CODEWITHSUNDEEP
androidhttpresponse
tyczj
tyczj

Reputation: 73753

Log response of HttpResponse then process

How can I log a response from HttpResponse and then process the response after.

for example when I do this

HttpEntity entity = resp.getEntity();
String xml = EntityUtils.toString(entity);
InputStream is = entity.getContent();

i get the exception java.lang.IllegalStateException: Content has been consumed

because I wrote the entity to a string. I only want to use that string for debugging and then use the InputStream to process all the content in the response

Upvotes: 0

Views: 105

Answers (2)

Nikola Despotoski
Nikola Despotoski

Reputation: 50538

HttpEntity entity = resp.getEntity();
InputStream is = entity.getContent();
String asString = getString(is);
Log.i(TAG,""+asString);

The point is to avoid closing the InputStream with is.close() or flushing it with is.flush(), to have it later for processing. The exception is thrown because the content is converted as InputStream by calling EntityUtils.toString(entity);

public static String getString( InputStream is) throws IOException {
        int ch;
        StringBuilder sb = new StringBuilder();
        while((ch = is.read())!= -1)
            sb.append((char)ch);
        return sb.toString();
    }

Don't forget to close the stream after you are done with the processing.

Upvotes: 1

RogueBaneling
RogueBaneling

Reputation: 4471

Here is an example of a code block that can be used to convert an HttpResponse into a String:

String output = inputStreamToString(httpResponse.getEntity().getContent()).toString();

public static StringBuilder inputStreamToString(InputStream is){
    String line;
    StringBuilder sb = new StringBuilder();
    BufferedReader rd = new BufferedReader(new InputStreamReader(is));
    try {
        while ((line = rd.readLine()) != null) {
            sb.append(line);
        }
    } catch (IOException e) {
        e.printStackTrace();
    }
    return sb;
}

Upvotes: 0

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