Question

我无法将我的名字更新为数据库，但却发出错误：

Notice: Undefined index: first_name in C:\Users\Raj\PhpstormProjects\new linkvessel\change_name.php on line 178

Notice: Undefined index: last_name in C:\Users\Raj\PhpstormProjects\new linkvessel\change_name.php on line 178

我已将first_name和last_name两列创建到数据库中，请问我做错了什么？请帮忙

这是我的代码：

<body>

<div id="left_box"><br>
    &nbsp;&nbsp;<img src="Images/general_setting.png" height="18" width="18"><a href="general_settings.php" style="text-decoration: none; color: #000000; font-family: Arial";> General</a><br><br>
    &nbsp;&nbsp;<img src="Images/photo_setting.png" height="18" width="18"><a href="photo_settings.php" style="text-decoration: none; color: #000000; font-family: Arial";> Photos</a><br><br>
</div>

<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>
        <label><strong>First name: </strong></label>
        <input type="text" name="first_name">
        <label><strong>Last name: </strong></label>
        <input type="text" name="last_name">
        </form>

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

        $result = mysqli_query($con,"SELECT * FROM admin");

        while($row = mysqli_fetch_array($result))
        {

        }

        mysqli_query($con,"UPDATE admin SET first_name='$_POST[first_name]' AND  last_name='$_POST[last_name]' WHERE id='$row[id]' ");

        mysqli_close($con);
        ?>
    </div>
</div>
</body>

EDITED

完整源代码：http://pastebin.com/TparWavH

Answer 1

<form method="POST" action="where you want to put the action">
        <label><strong>First name: </strong></label>
        <input type="text" name="first_name">
        <label><strong>Last name: </strong></label>
        <input type="text" name="last_name">
        </form>

===============================================

使用选择查询获取要更新的用户的ID，然后运行此

1）RUN SELECT QUERY

2）检查用户存在

3）如果用户存在FETCH他的ID并运行更新查询。

$update =mysqli_query($con,"UPDATE admin SET first_name='$_POST[first_name]' AND  last_name='$_POST[last_name]' WHERE id='ID' ");

* ID =您要更新的ID

Answer 2

您的代码存在许多问题，但纯粹是让它运行起来＆＃39;你需要做两件事。

首先，将表单更改为POST类型。

接下来，您可能想要更改以下内容：

 while($row = mysqli_fetch_array($result))
        {

        }

        mysqli_query($con,"UPDATE admin SET first_name='$_POST[first_name]' AND  last_name='$_POST[last_name]' WHERE id='$row[id]' ");

要：

 while($row = mysqli_fetch_array($result))
        {
            mysqli_query($con,"UPDATE admin SET first_name='$_POST[first_name]' AND  last_name='$_POST[last_name]' WHERE id='$row[id]' ");
        }

您当前的代码不会遍历语句，实际上只执行一次。

我不太确定你通过查询“管理员”来实现的目标。尽管如此 - 目前，它会遍历所有结果并使用您的新名字/姓氏更新最后一行？

无法将名称编辑到数据库中

2 个答案: