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Home/ Questions/Q 6862653
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Editorial Team
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Editorial Team
Asked: May 27, 20262026-05-27T02:41:15+00:00 2026-05-27T02:41:15+00:00

I know there are a lot of topics with the same title. But mostly

  • 0

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)

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-27T02:41:16+00:00Added an answer on May 27, 2026 at 2:41 am

    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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