Android Intent:“找不到处理Intent的活动”错误

时间:2015-08-24 09:26:51

标签: android android-intent

在模拟器上打开第一个活动没有问题,但问题是当按下按钮时

        import android.app.Activity;
        import android.content.Intent;
        import android.os.Bundle;
        import android.util.Log;
        import android.view.Menu;
        import android.view.MenuItem;
        import android.view.View;
        import android.view.View.OnClickListener;
        import android.widget.Button;
        import android.widget.Toast;

public class MainlayoutActivity extends Activity {

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.mainlayout);
        final Button man = (Button) findViewById(R.id.getDATA);
        man.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                Intent i = new Intent(MainlayoutActivity.this, second.class);
                startActivityForResult(i, 77);
            }
        });
    }

    public void onActivityResult(int requstCode, int resultCode, Intent data) {
        if ((requstCode == 77) && (resultCode == Activity.RESULT_OK)) {
            String gett = data.getExtras().getString("TEXT value");
            Toast.makeText(this, gett, Toast.LENGTH_LONG).show();
        }
    }
}

问题是在模拟器上启动第二个活动

public class second extends Activity {
    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.seconlayout);
        final Button btn = (Button) findViewById(R.id.GDATA);
        final EditText ed = (EditText) findViewById(R.id.value);

        btn.setOnClickListener(new OnClickListener() {

            @Override
            public void onClick(View v) {
                String word = ed.getText().toString();
                Intent i = new Intent();
                i.putExtra("TEXT value", word);
                setResult(Activity.RESULT_OK, i);
                finish();
            }
        });
      }
    }   

清单文件

<? xml version = "1.0"encoding="utf-8"?>
<manifest xmlns:android="http://schemas.android.com/apk/res/android"
        package="com.example.l9"
        android:versionCode="1"
        android:versionName="1.0">

<uses-sdk
        android:minSdkVersion="15"
        android:targetSdkVersion="15"/>

<application
android:allowBackup="true"
        android:icon="@drawable/ic_launcher"
        android:label="@string/app_name"
        android:theme="@style/AppTheme">
<activity
android:name="com.example.l9.MainlayoutActivity"
        android:label="@string/app_name">
<intent-filter>
<action android:name="android.intent.action.MAIN"/>

<category android:name="android.intent.category.LAUNCHER"/>
</intent-filter>
</activity>
<activity android:name="com.example.l9.second">
<intent-filter>
<action android:name="net.abdullaheid.action.ToDATA"/>
<category android:name="android.intent.categor.DEFAULT"/>
</intent-filter>

</activity>
</application>

</manifest>

1 个答案:

答案 0 :(得分:0)

import redis
r_server = redis.Redis("localhost")
r_server.rpush("requests", "Adam")
r_server.rpush("requests", "Bob")
r_server.rpush("requests", "Carol")

print r_server.lrange("requests", 0, -1)
print r_server.llen("requests")
print r_server.lindex("requests", 1)

如果您不想从此活动转移,则此参数无效。您需要使用参数创建Intent Intent intent = new Intent();

所以,你得到错误,因为你的Intent()构造函数是空的。