Question

我正在尝试传递事件ID的值，因此当用户按下编辑按钮时，它会传递事件ID并编辑他们按下的特定事件。

数据库布局：userid，eventname，venue，date，name，eventid

尝试使用以下方法编辑数据库中的事件：

WHERE eventid='12' ";

通过id号码调用它，然后附加到eventid等于该数字的数据库

file：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>

file：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();
    }

Answer 1

您还需要在html表单中添加eventid

<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>

所以这也会被提交，你可以像这样得到它

<?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' ";

将id变量传递给数据库以编辑信息的问题

1 个答案: