Question

我遇到了PHP和MYSQL的问题。我有一个HTML表单，提交时运行以下PHP脚本。问题是以下PHP代码将数据插入数据库两次。我认为这与以下PHP而不是数据库有关：

<?php

$first_name = $_POST['firstname'];
$last_name = $_POST['lastname'];
$display_name = $_POST['displayname'];
$email = $_POST['email'];
$password = $_POST['password'];
$add_line1 = $_POST['addline1'];
$add_line2 = $_POST['addline2'];
$city = $_POST['city'];
$county = $_POST['county'];
$postcode = $_POST['postcode'];

$sql = "INSERT INTO members (memberID, 
memberPassword, 
memberFirstName, 
memberLastName,
memberAddressLine1, 
memberAddressLine2, 
memberCity,
memberCounty, 
memberPostcode, 
memberDisplayName) 
VALUES ('$email', 
'$password', '$first_name', '$last_name',
 '$add_line1', '$add_line2','$city',
 '$county', '$postcode', '$display_name')";

if (!mysqli_query($conn,$sql))
{
     die('Error: ' . mysqli_error($conn));
}
mysqli_query($conn,$sql);
echo 'Guest Added';
mysqli_close($conn);

?>

Answer 1

if (!mysqli_query($conn,$sql))
{
     die('Error: ' . mysqli_error($conn));
}
mysqli_query($conn,$sql);

您的代码中有mysqli_query($conn,$sql);两次。一旦进入if（），一旦进入外部。其中每个都将插入到您的数据库中。

这里需要注意的是if中的mysqli_query被评估 - 也就是说，它运行并且if语句在函数调用的结果上执行。因此，您无需再次调用它。

Answer 2

Tushar指出了两个mysqli查询，他是对的，除此之外，现在的代码会导致安全问题，因为它允许sql注入...

请修改您的代码，如下所示：

 $first_name   = mysqli_escape_string($conn, $_POST['firstname']);
 $last_name    = mysqli_escape_string($conn, $_POST['lastname']);
 $display_name = mysqli_escape_string($conn, $_POST['displayname']);
 $email        = mysqli_escape_string($conn, $_POST['email']);
 $password     = mysqli_escape_string($conn, $_POST['password']);
 $add_line1    = mysqli_escape_string($conn, $_POST['addline1']);
 $add_line2    = mysqli_escape_string($conn, $_POST['addline2']);
 $city         = mysqli_escape_string($conn, $_POST['city']);
 $county       = mysqli_escape_string($conn, $_POST['county']);
 $postcode     = mysqli_escape_string($conn, $_POST['postcode']);

PHP在MYSQL数据库中插入两个重复的行

2 个答案: