CODEWITHSUNDEEP
phpmysqlmysqli
Philbert McPleb
Philbert McPleb

Reputation: 325

mysqli insert_id not working

Can someone take a look at this and tell me why it's not working. This is part of a class that I'm working on. The insert query works fine and if I call $char->db->insert_id; from outside the class it works ($char being the class)

function createChar($charName, $charRace, $charClass) {
    //put userId, charName and charRace into database
    $sql="insert into characters (userID, name, race, class) values ('$this->userID', '$charName', '$charRace', '$charClass') ";
    $result=$this->db->query($sql);
    if($this->db->affected_rows == '1')
        {
            return '1';
        }
        else
        {
            return '0';
        }

    //get last insert id     
    $this->charID=$this->db->insert_id;

}

Upvotes: 0

Views: 3200

Answers (3)

lightster
lightster

Reputation: 1220

Try moving $this->charID = $this->db->insert_id; before the if/else, since both will return before getting to the insert_id line.

function createChar($charName, $charRace, $charClass) {
    //put userId, charName and charRace into database
    $sql = "insert into characters (userID, name, race, class) values ('$this->userID', '$charName', '$charRace', '$charClass') ";
    $result = $this->db->query($sql);

    //get last insert id     
    $this->charID = $this->db->insert_id;

    if($this->db->affected_rows == '1')
    {
        return '1';
    }
    else
    {
        return '0';
    }
}

Upvotes: 2

user319198
user319198

Reputation:

Try below.

quote values properly: 

  $sql="insert into characters (userID, name, race, class) values 
    ('".$this->userID."', '".$charName."', '".$charRace."','".$charClass."') ";
  $result=$this->db->query($sql);
  $this->charID=$this->db->insert_id;

Upvotes: 0

Christopher Pelayo
Christopher Pelayo

Reputation: 802

It can't reach the assigning of the insert_id because of the return statement try to put it on the part before the return happens.

Upvotes: 2

Related Questions