Reputation: 2085
I got struck because of IllegalStateException
in the following code. Can anybody please help me? Code:
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.security.MessageDigest;
import java.security.NoSuchAlgorithmException;
import java.util.ArrayList;
import java.util.List;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
import org.apache.http.ParseException;
import org.apache.http.client.ClientProtocolException;
import org.apache.http.client.HttpClient;
import org.apache.http.client.entity.UrlEncodedFormEntity;
import org.apache.http.client.methods.HttpPost;
import org.apache.http.impl.client.DefaultHttpClient;
import org.apache.http.message.BasicNameValuePair;
import org.apache.http.protocol.HTTP;
import org.apache.http.util.EntityUtils;
import org.json.JSONObject;
import android.app.Activity;
import android.os.Bundle;
import android.telephony.gsm.GsmCellLocation;
import android.util.Log;
import android.view.View;
import android.view.View.OnClickListener;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;
public class Login extends Activity {
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
setContentView(R.layout.login);
Button bt = (Button) findViewById(R.id.logbt);
final EditText user = (EditText) findViewById(R.id.loguser);
final EditText pw = (EditText) findViewById(R.id.logpw);
bt.setOnClickListener(new OnClickListener() {
@Override
public void onClick(View v) {
if (user.getText().toString() != "" && pw.getText().toString() != "") {
Thread t = new Thread() {
public void run() {
try {
HttpClient client = new DefaultHttpClient();
String postURL = "http://surfkid.redio.de/login";
HttpPost post = new HttpPost(postURL);
ArrayList<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("username", user.getText().toString()));
params.add(new BasicNameValuePair("password", md5(pw.getText().toString())));
UrlEncodedFormEntity ent = new UrlEncodedFormEntity(params, HTTP.UTF_8);
post.setEntity(ent);
HttpResponse responsePOST = client.execute(post);
HttpEntity resEntity = responsePOST.getEntity();
final JSONObject jObject = new JSONObject(EntityUtils.toString(resEntity));
Log.e("XXX", EntityUtils.toString(resEntity));
} catch (Exception e) {
Log.e("XXX", e.toString());
}
}
};
t.start();
// Log.e("XXX",s);
}
}
});
}
private String md5(String in) {
MessageDigest digest;
try {
digest = MessageDigest.getInstance("MD5");
digest.reset();
digest.update(in.getBytes());
byte[] a = digest.digest();
int len = a.length;
StringBuilder sb = new StringBuilder(len << 1);
for (int i = 0; i < len; i++) {
sb.append(Character.forDigit((a[i] & 0xf0) >> 4, 16));
sb.append(Character.forDigit(a[i] & 0x0f, 16));
}
return sb.toString();
} catch (NoSuchAlgorithmException e) {
e.printStackTrace();
}
return null;
}
}
Logcat message:
01-18 18:39:53.383: ERROR/XXX(7113): java.lang.IllegalStateException: Content has been consumed
Upvotes: 33
Views: 35155
Reputation: 3759
it's also happens if you are writing the consuming statement in the Expressions of the debugger!!!
(e.g if you are doing "watch" to something like EntityUtils.toString(resEntity)
)
Upvotes: 7
Reputation: 6368
I just dealt with a case of a null check on the entity causing it to be flagged as "consumed". Hope my headache will help someone else out there.
Vikas Patidar's answer helped me figure out the key to the riddle, so many thanks
Upvotes: 1
Reputation: 43349
You can consume Content
only at once from an Entity
in the line :
final JSONObject jObject = new JSONObject(EntityUtils.toString(resEntity));
you have consumed content and again you are using the same at here:
Log.e("XXX",EntityUtils.toString(resEntity));
That why it is causing IllegalStateException: Content has been consumed
So the solution is here:
String _response=EntityUtils.toString(resEntity); // content will be consume only once
final JSONObject jObject=new JSONObject(_response);
Log.e("XXX",_response);
Upvotes: 106
Reputation: 48587
First, this has to be a mistake that every single new android programmer makes and it's asked here every single day. You have
user.getText().toString()!= ""&& pw.getText().toString()!= ""
This doesn't do what you want it to. You need
!user.getText().toString().equals("")&& !pw.getText().toString().equals("")
Also, you need to print the stacktrace. In your exception, you need
e.printStackTrace()
instead of logging
e.toString()
Upvotes: 3