由于以下代码中的IllegalStateException
,我感到震惊。有人可以帮帮我吗?代码:
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消息:
01-18 18:39:53.383:ERROR / XXX(7113): java.lang.IllegalStateException: 内容已被消费
答案 0 :(得分:106)
您只能从Content
Entity
在该行:
final JSONObject jObject = new JSONObject(EntityUtils.toString(resEntity));
您已经消费了内容,并且您在此处再次使用相同内容:
Log.e("XXX",EntityUtils.toString(resEntity));
导致IllegalStateException: Content has been consumed
所以解决方案就在这里:
String _response=EntityUtils.toString(resEntity); // content will be consume only once
final JSONObject jObject=new JSONObject(_response);
Log.e("XXX",_response);
答案 1 :(得分:5)
如果您在调试器的表达式中编写消费语句,也会发生这种情况!
(例如,如果你正在“观察”EntityUtils.toString(resEntity)
)
答案 2 :(得分:3)
首先,这必须是每个新的Android程序员制作的错误,并且每天都在这里询问。
user.getText().toString()!= ""&& pw.getText().toString()!= ""
这不符合您的要求。你需要
!user.getText().toString().equals("")&& !pw.getText().toString().equals("")
此外,您需要打印堆栈跟踪。在您的例外中,您需要
e.printStackTrace()
而不是记录
e.toString()
答案 3 :(得分:1)
我刚刚处理了一个对实体进行空检查的情况,导致它被标记为“已消耗”。希望我的头痛会帮助那里的其他人。
Vikas Patidar的回答帮助我找出了谜语的关键,非常感谢