PHP，MySQL，PDO事务 - try块中的代码是否在commit（）处停止？

Question

I'm pretty new to transactions.

Before, what I was doing was something like:

Code Block 1

$db = new PDO(...);

$stmt = $db->prepare(...);

if($stmt->execute()){
    // success
    return true;
}else{
    // failed
    return false;
}

But in an attempt to group multiple queries into a single transaction, I'm now using something like:

Code Block 2

$db->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

$db->beginTransaction();

try{
    $stmt = $db->prepare(... 1 ...);
    $stmt->execute();

    $stmt = $db->prepare(... 2 ...);
    $stmt->execute();

    $stmt = $db->prepare(... 3 ...);
    $stmt->execute();

    $db->commit();

    return true;
}catch(Exception $e){
    // Failed, maybe write the error to a txt file or something
    $db->rollBack();
    return false;
}

My question is: If the transaction fails for whatever reason, does the code stop at $db->commit(); and jump to the catch block? Or would the return true; run first, and then it would try to go to the catch? 'Cause if that's the case, then I've already returned, and so it wouldn't go to the catch. AND it would have returned the wrong value.

Do I still need to include something like:

Code Block 3

if($stmt->commit()){
    return true;
}

or is it sufficient the way I have it written in Code Block 2?

Answer 1

If the transaction fails for whatever reason, the code does stop at the very line where error occurred end then the execution jumps directly to the catch block. So it is sufficient the way you have it written in Code Block 2.

Note that you should always re-throw the Exception after rollback. Otherwise you will never have an idea what was a problem. So it should be

try{
    $stmt = $db->prepare(... 1 ...);
    $stmt->execute();

    $stmt = $db->prepare(... 2 ...);
    $stmt->execute();

    $stmt = $db->prepare(... 3 ...);
    $stmt->execute();

    $db->commit();

    return true;
}catch(Exception $e){
    $db->rollBack();
    throw $e;
}