How to send a referer request in a url for a webView

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渐次进展 2020-12-06 07:00

I need to display a web page in my Android app which is looking for a referer to bypass the security. I\'m new to Android so I know how to display the web page in a web view

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  • 2020-12-06 07:30

    You will need to use Intent Filters to capture and modify WebView requests.

    Assuming that you need to specify doamin.com/page.html as referrer

    1. Setup intent filters to capture all http requests in WebView
    2. If request is for "doamin.com/page.html", return pre-defined page that has a refresh tag to send user to "http://www.mywebsite.com"
    3. domain.com/page.html would be sent as a referrer to mywebsite.com

    In newer APIs you can specify headers in loadUrl itself.

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  • 2020-12-06 07:33

    For which API-level do you need that function?

    Since API Level 8 there is a second loadUrl function:

      public void loadUrl (String url, Map<String, String> extraHeaders)
    

    With the extraHeaders you should be able to send a referrer.


    EDIT:

    Here is a complete working example:

      String url = "http://www.targetserver.tld/";
    
      Map<String, String> extraHeaders = new HashMap<String, String>();
      extraHeaders.put("Referer", "http://www.referer.tld/login.html");
    
      WebView wv;
      wv = (WebView) findViewById(R.id.webview);
      wv.loadUrl(url, extraHeaders);
    
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