Question

所以我正在建立一个网站，并在上面放了一个个人简介。然后在下面是一个编辑按钮，它会将您发送到另一个页面，除了某些内容是可编辑的以外，其他页面看起来几乎相同。

当您单击保存按钮时，我有一部分可以将您发送回另一页，但它被注释掉了，因为无论您单击编辑按钮如何，它都会卡在另一页上。您可以在其中输入文字和内容，只是不会保存文字并正确地将您发送回去。另外，即使现在在我的服务器中，该简历也只显示空白，就像它删除了所有文本一样。

<div class = "profileLeftSideContent" >
    <textarea rows="15" cols="40" name="bio" id="bio">
    <?php
$get_info = mysqli_query($connection, "SELECT first_name, last_name, 
bio FROM users WHERE username = '$username'");
    $get_row = mysqli_fetch_assoc($get_info);
    $db_firstname = $get_row['first_name'];
    $db_last_name = $get_row['last_name'];
    $db_bio = $get_row['bio'];
    echo $db_bio;
    ?>
        </textarea>
</div>


<form method="post" class="budon" action="" onsubmit="return false">

<input type="submit" name="test" id="test" value="Save"/><br/>

</form>

<?php


    $get_info = mysqli_query($connection, "SELECT first_name, 
last_name, bio FROM users WHERE username = '$username'");
    $get_row = mysqli_fetch_assoc($get_info);
    $db_firstname = $get_row['first_name'];
    $db_last_name = $get_row['last_name'];
    $db_bio = $get_row['bio'];
    $clided = "false";

    $bio = @$_POST['bio'];

$clunk = "true";

function click(){

    $clided = "true";



}
if($clided){

        //header("Location: profile.php ");

    $info_submit_query = mysqli_query($connection, "UPDATE users 
SET bio='$bio' WHERE username = '$username'");
    $clided = "false";




}   

if(array_key_exists('test',$_POST)){
   click();
    }
?>

Here is u: $username = mysqli_real_escape_string($connection, $_GET['u']);


This is what click() does: function click(){

                 $clided = "true";      
             }
if($clided){

        //header("Location: profile.php ");

    $info_submit_query = mysqli_query($connection, "UPDATE users SET 
  bio='$bio' WHERE username = '$username'");
    $clided = "false";




}

我希望用户在文本框中键入内容，点击“保存”，然后发送回数据库中包含新生物的其他页面。

我正在尝试使网站的简历可编辑，但我无法获取

0 个答案: