Trying to reach the returned ID I get after I insert value but mysqli_insert_id() returns null

Question

<?php 

 $value = json_decode(file_get_contents('php://input'));
 $mysql_pekare= new mysqli ("serv", "user","pass", "db");

 if(!empty($value)) {
 $stmt = $mysql_pekare->prepare("INSERT INTO users (`username`, `password`)  VALUES(?,?)"); 

 $stmt->bind_param("ss", $value->username, $value->password);
 $stmt->execute();

     if(!empty($stmt)) {

        $contacts = array(); 
        $id = mysqli_insert_id(); 
        $contact = array("objectId" =>  ($id));
        array_push($contacts, $contact);
        echo json_encode(array('results' => $contacts), JSON_PRETTY_PRINT);
     }

 $stmt->close();
 $mysql_pekare->close();

 }
 ?>

Now when I insert my info from my frontend the value (username + password) gets added into MySQL just fine with a unique ID but I do not get the returned ID, currently I get it like this:

   {
  "results": [
      {
  "objectId": null
      }
     ]
   }

**Never store plain text passwords!** Please use PHP's [built-in functions](http://jayblanchard.net/proper_password_hashing_with_PHP.html) to handle password security. If you're using a PHP version less than 5.5 you can use the `password_hash()` [compatibility pack](https://github.com/ircmaxell/password_compat). Make sure you ***[don't escape passwords](http://stackoverflow.com/q/36628418/1011527)*** or use any other cleansing mechanism on them before hashing. Doing so *changes* the password and causes unnecessary additional coding. — Jay Blanchard, Jun 24 '16 at 17:43

score 2 · Accepted Answer · answered Jun 24 '16 at 17:42

2

Your $id is most likely null or an error not sure...

//This is for procedural and needs a link
$id = mysqli_insert_id(); 
//should be
$id = mysqli_insert_id($mysql_pekare); 
//$mysql_pekare I am assuming is your connection...

//however you are using oo (or I think) so should be
$stmt->insert_id

answered Jun 24 '16 at 17:42

nerdlyist

2,842
2
20
32

*small rant in head* yay! – nerdlyist Jun 24 '16 at 17:48

score 2 · Answer 2 · answered Jun 24 '16 at 17:45

2

Based on an example in php.net this is what you want to change in your code

$id = $stmt->insert_id();

answered Jun 24 '16 at 17:45

Works like a charm. – medvedo Jun 24 '16 at 17:46
Someone answered before you but I will give you an upvote – medvedo Jun 24 '16 at 17:48
@medvedo Had not seen that when posting answer but you are right thanks :) – Jun 24 '16 at 17:50

score 0 · Answer 3 · answered Jun 24 '16 at 17:48

0

You forgot to add the database as parameter, mysqli_insert_id($link); But, anyway you mix up class style and procedural style.

Use $id = $mysql_pekare->insert_id;

answered Jun 24 '16 at 17:48

lexx9999

736
3
9

Trying to reach the returned ID I get after I insert value but mysqli_insert_id() returns null

3 Answers3