Question

我遇到了问题，我不知道自己做错了什么。我想将数据库中的值输出到表单中，我得到的只是一个空白页面。值应该在每个用户的表单内，因此我使用用户ID。这是我做的：

<?php 
include 'inc/database.php'; 

$memberID = "SELECT id FROM admin";
$result_member_ID = mysqli_query ($connect, $memberID);

$get_admin = "SELECT * FROM admin WHERE id = '$result_member_ID'";
$result_get_admin = mysqli_query($connect, $get_admin);
$_SESSION['data'] = mysqli_fetch_assoc($result_get_admin);
?>

现在在我的表单中，字段看起来像这样：

<form method="post" action="edit_mitarbeiter.php" class="form-horizontal form-label-left">
<div class="form-group">
  <label class="control-label col-md-3 col-sm-3 col-xs-12">Benutzername:  </label>
   <div class="col-md-9 col-sm-9 col-xs-12">
    <input type="text" name="username" class="form-control" placeholder="Geben Sie einen Benutzernamen an." value="<?php echo $_SESSION['data']['username']; ?>">
</div>
</div>
</form>

现在我得到的是一个空白页面，因为我的错误在这里：

$memberID = "SELECT id FROM admin";
$result_member_ID = mysqli_query ($connect, $memberID);

这个变量有什么问题？我想从用户那里获取ID，之后我选择id = $result_member_ID的所有用户！

有人能告诉我我做错了吗？

编辑：这是我目前获取记录的php代码。我收到一条错误消息，我将在下面发布。代码：

include 'inc/database.php'; 

$memberID = "SELECT id FROM admin";
$result_member_ID = mysqli_query ($connect, $memberID);

$get_admin = "SELECT * FROM admin WHERE id = '$result_member_ID'";
$result_get_admin = mysqli_query($connect, $get_admin);
while($dataarray =  mysqli_fetch_assoc($result_get_admin))  //fetch multiple data
{
$_SESSION['data'] = $dataarray ['id'];
$_SESSION['name'] = $dataarray ['username'];
}

错误讯息：可捕获的致命错误：第23行的/var/customers/webs/ccsolution/work/schulkantine/admin/edit_mitarbeiter.php中无法将类mysqli_result的对象转换为字符串

第23行：

$get_admin = "SELECT * FROM admin WHERE id = '$result_member_ID'";

编辑2：现在它正在工作，正确的数据显示在输入字段内。现在的问题是，只要我点击提交按钮，我就会收到一条成功消息，但数据库中的数据没有改变。请查看我的代码现在的样子：

<?php 
include 'inc/database.php'; 

$result_member_ID = $_GET['id'];  //receive the id from page

$get_admin = "SELECT * FROM admin WHERE id = '$result_member_ID'";
$result_get_admin = mysqli_query($connect, $get_admin);
$_SESSION['data'] = mysqli_fetch_assoc($result_get_admin);

// Check if form is submitted
if (isset ($_POST['editmitarbeiter'])) {
    $update_username = mysqli_real_escape_string ($connect, $_POST['username']);
    $update_password = mysqli_real_escape_string ($connect, $_POST['password']);
    $update_email = mysqli_real_escape_string ($connect, $_POST['email']);
    $update_vorname = mysqli_real_escape_string ($connect, $_POST['vorname']);
    $update_nachname = mysqli_real_escape_string ($connect, $_POST['nachname']);
    $update_telefon = mysqli_real_escape_string ($connect, $_POST['telefon']);


    $sql = mysqli_query ($connect, "UPDATE `admin` SET 
     username='".$update_username."', password='".$update_password."', email='".$update_email."', vorname='".$update_vorname."', nachname='".$update_nachname."', telefon='".$update_telefon."'  WHERE id = $result_member_ID");

if (mysqli_affected_rows($connect) == 0) //<--
{
    die('Could not update data: ' . mysqli_error($connect));
} else {
    $msg_success= '<strong>Gratulation!</strong> Der Mitarbeiter wurde erfolgreich hinzugefügt. Zur <a href="mitarbeiter.php"><span style="color:#fff;">Übersicht aller Mitarbeiter >></span></a>';
}       

}       
?>

为什么更改未保存在数据库中的任何想法？

谢谢，克里斯

Answer 1

<?php 
include 'inc/database.php'; 

$result_member_ID = $_GET['id'];  //receive the id from page

    $get_admin = "SELECT * FROM admin WHERE id = '$result_member_ID'";
        $result_get_admin = mysqli_query($connect, $get_admin);
        while($dataarray =  mysqli_fetch_assoc($result_get_admin))  //fetch multiple data
        {
        $_SESSION['data'] = $dataarray ['id'];
        $_SESSION['name'] = $dataarray ['username'];
        }

接收多个数据时必须使用while循环。无法在一行中获取一组数据。

更新数据库时遇到的第二个问题失败，查询中的$result_member_ID原因位于if (isset ($_POST['editmitarbeiter']))内，因此未获得该值

所以解决方案是

添加这样的隐藏输入并将其放入表单

<input type="hidden" value="<?php echo $result_member_ID;?>" name="member_Id">

在表单与其他输入一起提交时发布

$member_Id = mysqli_real_escape_string ($connect, $_POST['member_Id']);

将查询更改为此

 $sql = mysqli_query ($connect, "UPDATE `admin` SET 
 username='$update_username',
 password='$update_password',
 email='$update_email',
 vorname='$update_vorname',
 nachname='$update_nachname',
 telefon='$update_telefon'
 WHERE id = '$member_Id'");

Answer 2

不确定PHP但是你检查了你的查询;根本没有意义。

$memberID = "SELECT id FROM admin";
$result_member_ID = mysqli_query ($connect, $memberID);

$get_admin = "SELECT * FROM admin WHERE id = '$result_member_ID'";

这应该在下面，您应该使用IN运算符

SELECT * FROM admin WHERE id IN ('$result_member_ID'")

但这与说

相同

SELECT * FROM admin WHERE id IN ( select id from admin)

（OR）就在下面，因为您选择了每条记录

SELECT * FROM admin

PHP显示当前用户ID的值

2 个答案: