Question

我的php页面一直出现错误。当我提交表单时，它表示我没有授权查看php页面，当我把它放入sql查询测试时，它说它是错误＃1064。我一直收到此错误消息，我不知道为什么。在查询测试中，我只是在之间放置信息。有人可以帮忙解决这个问题吗？

<html>
<head>
<meta charset="utf-8">
<title>Enquiry</title>
</head>

<body> 
<?php

$name=$_POST['name'];
$email=$_POST['email'];
$phone = $_POST['phone'];
$radio = $_POST['radio'];
$enquiry = $_POST['enquiry'];

$user="root";
$password="";
$database="test";

mysql_connect('localhost',$user,$password) or die("Unable to connect to server"); 
mysql_select_db($database) or die("Unable to select database"); 
$query = "INSERT INTO Enquiry VALUES ('','$name','$email','$phone','$radio','$enquiry')";
mysql_query($query); 
mysql_close(); 

?>
</body>
</html>

Answer 1

假设数据库中的第一列是自动递增的主键，则需要将SQL更改为：

$sql = "
    INSERT INTO Enquiry (
        name,
        email,
        phone,
        radio,
        enquiry
    ) VALUES (
        '$name',
        '$email',
        '$phone',
        '$radio',
        '$enquiry'
    )
    ";

Answer 2

<?php
$con= mysql_connect("localhost","root","") or die('Unable to connect to server:'mysql_error());
mysql_select_db("test",$con);
if(isset($_POST['submit']))  
{

$name=test_input($_POST['name']);
$email=test_input($_POST['email']);
$phone = test_input($_POST['phone']);
$radio = test_input($_POST['radio']);
$enquiry = test_input($_POST['enquiry']);
$sql = "
"INSERT INTO `Enquiry`"." (
    name,
    email,
    phone,
    radio,
    enquiry
)". "VALUES". "(
    "$name",
    "$email",
    "$phone",
    "$radio",
    "$enquiry"
)
";
 $res=mysql_query($sql);

if(!$res){
die('Could not enter data: ' . mysql_error());
}
echo 'your successfull msg goes here';

function test_input($data)
{
$data = trim($data);
$data = stripslashes($data);
$data = mysql_real_escape_string($data);
return $data;
}


?>

假设primaryid是自动递增的。试试这段代码

PHP插入错误1064

2 个答案: