php while-loop only updating first row results

Question

I want the cronjob function to update the money of all allusers2 by 1, instead only the first row is increased and increases by 9. (there are 10 users). I tried swapping the while loop for a for loop and had the same results. I tried turning $database->fetch into mysqli_fetch_array with the same results and turning it into mysqli_fetch_all gives me "Notice: Undefined index: //variable name\" for all database variables.

class allusers2
{
    public $id;
    public $level;
    public $money;
    private $database;

    // Methods (functions)
    public function __construct($allusers_id, $database)
    {
        $this->database = $database;
        $allusers_id = (int)$allusers_id;
        $result = $this->database->query("SELECT * FROM `users` WHERE `id`='$allusers_id'");
        if($this->database->num_rows($result) == 0) {
            throw new Exception("allusers does not exist!");
        }

        $allusers = $this->database->fetch($result);

        $this->id = $allusers['id'];
        $this->level = $allusers['level'];
        $this->money = $allusers['money'];

    }

    public function update()
    {

        $this->database->query("UPDATE `users` SET
            `level` = '{$this->level}',
            `money` = '{$this->money}'
            WHERE `id`='{$this->id}'");
    }
}

function cronjob()
{
    global $database;
    global $player;
    global $self_link;
    require('allusers2.php');
    $result = $database->query("SELECT id FROM `users`");
    $allusers_id = $database->fetch($result);
    $allusers2 = new Allusers2($allusers_id, $database);
    while($allusers_id = $database->fetch($result)) {

        $allusers2->money += 1;
        $allusers2->update();
    }
}

you should start by properly indenting your code so its readable. — Cfreak
– Cfreak, Commented Aug 10, 2016 at 19:46
"I want the cronjob function to update the money of all allusers2 by 1" - UPDATE users SET money = money + 1 and done. No Select, no loops necessary. (Assuming that by "all" you actually meant all.) — C3roe
– C3roe, Commented Aug 10, 2016 at 19:51
Rather than editing your answer to mark it as resolved, please consider accepting one of the answers that solved your problem. — Don't Panic
– Don't Panic, Commented Aug 10, 2016 at 20:41

C3roe · Accepted Answer · 2016-08-10 20:39:04Z

1

If you really want to update all records, and increment the money by the same amount, then you don't need to select the records first and loop over them - the database can do that for you in one go:

UPDATE users SET money = money + 1

answered Aug 10, 2016 at 20:39

C3roe

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Comments

aynber · Accepted Answer · 2016-08-10 19:53:33Z

0

You're only ever updating the first user:

$result = $database->query("SELECT id FROM `users`");
$allusers_id = $database->fetch($result); // Extract the first user
$allusers2 = new Allusers2($allusers_id, $database); // Create object for first user
while($allusers_id = $database->fetch($result)) {

    $allusers2->money += 1; // Keep updating that first user
    $allusers2->update();
}

You need to not do the first fetch, and move the Allusers2 into the loop:

   while($allusers_id = $database->fetch($result)) {

        $allusers2 = new Allusers2($allusers_id, $database); // Now you're creating a new object for each user
        $allusers2->money += 1;
        $allusers2->update();
    }

answered Aug 10, 2016 at 19:53

aynber

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1 Comment

Kelly Ward Over a year ago

Tried that, it has the same problem except increases the first row's money by 10 instead of 9.

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