Question

我已经创建了一个表单来更新名称到数据库，我已经在数据库中创建了两个表first_name和last_name，看看曾经这里是我的代码：

$sql="
CREATE TABLE admin
(
id INT PRIMARY KEY AUTO_INCREMENT,
first_name VARCHAR(100),
last_name VARCHAR(100),
)";

现在我创建了更新表单：

<div class="box">
    <h1 style="font-family: consolas">Change your name</h1><hr>
    <div id="change_name">
        <label><strong>Your current name: </strong></label>
        <?php
        include('change_setting_db.php');

        while($row = mysqli_fetch_array($result))
        {
            echo $row['first_name']." ".$row['last_name'];
        }
        ?><br><br>

        <form method="post" action="do_update_name.php">
        <label><strong>First name: </strong></label>
        <input type="text" name="first_name">
        <label><strong>Last name: </strong></label>
        <input type="text" name="last_name">
            <input type="submit" value="Submit">
        </form>

    </div>

</div>

这是我的do_update_name.php文件：

<?php
$firstname=$_POST['first_name'];
$lastname=$_POST['last_name'];
$id=$_GET['id'];

$con=mysqli_connect("localhost","root","Bhawanku", "members");
// Check connection
if (mysqli_connect_errno())
{
    echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

$update =mysqli_query($con,"UPDATE admin SET first_name='$firstname' AND  last_name='$lastname' WHERE id='$id' ");
if($update){
    echo "Successfully created!!";
}
?>

现在问题是，当我输入新名称并提交时，它显示错误：

Notice: Undefined index: id in C:\Users\Raj\PhpstormProjects\...\new do_update_name.php on line 4
Successfully created!!

Answer 1

<form method="post" action="do_update_name.php">
        <label><strong>First name: </strong></label>
        <input type="text" name="first_name">
        <label><strong>Last name: </strong></label>
        <input type="text" name="last_name">
            <input type="submit" value="Submit">
        </form>

您必须在此处添加ID，以便对方可以选择$ _GET参数[＆＃39; id＆＃39;] 例如：

<form method="post" action="do_update_name.php?id=1">
        <label><strong>First name: </strong></label>
        <input type="text" name="first_name">
        <label><strong>Last name: </strong></label>
        <input type="text" name="last_name">
            <input type="submit" value="Submit">
        </form>

Answer 2

输入类型为隐藏ID的HTML表单

<div class="box">
    <h1 style="font-family: consolas">Change your name</h1><hr>
    <div id="change_name">
        <label><strong>Your current name: </strong></label>
        <?php
        include('change_setting_db.php');

        while($row = mysqli_fetch_array($result))
        {
            $id = $row['id'];
            $first_name = $row['first_name'];
            $last_name = $row['last_name'];
            echo $id.") ".$first_name." ".$last_name;
        }
        ?>
        <br>
        <br>
        <form method="post" action="do_update_name.php">
        <input type="hidden" name="id" value="<?php echo $id;?>">
        <label><strong>First name: </strong></label>
        <input type="text" name="first_name" value="<?php echo $first_name;?>">
        <label><strong>Last name: </strong></label>
        <input type="text" name="last_name" value="<?php echo $last_name;?>">
        <input type="submit" value="Submit">
        </form>

    </div>
</div>

带有$ _POST ['id']的PHP代码

<?php
$firstname=$_POST['first_name'];
$lastname=$_POST['last_name'];
$id=$_POST['id'];

…
?>

如何从数据库中调用`id`？

2 个答案: