PHP生成的表单名称不发布？

Question

I am trying to make a messaging system for my site. I have this code in my handler php file (which processes the sending of msg):

if($actions=="verstuur"){
    //read variables
    $naar = $_POST['naar'];//varchar in database. the post = 'ikdekker
    $van = $username;//varchar in database. the username = 'user1'
    $status = "0";//int in database

    if ($_POST['admin'] != ""){
        $admin = $_POST['admin'];}else{
        $admin=2; //int in database. the post = '0 or 1'
    }

    $onderwerp = $_POST['onderwerp'];//varchar in database. the post = 'example'
    $bericht = $_POST['berichtl'];//varchar in database. the post = 'example message'
    $tijd=date("Y-m-d H:i:s");//timestamp in database. the post = '2014-18-1 20:20:20'
    // enter is enter
    $bericht=nl2br($bericht);
    $bericht=eregi_replace("\n","",$bericht);
    mysql_query($conn, "INSERT INTO pm (van,naar,status,admin,onderwerp,tijd,bericht)         VALUES ('$van','$naar','$status','$admin', '$onderwerp','$tijd','$bericht')");
    echo"<script>";
    echo"alert('". $actions . "');";
    echo "</script>";
}

but it keeps giving me this error:

Warning: mysql_query() expects parameter 1 to be string, resource given in /home/deb70377/domains/cowboycombat.nl/public_html/misc/php/pm_verwerk.php on line 34

cant figure it out, I normally do the html..

Answer 1

You should not use mysql* functions as they are deprecated. Use mysqli* ones:

mysqli_query($conn, "INSERT INTO pm (van,naar,status,admin,onderwerp,tijd,bericht)         VALUES ('$van','$naar','$status','$admin', '$onderwerp','$tijd','$bericht')");

Also consider using prepared statements http://lt1.php.net/pdo.prepared-statements

UPADTE

to get the $conn variable you write:

$conn = mysqli_connect('localhost', 'my_user', 'my_password', 'my_db');

Answer 2

首先查询，然后查询$ conn变量