CODEWITHSUNDEEP
phpsql
Karem
Karem

Reputation: 18103

PHP: output count()

How do i work with the COUNT()

$sql = $connect->prepare("SELECT COUNT() FROM discos_events e INNER JOIN discos_events_guests eg ON (e.ID = eg.eID) INNER JOIN users u ON (eg.uID = u.id) WHERE e.dID =:id");
$sql->bindValue(":id", $cID);

$sql->execute();

...then what? echo $sql["count"]; ? to output the count?

Upvotes: 1

Views: 250

Answers (4)

marcelog
marcelog

Reputation: 7180

you need to execute() the query, so:

$result = $sql->execute(array('id' => $cId)); // just to illustrate that you can use this instead of bindParam
if ($result) {
    $row = $sql->fetch();
}

Upvotes: 1

Spudley
Spudley

Reputation: 168685

Your query needs to assign the count value to a name, like so:

SELECT COUNT() n FROM discos_events ...

Then you can reference the name n in your PHP array:

echo $sql["n"];

You can, of course, call it 'count' or any other name you prefer, but be careful of using names which are reserved words in SQL (such as 'count'). If you want to give it a reserved name, you need to enclose it in backtick characters so that SQL recognises that it's a name you want to use rather than its own reserved word.

Upvotes: 0

Maxim Krizhanovsky
Maxim Krizhanovsky

Reputation: 26699

after execute, you have to store_result() and fetch() As @Michael suggests, you may alias the count(), to get it in more reatable form.

Upvotes: 0

Michael Berkowski
Michael Berkowski

Reputation: 270627

You need an alias name for your COUNT() column:

$sql = $connect->prepare("SELECT COUNT() AS num_events FROM discos_events e INNER JOIN discos_events_guests eg ON (e.ID = eg.eID) INNER JOIN users u ON (eg.uID = u.id) WHERE e.dID =:id");
$sql->bindValue(":id", $cID);

// Fetch the results and then access the alias
$sql->execute();
$result = $sql->fetch();
echo $result['num_events'];

Upvotes: 2

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