PHP正在MYSQL数据库中插入两个重复的行
我在PHP和MYSQL方面遇到了问题。我有一个HTML表单,它在提交时运行下面的PHP脚本。我认为这与以下PHP有关,而与数据库无关:PHP正在MYSQL数据库中插入两个重复的行,php,html,mysql,Php,Html,Mysql,我在PHP和MYSQL方面遇到了问题。我有一个HTML表单,它在提交时运行下面的PHP脚本。我认为这与以下PHP有关,而与数据库无关: <?php $first_name = $_POST['firstname']; $last_name = $_POST['lastname']; $display_name = $_POST['displayname']; $email = $_POST['email']; $password = $_POST['password']; $add_li
<?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($conn);
?>
您有mysqli\u查询($conn,$sql)代码>在代码中输入两次。一次在if()中,一次在外部。每一个都将插入到您的数据库中
这里需要注意的是,if中的mysqli\u查询
是经过计算的,也就是说,它是运行的,if语句根据函数调用的结果执行。因此,您不需要再次调用它。Tushar指出了mysqli的两个查询,他是对的,此外,现在的代码将给您带来安全问题,因为它允许sql注入
请按如下方式修改您的代码:
$first_name = mysqli_escape_string($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, $_POST['postcode']);
对,<代码>mysqli_查询($conn,$sql)
可以被删除,只需将else{}
添加到if
-语句中,并在其中回显“Guest Added”消息,作为成功的标志。感谢您的帮助-在评估时没有意识到它运行了。很好的建议-PHP新手和关于安全性的良好实践。
$first_name = mysqli_escape_string($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']);