将数据从表单存储到数据库中

问题描述

当我运行它时它不会显示任何错误，但是它不会在表客户端的 bookstore_hw3 数据库中存储任何内容。

It does not show any error when I run it but it does not store anything on the bookstore_hw3 database on the table clients.

表客户端上还有一列称为 client_id ，并且会自动递增。这是问题吗？

There is one more column on the table clients called client_id and is auto incremented. Is this the problem?

有人可以查看此代码并提出问题所在吗？

Can someone review this code and suggest what the problem is?

这是HTML表单：

<form  action="clientData.php" method="post">
First Name: <input type='text' id='client_fname' /><br />
Last Name: <input type='text' id='client_lname' /><br />
City: <select id='client_city'>
    <option>Please Choose</option>
    <option>Prishtine</option>
    <option>Mitrovice</option>
    <option>Peje</option>
    <option>Gjakove</option>
        <option>Ferizaj</option>
        <option>Prizren</option>
</select><br />
Gender: <select id='client_sex'>
    <option>Please Choose</option>
    <option>F</option>
    <option>M</option>
</select><br />
Username(3-10 characters): <input type='text' id='client_username' /><br />
Password(3-10 characters): <input type='password' id='client_pass' /><br />
<input type='submit' value='Submit' />
<input type="reset" value="Clear" />
</form>

这是PHP代码：

<?php
include('db_login.php');
// Connect
$connection = mysql_connect($db_host, $db_username, $db_password);
if (!$connection){
  die("Could not connect to the database: <br />". mysql_error( ));
}

// Select the database
$db_select = mysql_select_db($db_database);
if (!$db_select){
  die ("Could not select the database: <br />". mysql_error( ));
}
$fname = isset($_POST['client_fname']);
$lname = isset($_POST['client_lname']);
$city = isset ($_POST['client_city']);
$sex = isset($_POST['client_sex']);
$username = isset ($_POST['client_username']);
$pass = isset($_POST['client_pass']);

$sql = "INSERT INTO clients (client_fname, client_lname, client_city, client_sex, client_username, client_pass) VALUES ('$fname','$lname','$city','$sex','$username','$pass')";

mysql_close();

echo "Data stored on database.";
?>

这是登录代码：

<?php
$db_host='localhost';
$db_database='bookstore_hw3';
$db_username='root';
$db_password='';
?>

推荐答案

第一个问题是HTML代码。 PHP可以通过属性 name 而不是 id 来识别HTML表单中的数据，因此您应该以这种方式修改HTML标记：

The first problem is in the HTML code. PHP recognizes the data from the HTML form by the attribute name not id, so you should modify your HTML tags this way:

<input type='text' id='client_fname' name='client_fname' />

第二个问题是您的PHP代码。
替换此部分：

The second problem is with your PHP code.Replace this part:

$fname = isset($_POST['client_fname']);
$lname = isset($_POST['client_lname']);
$city = isset ($_POST['client_city']);
$sex = isset($_POST['client_sex']);
$username = isset ($_POST['client_username']);
$pass = isset ($_POST['client_pass']);

其中：

$fname = isset($_POST['client_fname']) ? $_POST['client_fname'] : null;
$lname = isset($_POST['client_lname']) ? $_POST['client_lname'] : null;
$city = isset ($_POST['client_city']) ? $_POST['client_city'] : null;
$sex = isset($_POST['client_sex']) ? $_POST['client_sex'] : null;
$username = isset ($_POST['client_username']) ? $_POST['client_username'] : null;
$pass = isset ($_POST['client_pass']) ? $_POST['client_pass'] : null;

isset（）仅检查变量存在并返回true / false。在我的代码中，语句 isset（foo）？ foo：bar 检查变量是否存在，如果存在，则返回其内容；如果不存在，则返回 null 。

isset() only checks if the variable exists and returns true/false. In my code, the statement isset(foo) ? foo : bar checks if the variable exists and if yes, its content is returned, if no, then null is returned.

第三个问题是您没有在数据库上执行SQL查询。因此，还要在此添加：

The third problem is that you are not executing your SQL query on the database. So add there also this:

mysql_query($sql);

此外，您的PHP代码容易受到SQL注入的影响，应予以修复（您可以在这篇文章：）

Also your PHP code is vulnerable to SQL injection and should be fixed (you can read more about it in this post: How can I prevent SQL injection in PHP?)

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