Question

我尝试使用此表单编辑我输入数据库的数据。

<form method="POST">
                        <table align ="center">
                        <br>
                            <tr><td>First Name: </td> <td><input type="text" name="first" pattern="[A-Za-z]{1,}" 
                            title="A-Z only" required value="<?php echo $query2['fname']; ?>"></td> </tr>

                            <tr><td>Middle Initial: </td><td><input type="text" name="middle" 
                            value="<?php echo $query2['middleinitial'] ?>"></td></tr>

                            <tr><td>Last Name:</td><td> <input type="text" name="last"
                            value="<?php echo $query2['lname'] ?>"></td></tr>

                            <tr><td>Contact Number:</td><td> <input type="tel" name="contact" maxlength="11" 
                            value="<?php echo $query2['contactnum'] ?>"></td></tr>

                            <tr> <td>Province: </td><td> <input type="text" name="provincee" 
                            value="<?php echo $query2['province'] ?>"></td></tr>

                            <tr> <td>City: </td><td> <input type="text" name="cityy" 
                            value="<?php echo $query2['city'] ?>"></td></tr>

                            <tr> <td>Username:</td><td> <input type="email" name="usernamee" 
                            value="<?php echo $query2['username'] ?>"></td></tr>

                            <tr><td>Password:</td><td> <input type="password" name="pass" 
                            value="<?php echo $query2['password'] ?>"></td></tr>

                            <tr><td><br> <input type="Submit" name="submitt"></td></tr>
                        </table>
                        </form>

这是我的PHP代码。

if(isset($_GET['id'])){

                            $userid=$_GET['id'];
}

                            if(isset($_POST['submitt'])){
                            $firstname= mysqli_real_escape_string($mysqli,$_POST['first']);
                            $middleinitial= mysqli_real_escape_string($mysqli,$_POST['middle']);
                            $lastname= mysqli_real_escape_string($mysqli,$_POST['last']);
                            $contactnumber= mysqli_real_escape_string($mysqli,$_POST['contact']);
                            $province= mysqli_real_escape_string($mysqli,$_POST['provincee']);
                            $city = mysqli_real_escape_string($mysqli,$_POST['cityy']);
                            $username= mysqli_real_escape_string($mysqli,$_POST['usernamee']);
                            $password= mysqli_real_escape_string($mysqli,$_POST['pass']); 

                         $query=mysqli_query($mysqli, "UPDATE usertbl SET fname='$firstname', 
                         middleinitial='$middleinitial', lname='$lastname',contactnum = '$contactnumber',
                         province = '$province',city= '$city' ,username = '$username' ,
                         password = '$password' where userid='$userid'");


                          if($query){

                              header('location:employees.php');

                          }

                        $query1=mysqli_query($mysqli,"SELECT * FROM usertbl WHERE userid='$userid'");
                        $query2 = mysqli_fetch_array($query1);
                            }

继续显示的错误是表单上信息的输入类型值的未定义变量。我试图从数据库中获取数据，但我似乎无法得到它。这似乎是什么问题？

这些是错误：
注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 372

注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 375

注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 378

注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 381

注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 384

注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 387

注意：未定义的变量： C：\ xampp \ htdocs \ thesis2 \ admin \ pages \ editemployee.php 中的查询2 390

Answer 1

我认为问题在于您的评估顺序。您尝试仅在提交表单时进行查询（以及重定向之后 - 这没有任何意义，因为在您之前进行重定向之前永远不会到达查询）。

我就是这样做的

<?php
$userid = isset($_GET['id']) ? $_GET['id'] : null;

if (isset($_POST['submitt'])) {
    $firstname = trim($_POST['first']);
    $middleinitial = trim($_POST['middle']);
    $lastname = trim($_POST['last']);
    $contactnumber = trim($_POST['contact']);
    $province = trim($_POST['provincee']);
    $city = trim($_POST['cityy']);
    $username= trim($_POST['usernamee']);
    $password= $_POST['pass']; // leave it as is

    $stmt = mysqli_prepare($mysqli, "
        UPDATE usertbl 
        SET fname = ?, middleinitial = ?, lname = ?, contactnum = ?, province = ?, city = ?, username = ?, password = ?
        WHERE userid = ?
    ");

    mysqli_stmt_bind_param($stmt, 'ssssssssd', 
        $firstname, $middleinitial, $lastname, $contactnumber, $province, $city, $username, $password, $userid);

    $updated = mysqli_stmt_execute($stmt);

    mysqli_stmt_close($stmt);

    // redirect if user is updated
    if ($updated) {
        header('Location: employees.php');
        exit;
    }
}

// note: the next lines of code appear OUTSIDE the previous if statement (1)

// cheating here as you should definitely be using prepared statements here also!
if ($result = mysqli_query($mysqli, "SELECT * FROM usertbl WHERE userid = '$userid'")) {
    $query2 = mysqli_fetch_array($result);
}
?>

<!-- HTML form -->

<强>侧面备注：

没有理由真正使用mysqli_real_escape_string。您应该使用prepared statements。
Always exit after doing redirecting.
我通常不使用MySQLi，因为它们的语法非常丑陋。使用PDO更简单，更清洁。

将数据从数据库获取到表单的输入类型，但是存在未定义变量

1 个答案: