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Home/ Questions/Q 8068415
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Editorial Team
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Editorial Team
Asked: June 5, 20262026-06-05T12:40:42+00:00 2026-06-05T12:40:42+00:00

I have the following code to perform a GET request on the following URL:

  • 0

I have the following code to perform a GET request on the following URL:

http://rt.hnnnglmbrg.de/server.php/someReferenceNumber

However, here is my output from Logcat:

java.io.FileNotFoundException: http://rt.hnnnglmbrg.de/server.php/6

Why does it return 404 when the URL is clearly valid?

Here is my connect code:

/**
 * Performs an HTTP GET request that returns base64 data from the server
 * 
 * @param ref
 *            The Accident's reference
 * @return The base64 data from the server.
 */
public static String performGet(String ref) {
    String returnRef = null;
    try {
        URL url = new URL(SERVER_URL + "/" + ref);
        HttpURLConnection con = (HttpURLConnection) url.openConnection();
        con.setRequestMethod("GET");

        BufferedReader reader = new BufferedReader(new InputStreamReader(con.getInputStream()));

        StringBuilder builder = new StringBuilder();
        String line;
        while ((line = reader.readLine()) != null) {
            builder.append(line);
        }

        returnRef = builder.toString();

    } catch (IOException e) {
        e.printStackTrace();
    }
    return returnRef;
}
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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-06-05T12:40:43+00:00Added an answer on June 5, 2026 at 12:40 pm

    When you request the URL, it actually return HTTP code 404 which mean not found. If you have control to the PHP script, set the header to 200 to indicate file is found.

    enter image description here

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