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php - SQL 语法错误,连接问题简单形式 PHP

转载 作者:行者123 更新时间:2023-11-29 02:04:15 26 4
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我有一个简单的表单,可以在我的服务器上的一个表中插入数据。我已经设置了一个特殊的用户来处理这个,只有插入权限。我收到连接和语法错误。

这是我的表格:

<form id="form1" name="form1" method="post" action="mailform.php" onsubmit="return validateForm();">

<input type="text" id="First" maxlength="100" autocorrect placeholder="First name" />
<input type="text" id="Last" maxlength="100" autocorrect placeholder="Last name" />
<input type="text" id="Email" maxlength="100" autocorrect placeholder="Email address" />
<select name="SalesPerson">
<option value="SP1">SP1</option>
<option value="SP2">SP2</option>
<option value="SP3">SP3</option>
</select>
<select name="Show">
<option value="Show1">Show1</option>
<option value="Show2">Show2</option>
</select>

<button type="submit" id="submit" class="oneup">Submit</button>

</form>

在 mailform.php 我们有:

<?php

$name = "xxx_xxx";
$name = mysql_real_escape_string($name);
$SQL = "SELECT * FROM users WHERE username = '$name'";

$con = mysql_connect("localhost","xxx_xxx","xxxxxxxxx");
if (!$con)
{
die('Could not connect: ' . mysql_error());
}

mysql_select_db("xxx_x", $con);

$sql="INSERT INTO email_signup (First, Last, Email, SalesPerson, Show)
VALUES
('$_POST[First]','$_POST[Last]','$_POST[Email]','$_POST[SalesPerson]','$_POST[Show]')";

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

mysql_close($con)
?>

这是错误 -

Warning: mysql_real_escape_string() [<a href='function.mysql-real-escape-string'>function.mysql-real-escape-string</a>]: Access denied for user 'xxx'@'localhost' (using password: NO) in <b>.../mailform.php</b> on line 28

Warning: mysql_real_escape_string() [<a href='function.mysql-real-escape-string'>function.mysql-real-escape-string</a>]: A link to the server could not be established in <b>.../mailform.php</b> on line 28
Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'Show)
VALUES
('','','','SP1','Show1')' at line 1

知道为什么我会遇到连接问题吗?我在另一个地方设置了几乎相同的表格,效果很好。

最佳答案

首先建立连接,然后运行 ​​mysql_real_escape_string(),然后是查询。 mysql_real_escape_string() 实际上连接到数据库以让它转义您的字符串。如果您没有连接器,它将无法工作

关于php - SQL 语法错误,连接问题简单形式 PHP,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/9367376/

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