Question

我已经更新了Mamp服务器，现在我的php版本是php7.2.1。更新后，出现以下错误：

不建议使用：parse_str（）：在没有结果的情况下调用parse_str（）不推荐使用 /Applications/MAMP/htdocs/projects/functions/checklogin.php，第3行

来自此代码：

<?php 
if(isset($_COOKIE[$cookie_name])){
parse_str($_COOKIE[$cookie_name]); 
$result = mysqli_query($db, "SELECT * FROM `users` WHERE uid='$u' AND username='$u_name'");
if(mysqli_num_rows($result) == '0'){
    header("Location: ".$base_url."logout.php");    
}
}else{
  header("Location: ".$base_url."logout.php");      
}

?>

能帮我解决这个问题吗？

Answer 1

什么是php文档说：

http://php.net/manual/en/migration72.deprecated.php

Without the second argument to parse_str(), the query string parameters would populate the local symbol table. Given the security implications of this, using parse_str() without a second argument has now been deprecated. The function should always be used with two arguments, as the second argument causes the query string to be parsed into an array.

您可以写parse_str($_COOKIE[$cookie_name], $myArray);代替parse_str($_COOKIE[$cookie_name]);来解决此问题

并在此行之后使用extract($myArray);；

<?php 
if(isset($_COOKIE[$cookie_name])){
parse_str($_COOKIE[$cookie_name], $myArray);
extract($myArray); 
$result = mysqli_query($db, "SELECT * FROM `users` WHERE uid='$u' AND username='$u_name'");
if(mysqli_num_rows($result) == '0'){
    header("Location: ".$base_url."logout.php");    
}
}else{
  header("Location: ".$base_url."logout.php");      
}

?>

此外，您应该使用prepare语句来保护您的网站。

parse_str（）需要在php7上安装

1 个答案: