将id变量传递给数据库以编辑信息的问题

我正在尝试传递事件ID的值，因此当用户按下编辑按钮时，它将传递事件ID并编辑它们按下的特定事件编辑。将id变量传递给数据库以编辑信息的问题

数据库布局：

WHERE eventid='12' ";

通过识别号码打电话，然后附加到：用户ID，事件名称，地点，日期，名称，使用EVENTID

试图编辑在DATABSE事件在事件ID等于这个数字

文件数据库：editevent.php：

<?php 

    include 'dbh.php'; 
    include_once 'header.php'; 
    session_start(); 

    if(!isset($_SESSION['u_uid'])){ 
     header("Location:signup.php"); 
    } 

?> 

<section class="main-container"> 

<h1>test</h1> 

<?php 

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

    echo "<form method='POST' action='includes/editevents.inc.php'> 
     <input type ='text' name='eventname' placeholder='event name'> 
     <input type ='text' name='venue' placeholder='event name'> 
     <input type ='date' name='date' placeholder='event date'> 
     <button type='submit' name='eventsubmit'>Submit</button> 
     </form>"; 
?> 

</section>

文件：editevents.inc.php：

<?php 

include 'dbh.inc.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 eventid='12' "; 
     mysqli_query($conn, $sql); 
     header("Location: ../members.php?event=success"); 
     exit(); 
    } else { 
     header("Location: ../signup.php"); 
     exit(); 
    }

来源

2017-10-07 d joe

您需要将事件ID添加到您的html表单以及

<input type ='text' name='eventname' placeholder='event name'> 
     <input type ='text' name='venue' placeholder='event name'> 
     <input type ='date' name='date' placeholder='event date'> 

     <!-----Include event id of the event you want to update ----> 

     <input type ='hidden' name='eventid' value="$eventid"> 
    <button type='submit' name='eventsubmit'>Submit</button>

因此，这被提交，以及，你会得到像这样

<?php 
    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']); 
      $eventid =mysqli_real_escape_string($conn, $_POST['eventid']); 


       $sql = "UPDATE events SET eventname='$eventname' WHERE eventid='$eventid' ";

来源

2017-10-07 22:50:44

将id变量传递给数据库以编辑信息的问题

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