php - 使用mysql编辑数据库中的数据时出错

Question

这是我的代码，它显示当前事件并允许用户更改事件的日期、名称或地点。

出于某种原因，我不断收到 500 错误。我认为这是由于信息被传入和传出数据库。

数据库设置为:userid ,eventname, venue, date, name ,eventid... respectivley

        <div class="current events">
            <h1>Your Current Events:</h1>
            <?php

            $sql = "SELECT * FROM events WHERE userid='{$_SESSION['u_id']}';";
            $result = mysqli_query($conn, $sql);
            if (mysqli_num_rows($result) > 0){
                while ($row = mysqli_fetch_assoc($result)){
                    echo "<b>Event name: </b>";
                    echo " ";
                    echo $row['eventname'];
                    echo " ";
                    echo "<b>Event Venue: </b>";
                    echo " ";
                    echo $row['venue'];
                    echo " ";
                    echo "<b>Event Date: </b>";
                    echo " ";
                    echo $row['date'];
                    echo "
                        <form method='POST' action='editevent.php'>
                            <input type='hidden' name='eventname' value='" .$row['eventname']. "'>
                            <input type='hidden' name='venue' value='" .$row['venue']. "'>
                            <input type='hidden' name='date' value='" .$row['date']. "'>
                            <input type='hidden' name='name' value='" .$row['name']. "'>
                            <button>Edit</button>

                        </form>


                    ";

                }
            }else{
                echo "No Upcoming Events";
            }

            ?>
        </div>

然后我的 includes 目录中有另一个文件允许更改信息。

<?php

    session_start();
    if (isset($_POST['eventsubmit'])) {
        $eventname = $_POST['eventname'];
        $venue = $_POST['venue'];
        $date = $_POST['date'];
        $name = $_POST['name'];

        $eventname = mysqli_real_escape_string($conn, $_POST['eventname']);
        $venue = mysqli_real_escape_string($conn, $_POST['venue']); 
        $date = mysqli_real_escape_string($conn, $_POST['date']); 
        $name = mysqli_real_escape_string($conn, $_POST['name']);   


            $sql = "UPDATE events SET eventname='$eventname' WHERE userid='2' ";
            mysqli_query($conn, $sql);
            header("Location: ../members.php?event=success");
            exit();



    } else {
        header("Location: ../signup.php");
        exit();
    }

}

Answer 1

我在第二个 php 文件中检查了你的代码，你多放了一个 请删除它。