CODEWITHSUNDEEP
phpmysqlcount
Bate Wes
Bate Wes

Reputation: 39

Incorrect count

So I have a count for a database and I'm currently struggling. It tells me there's 1 result no matter how many are there. I am trying to get the code to count the number of teams with the league set to 5, and there are definitely more than one, but it continuously, the code I have written, returns the number 1. Here is the code:

function inactive_team_search() {
    $count = mysql_query("SELECT * FROM `teams` where `league`=5");
    echo count($count);
}

Any help is much appreciated! Probably a silly error where I've been coding for too long but just can't figure it out!

Upvotes: 0

Views: 58

Answers (3)

Nanhe Kumar
Nanhe Kumar

Reputation: 16297

function inactive_team_search() {
    $result = mysql_query("SELECT * FROM `teams` where `league`=5");
    echo mysql_num_rows($result);
}

Upvotes: 0

Chris Forrence
Chris Forrence

Reputation: 10084

Try this as your query instead; this will actually return the count of teams with a league of 5.

SELECT COUNT(*) FROM `teams` WHERE `league`=5

In addition, mysql_query returns its own data structure, which you need to interpret.

function inactive_team_search() {
    $query = "SELECT COUNT(*) inactiveCount FROM `teams` WHERE `league`=5";
    $result = mysql_query($query);
    if(!$result) { 
        echo 'Could not run query.';
    }
    else {
        $row = mysql_fetch_array($result); 
        echo $row['inactiveCount'];
    }
}

Finally, as an aside, mysql_query is deprecated starting in PHP 5.5, so it'd be good for you to switch to using either MySQLi or PDO.

Upvotes: 1

AmazingGrace
AmazingGrace

Reputation: 62

Use this code:

function inactive_team_search() 
{
    $result = mysql_query("SELECT COUNT FROM `teams` where `league`=5");
    $count = mysql_fetch_array($result);
    echo $count[0];
}

Upvotes: 1

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