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我正在嘗試傳遞事件ID的值,因此當用戶按下編輯按鈕時,它將傳遞事件ID並編輯它們按下的特定事件編輯。將id變量傳遞給數據庫以編輯信息的問題
數據庫佈局:
WHERE eventid='12' ";
通過識別號碼打電話,然後附加到:用戶ID,事件名稱,地點,日期,名稱,使用EVENTID
試圖編輯在DATABSE事件在事件ID等於這個數字
文件數據庫: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();
}