使用mysql編輯數據庫中的數據時出錯

Question

這是我的代碼，它顯示當前事件並讓用戶更改事件的日期，名稱或地點。

出於某種原因，我總是收到500錯誤。我認爲這是由於信息傳入和傳出數據庫。

數據庫設置爲：用戶ID，事件名稱，地點，日期，名字，事件ID ...... 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> 

然後我還有一個文件在我的包括目錄，允許更改信息。

<?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

我檢查你的代碼在第二個PHP文件中，你把一個額外的這個}請刪除它。

Answer 2

請使用mysqli_error指令，達到您得到

mysqli_query($conn, $sql)or die(mysqli_error($conn)); 

哪個錯誤，或者你可以用它來展示的php錯誤，如果有在PHP語法的錯誤頁面的開始

error_reporting(E_ALL); 
ini_set('display_errors', 1);