Job $this->release() with honoring the backoff

[hsleewis]

Sometimes I don't want to throw an exception because that ends up being noise in sentry.
Is there a way to manually srelease a job back to the queue while using the backoff?

Right now I do this:

$backoff = method_exists($this, 'backoff') ? $this->backoff() : ($this->backoff ?? 60);
$backoff = is_array($backoff) ? $backoff : [$backoff];

$idx = min(count($backoff) - 1, max(0, $this->attempts() - 1));

$this->release((int) ($backoff[$idx] ?? 60) + random_int(0, 6)); // add small jitter to prevent overloading on race conditions

Is there a better way?

[heathdutton]

Short answer: No, Laravel doesn't have a built-in method for this.

When you call $this->release() manually, you need to handle the backoff delay calculation yourself. The automatic backoff handling in Laravel is only triggered when a job fails (throws an exception).

You could:

// In your job class
protected function releaseWithBackoff(): void
{
    $delay = $this->getBackoffDelay();
    $this->release($delay);
}

protected function getBackoffDelay(): int
{
    $backoff = method_exists($this, 'backoff') ? $this->backoff() : [];

    if (is_array($backoff)) {
        // Use attempt - 1 since attempts() is 1-indexed
        return $backoff[$this->attempts() - 1] ?? end($backoff) ?: 0;
    }

    return (int) $backoff;
}

Or create a reusable trait:

trait ReleasesWithBackoff
{
    protected function releaseWithBackoff(?int $jitter = null): void
    {
        $delay = $this->calculateBackoffDelay();

        if ($jitter) {
            $delay += random_int(0, $jitter);
        }

        $this->release($delay);
    }

    protected function calculateBackoffDelay(): int
    {
        $backoff = method_exists($this, 'backoff') ? $this->backoff() : [];
        $attempt = $this->attempts();

        if (is_array($backoff)) {
            return $backoff[$attempt - 1] ?? end($backoff) ?: 0;
        }

        return (int) $backoff;
    }
}