我試圖使用表單更新數據庫中的某些文件,出於某種原因它只是刷新頁面而不做任何事情。下面是我使用的代碼:無法更新數據庫中的字段
<?php
$id = $_GET["id"];
$select = "SELECT * FROM blog_members WHERE memberID = '$id'";
$result = mysqli_query($link, $select);
$row=mysqli_fetch_assoc($result);
$id2=$row['memberID'];
$username=$row['username'];
$password = $row['password'];
$email = $row['email'];
?>
<form id="form" action="#" method="post">
<br><table cellspacing='0' cellpadding='0'>
<tr><td>Username:</td>
<td><input name="username" type="text" id="username" placeholder="<?php echo $username; ?>" size="25" /></td></tr>
<tr><td>Password:</td>
<td><input name="password" type="text" id="password" placeholder="<?php echo $password; ?>" size="25" /></td></tr>
<tr><td>Email:</td>
<td><input name="email" type="text" id="email" placeholder="<?php echo $email; ?>" size="25" /></td></tr>
</table>
<br>
<input type="submit" name="Edit" value="Save" />
</center></form>
<?php
if(isset($POST['Edit'])){
$username2 = $_POST['username'];
$password2 = $_POST['password'];
$email2 = $_POST['email'];
$edit = mysqli_query($link, "UPDATE blog_members SET username='$username2',password='$password2',email='$email2' WHERE memberID='$id2'");
$result1 = mysqli_query($link,$edit);
if(!$result){
echo mysqli_error($link);
}else{
echo "Changes have been saved successfully!";
echo "<meta http-equiv=\"refresh\" content=\"2;URL=view-users.php\">";
}
}
?>
我想查詢$編輯它,因爲頁面只是被刷新,並且不更新從DATABSE領域不是正常工作。我覺得我失去了一些東西。有什麼建議麼?
爲什麼兩個時間更新查詢? –
表單中沒有任何操作? – Thorin
提交後可能會重置id。 –