File_put_contents未创建或未添加到文本文件

Question

我在使用某些PHP代码时遇到了麻烦，作为一个项目，我认为我会构建一个简单的SQL Injection脚本。我找到了这个GitHub存储库作为参考，并计划进行扩展： https://github.com/fakhrizulkifli/Website-Vulnerability-Scanner-v1.0/blob/master/scanner.php

我一直在尝试运行此功能

function sql($site = '', $full = '0') {
    $sql_error = array(
        'You have an error in your SQL',
        'Division by zero in',
        'supplied argument is not a valid MySQL result resource in',
        'Call to a member function',
        'Microsoft JET Database','ODBC Microsoft Access Driver',
        'Microsoft OLE DB Provider for SQL Server',
        'Unclosed quotation mark',
        'Microsoft OLE DB Provider for Oracle',
        '[Macromedia][SQLServer JDBC Driver][SQLServer]Incorrect',
        'Incorrect syntax near'
    );
        if($full == '0') {
        fwrite(STDOUT, "\n-SITE: ");
        $site = trim(fgets(STDIN));
        } else {
            $site = $site;
        }

        $request = parse_url($site);
        print "[-] URL : $request[host]\n";
        print "[-] Path: $request[path]\n";
        print "[-] Try connect to host\n";
        $url = "".$request['scheme']."://".$request['host'].$request['path']."";
        if(con_host($url))
        {
            print "[-] Connect to host successful\n";
            print Get_Info($url);
            print "[-] Finding link on the website\n";
            print "[+] Found link : ".count(find_link($url))."\n";
            print "[-] Finding vulnerable...\n";
            if(is_array(find_link($url)))
            foreach(find_link($url) as $link) {
                $file = explode("/", $request['path']);
                $request['path'] = preg_replace("/".$file[count($file)-1]."/", "", $request['path']);
                if(!preg_match("/$request[host]/", $link)) { $link = "http://$request[host]/$request[path]$link"; }
                $link = preg_replace("/=(.+)/", "=1'", $link);
                foreach($sql_error as $error) {
                    if(preg_match("/$error/", con_host($link))) {
                        print "[+] SQL Injection vulnerable : $link\n";
                        $save[] = $link;
                    }
                }
            }
            print "[+] Done\n";
            if(is_array($save)) {
               foreach($save as $link) {
               $save = @file_put_contents('vulnerable.log', "".$link."\r\n",FILE_APPEND);
               }}
               print "[-] See 'vulnerable.log' for vulnerable list\n";
    }
  }

当我运行此功能时，似乎没有创建文件'vulnerable.log'，我认为这可能是由于权限错误所致，但我将文件权限设置为chmod 755 scan.php。我还尝试过手动创建文件'vulnerable.log'，以查看是否将结果添加进来，但这也行不通。

在这种情况下的任何帮助或指导，将不胜感激。