Laravel 将URL参数作为命令行脚本参数传递安全注意事项

我正在编写一个web应用程序，需要将一些url参数作为命令行参数传递给另一个脚本。我应该如何避免安全问题？我正在使用Symphony的进程来执行bash命令，并使用Laravel来构建应用程序

下面是一些代码，请查看

buildScreenshotCommand

以了解我是如何构建命令字符串的，还请注意，$urlRequest是使用Laravel的

Request$Request

类填充的：

<?php

namespace App\Logic;

use App\Logic\TimeHelper;
use App\UrlRequest;
use Illuminate\Support\Facades\Storage;
use Symfony\Component\Process\Exception\ProcessFailedException;
use Symfony\Component\Process\Process;

class Screenshot {

    static function take(UrlRequest $urlRequest)
    {
        $name = self::generateName($urlRequest);
        $command = self::buildScreenshotCommand($name, $urlRequest);

        $startTime = TimeHelper::milliseconds();

        $process = new Process($command);
        $process->run();

        $endTime = TimeHelper::milliseconds();

        if (!$process->isSuccessful())
        {
            throw new ProcessFailedException($process);
        }

        $output = $process->getOutput();

        if (trim($output) === '')
        {
            $urlRequest->successful = 1;

            $file = self::uploadToS3($name);
            $urlRequest->image_url = $file['url'];
            $urlRequest->file_size = $file['size'];
            $urlRequest->file_name = $name;
            $urlRequest->time_it_took_to_take_screenshot_ms = $endTime - $startTime;

            if ($urlRequest->save())
            {
                return $urlRequest;
            }
        }
        else
        {
            $urlRequest->error = $output;
            $urlRequest->save();
        }

        return false;
    }

    static function uploadToS3($name)
    {
        $name = 'screenshots/' . $name;

        Storage::disk('s3')->put($name, Storage::disk('local')->get($name), ['visibility' => 'public']); // upload to S3

        $fileSize = Storage::disk('local')->size($name);
        Storage::disk('local')->delete($name);

        return [
            'url' => Storage::disk('s3')->url($name),
            'size' => $fileSize
        ];
    }

    static function generateName($urlRequest)
    {
        $name = time() . rand(10000, 99999);
        $extension = '.png';

        if (isset($urlRequest->pdf) AND $urlRequest->pdf == 1)
        {
            $extension = '.pdf';
        }

        while (UrlRequest::where('file_name', '=', $name . $extension)->first())
        {
            $name = time() . rand(10000, 99999);
        }

        return $name . $extension;
    }

    static function buildScreenshotCommand($name, $urlRequest)
    {
        $command = 'cd ' . base_path() . ' && node puppeteer-screenshots-init.js ';
        $command .= "--url={$urlRequest->url} ";

        $fullPath = storage_path('app') . '/screenshots/' . $name;

        $command .= "--path={$fullPath} ";

        if (isset($urlRequest->pdf))
        {
            $command .= "--pdf=true ";
        }

        if (isset($urlRequest->viewport_width))
        {
            $command .= "--viewportWidth={$urlRequest->viewport_width} ";
        }

        if (isset($urlRequest->mobile))
        {
            $command .= '--mobile=true ';
        }

        if (isset($urlRequest->media_type_print))
        {
            $command .= '--mediaTypePrint=true ';
        }

        if (isset($urlRequest->user_agent))
        {
            $command .= '--userAgent="' . $urlRequest->user_agent . '" ';
        }

        $command .= '2>&1 &';

        return $command;
    }

}

---

我认为这是一个非常糟糕的主意。如果
$urlRequest->url
是：

 // urlencoded
 http%3A%2F%2Fgoogle.com%3F%3Brm+-Rf+%2F%3B

 http://google.com?;rm -Rf /;

您正在执行什么类型的输入清理？
目前没有执行清理，只有验证。您知道专门针对bash参数的好的净化包吗？另外，Symfony的过程是转义参数以防止安全问题，如下所述：另外，如果我尝试访问您建议的url，我会得到一个验证错误。这很危险，这就是我在这里问的原因。