query string. You can't pass a string to mysqli_num_rows() like you're doing it, it expects a result set identifier returned by mysqli_query(). In your case it's $query (you might wanna change your variable names to make more sense, BTW).

Upvotes: 0

crazyphoton
crazyphoton

Reputation: 623

$results_num is 0. $results is - SELECT a FROM b WHERE keywords LIKE '%abc%' try this

echo "Results is $results <br />";
echo "Results_num is $results_num <br />";

and everything will reveal itself :)

Upvotes: 1

Related Questions