CURL 响应发送到自定义 URL PHP/Laravel

Question

    public  function amazon_curl_request($url)
{
    $curl = curl_init();
    curl_setopt_array($curl, array(
        CURLOPT_URL => $url,
        CURLOPT_RETURNTRANSFER => true,
        CURLOPT_ENCODING => "",
        CURLOPT_MAXREDIRS => 10,
        CURLOPT_TIMEOUT => 0,
        CURLOPT_FOLLOWLOCATION => true,
        CURLOPT_HTTP_VERSION => CURL_HTTP_VERSION_1_1,
        CURLOPT_HTTPHEADER => array(
        ),
    ));
    $response = curl_exec($curl);
    curl_close($curl);
    return $response;
}

  public function amazon(Request  $request)
{
    $url = $request->input('link');
    $parsed_link = parse_url($url);
    $url = 'https://www.amazon.com/s?k=' .$url;
    $url=str_replace(' ', '+', $url);
    $response =  $this->amazon_curl_request($url);        
    return view('FrontEnd/amazon/result_page', compact('response'));
}

这里是路线Route::get('/amazon/search', [AmazonController::class , 'amazon'])->name('search.amazon');

我需要更改网址，默认情况下它会将我重定向到这里http://localhost:8000/s?k=bag

但我需要这个 URL 来显示响应 http://localhost:8000/amazon-search/bag

这是一个 Amazon Scraping 网站。

Answer 1

使用 Http，因为您正在使用 laravel 8 参考链接 https://laravel.com/docs/8.x/http-client

use Illuminate\Support\Facades\Http;

public function amazon(Request  $request)
{
    $url = $request->input('link');
    $parsed_link = parse_url($url);
    $url = 'https://www.amazon.com/s?k=' .$url;
    $url=str_replace(' ', '+', $url);
    $response =  Http::get($url)->body();   // body() function will return html of url   

    echo $response  // as it is already html of amazon or do some datamanipulation here 
    // return view('FrontEnd/amazon/result_page', compact('response'));
}