Question

我正在尝试将sender_id和receiver_id添加到我的变量中，但它会让我跟踪错误

警告：mysql_fetch_array（）期望参数1是资源，   在/opt/lampp/htdocs/Task/php/insertdatanew.php中给出的对象在线   36

警告：mysql_fetch_row（）期望参数1是资源，对象   在第42行的/opt/lampp/htdocs/Task/php/insertdatanew.php中给出

CODE：

 <?php
          $sender_id = 0;
          $receiver_id = 0;
          session_start();
    if(isset($_SESSION['login_user'])) {
        //header('location: chat.php');
    }
    else {
        header('location: chatlogin.php');
    }

    if(isset($_REQUEST['sender']) AND isset($_REQUEST['msg']) AND isset($_REQUEST['time']) AND isset($_REQUEST['receiver']) ){

        $msg = $_REQUEST['msg'];
        $time = $_REQUEST['time'];
        $sender = $_REQUEST['sender'];
        $receiver = $_REQUEST['receiver'];



        echo $sender;
        echo $receiver;
        echo $msg;
        echo $time;

    //echo $msg;

        if ( empty($_REQUEST['msg']) ) {
        }

        else{
            require_once 'dc_chat.php';
            $mysqli = new mysqli(DB_HOST, DB_USER, DB_PASS, DB_NAME);

            $result = $mysqli -> query("SELECT id from users where username LIKE '{$sender}'");

            $row = mysql_fetch_row($result); 
            $sender_id = $row[0]; 

            $result = $mysqli -> query("SELECT id from users where username LIKE '{$receiver}'");
            $row = mysql_fetch_row($result); 
            $receiver_id = $row[0]; 

            //echo $receiver_id;

            $sql = $mysqli -> query("INSERT  INTO `messages` (`sender_id`, `receiver_id`, `msg`, 'chattime') VALUES ('{$sender_id}', '{$receiver_id}', '{$msg}' ,'{$time}')");              

                if(! $sql ) {
                    echo 'N';
                }
                else {
                    echo 'Y';
                }
            }
        }


        else{
            echo "hello";
        }



?>

我从ajax得到$ msg，$ time，$ sender，$ receiver，数据类型是JSON

Answer 1

你正在混合mysql和mysqli函数。 mysql_fetch_row应该是mysqli_fetch_row。

将if (!$sql)替换为if ($mysqli->affected_rows == 0)以测试INSERT语句是否成功。

如果您怀疑SQL语句中有错误，请在查询行后使用echo $mysqli->error;将其放在屏幕上。

从数据库中检索int值

1 个答案: