Access lastInsertId variable outside a function

Question

so I have a simple code here that will insert a data and will return the last inserted id. Here is my code:

function newUser($fname, $age) {
  global $newUserLastID;
  $conn = new PDO('mysql:host=localhost;dbname=myDB', 'root', '');
  $conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
  $data = $conn->prepare("INSERT INTO accounts (fname, age) VALUES (?, ?)");
  $data->execute(array($fname, $age));
  $newUserLastID = $conn->lastInsertId('accounts');
}

And what I wanted to do is to run the function and get the global variable like:

newUser('JohnDoe', '22');
$someVar = $newUserLastID;

My problem is whenever I run this code, my CLI always crashes. Is there a way to fix this? I also get this on production server. I'm not getting any error besides this.

(by just running the function, my CLI crashes)

Are you positive `$someVar = $newUserLastID; ` is what's crashing it? if you comment out that line, it doesn't crash? — sqram, Aug 25 '15 at 01:40
No it isn't, just by running the function my CLI crashes @sqram — wobsoriano, Aug 25 '15 at 01:42
@FewFlyBy Your best bet, and first step, should be to run the body of the function line by line in the CLI and see which line crashes it. — sqram, Aug 25 '15 at 01:43
MySQL doesn't use sequences so you shouldn't be passing anything into `$conn->lastInsertId()` — Phil, Aug 25 '15 at 02:27

score 1 · Answer 1 · answered Aug 25 '15 at 02:24

Why not just return the new ID instead of using a global variable? So, the last line of your function would be:

return $newUserLastID

Then, when you call the function, you'd instead assign it to a variable, like this:

$myNewId = newUser('JohnDoe', '22');

I don't know if this is the cause of your error, but it might be a place to start.

Access lastInsertId variable outside a function

1 Answers1