mysql_insert_id() returns 0

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According to the manual mysql_insert_id returns:

The ID generated for an AUTO_INCREMENT column by the previous query on success, 0 if the previous query does not generate an AUTO_INCREMENT value, or FALSE if no MySQL connection was established.

Since it does not give you false and not the correct number it indicates that the queried table didn't generate an auto-increment value.

There are two possibilities I can think of:

  1. Your table doesn't have an auto_increment field
  2. Since you doesn't provide the link to the mysql_insert_id() but using a link with mysql_query() it might not be the correct table that's queried when retrieving the last inserted id.

Solution:

  1. Make sure it has an auto_increment field
  2. Provide the link aswell: $waarde = mysql_insert_id($this->db);
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user750079
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user750079

Updated on July 19, 2022

Comments

  • user750079
    user750079 almost 2 years

    I know there are a lot of topics with the same title. But mostly it's the query that's been inserted in the wrong place. But I think I placed it right. So the problem is, that I still get 0 even when the data is inserted in the db. Does someone knows an answer where I could be wrong?

    here's my code:

    mysql_query('SET NAMES utf8');
        $this->arr_kolommen = $arr_kolommen;
        $this->arr_waardes = $arr_waardes;
        $this->tabel = $tabel;
        $aantal = count($this->arr_kolommen);
        //$sql="INSERT INTO `tbl_photo_lijst_zoekcriteria` ( `PLZ_FOTO` , `PLZ_ZOEKCRITERIA`,`PLZ_CATEGORIE`)VALUES ('$foto', '$zoekje','$afdeling');";
        $insert = "INSERT INTO ".$this->tabel." ";
        $kolommen = "(";
        $waardes = " VALUES(";
        for($i=0;$i<$aantal;$i++)
        {
            $kolommen .=$this->arr_kolommen[$i].",";
            $waardes .="'".$this->arr_waardes[$i]."',";
        }
        $kolommen = substr($kolommen,0,-1).")";
        $waardes = substr($waardes,0,-1).")";
        $insert .=$kolommen.$waardes;   
        $result = mysql_query($insert,$this->db)  or die ($this->sendErrorToMail(str_replace("  ","",str_replace("\r\n","\n",$insert))."\n\n".str_replace(" ","",str_replace("\r\n","\n",mysql_error()))));
        $waarde = mysql_insert_id();
    

    Thanks a lot in advance, because I have been breaking my head for this one for almost already a whole day. (and probably it's something small and stupid)

  • user750079
    user750079 over 12 years
    The ID is an auto_increment field. But how could I solve it, if it's the 2nd problem?
  • Marcus
    Marcus over 12 years
    @user750079 try call mysql_insert_id with $this->db as I've mentioned in the answer
  • user750079
    user750079 over 12 years
    @ Marcus Sorry I overread it. But it worked. Thank you very much. I was so stupid to write instead of $this->db ; $insert, $result (with other words, all other vars, but forgot the $this). Thank you very much.
  • Buffalo
    Buffalo almost 11 years
    What do you mean by $this->db? my code is this: mysql_query("INSERT INTO mytable (myColumn) values ('myValue');"); $id = mysql_insert_id();
  • Sudhakaran Packianathan
    Sudhakaran Packianathan almost 4 years
    Please give the resource name also in case you get 0. eg. mysql_insert_id($connection)