2

请告诉我我的代码有什么问题！！！！

new-payment.php

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Process New Payment</title>
</head>

<body>
<h1>Please Input Payment Details</h1>
<fieldset>
    <legend>New Payment</legend>
    <form action="process-payment.php" method="post" />
<table>
<tr>
<td>Date:</td><td><input type="date" name="date" /><br /></td>
</tr>
<tr>
<td>Today's Charge:</td><td><input type="text" name="charge" /><br /></td>
</tr>
<tr>
<td>Today's Payment:</td><td><input type="text" name="payment" /><br /></td>
</tr>
<tr>
<td>Client Number:</td><td><input type="text" name="client_no" /><br /></td>
</tr>
<tr>
<td>Client Name:</td><td><input type="text" name="client_name" /><br /></td>
</tr>
<tr>
<td>Check Number:</td><td><input type="text" name="check_no" /><br /></td>
</tr>
<tr>
<td>Check Amount:</td><td><input type="text" name="check" /><br /></td>
</tr>
<tr>
<td>Cash Amount:</td><td><input type="text" name="cash" /><br /></td>
</tr>
<tr>
<td>Notes:</td><td><input type="text" name="notes" /><br /></td>
</tr>
<tr>
<td>Staff Initials:</td><td><input type="text" name="staff_initials" /><br /></td>
</tr>
</table>
<input type="submit" value="Process Payment">
    </form>
</fieldset>
<br />
</body>
</html>

流程-payment.php

<?php

define('DB_NAME', 'DBNAME');
define('DB_USER', 'USERNAME');
define('DB_PASS', '');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASS);

if (!$link) {
    dir('There was a problem when trying to connect to the host. Please contact Tech Support. Error: ' . mysql_error());    
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$link) {
    dir('There was a problem when trying to connect to the database. Please contact Tech Support. Error: ' . mysql_error());    
}

$date = $_POST['date'];
$charge = $_POST['charge'];
$payment = $_POST['payment'];
$client_no = $_POST['client_no'];
$client_name = $_POST['client_name'];
$check_no = $_POST['check_no'];
$check = $_POST['check'];
$cash = $_POST['cash'];
$notes = $_POST['notes'];
$staff_initials = $_POST['staff_initials'];

$sql = "INSERT INTO payments (date, charge, payment, client_no, client_name, check_no, check, cash, notes, staff_initials) VALUES ('$date', '$charge', '$payment', '$client_no', '$client_name', '$check_no', '$check', '$cash', '$notes', '$staff_initials')";

if (!mysql_query($sql)) {
    die('Error: ' . mysql_error()); 
}

?>

我不知道出了什么问题，但是当我按下处理付款时出现错误：

错误：您的 SQL 语法有错误；检查与您的 >MySQL 服务器版本相对应的手册，以获取在“支票、现金、笔记、员工姓名首字母”附近使用的正确语法 >VALUES ('2012-09-24', '$0.00', '$20.00', '46' at 1号线

4

3 回答 3

2

CHECK是MySQL 保留关键字。您必须将其括在反引号中才能将其用作列或表标识符。

$sql = "INSERT INTO payments (date, charge, payment, client_no, client_name, check_no, `check`, cash, notes, staff_initials) VALUES ('$date', '$charge', '$payment', '$client_no', '$client_name', '$check_no', '$check', '$cash', '$notes', '$staff_initials')";

请注意，您的脚本容易受到 SQL 注入的攻击。至少，您必须调用mysql_real_escape_string()这些输入变量中的每一个。

// As in:
$charge = mysql_real_escape_string($_POST['charge']);
于 2012-09-26T01:33:09.627 回答
1

改变

$sql = "INSERT INTO payments (date, charge, payment, client_no, client_name, check_no, check, cash, notes, staff_initials) VALUES ('$date', '$charge', '$payment', '$client_no', '$client_name', '$check_no', '$check', '$cash', '$notes', '$staff_initials')";

至

$sql = "INSERT INTO payments (date, charge, payment, client_no, client_name, check_no, check, cash, notes, staff_initials) VALUES ('".$date."', '".$charge."', '".$payment."', '".$client_no."', '".$client_name."', '".$check_no."', '".$check."', '".$cash."', '".$notes."', '".$staff_initials."')";

查找 MySQL PDO 而不是使用您正在使用的折旧连接代码可能会有所帮助。

于 2012-09-26T01:34:43.877 回答
0

尝试回显所有元素并查看是否某些变量可能为空白或为空。

我有同样的错误。我通过这种方式解决了

于 2016-08-24T18:04:34.053 回答