Question

我正在尝试为我的网站制作一个消息传递系统。我在我的处理程序php文件（处理msg的发送）中有这个代码：

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>";
}

但它一直给我这个错误：

警告：mysql_query（）期望参数1为字符串，资源在第34行的/home/deb70377/domains/cowboycombat.nl/public_html/misc/php/pm_verwerk.php中给出

无法弄清楚，我通常会做html ..

Answer 1

您不应该使用mysql *函数，因为它们已被弃用。使用mysqli *：

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

另请考虑使用预先准备好的陈述http://lt1.php.net/pdo.prepared-statements

UPADTE

获取你写的$ conn变量：

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

Answer 2

先查询，然后查询$ conn变量

php生成的表单名称未发布？

2 个答案: