Birrel Birrel - 6 months ago 11
PHP Question

PHP, MySQL, PDO Transactions - Does the code inside try block stop at commit()?

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

The execution is stopped when an exception is thrown.

The first return will not be reached but the catch statement will be executed.

You can even return the commit directly:

$dbh->beginTransaction();
try {
    // insert/update query
    return $dbh->commit();
} catch (PDOException $e) {
     $dbh->rollBack();
     return false;
}
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