Question

我想创建连接到数据库并插入/删除/显示内容的应用。非常简单，特别是如果PHP选择它。

首先我尝试谷歌它，我找到了一些解决方案，但大多数都没有为我工作，最后我尝试看起来很神奇。但是有一些问题，它会引发错误：

activity_main cannot be resolved or is not a field

package com.example.helloworld;

import java.io.BufferedReader;
import java.io.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import org.apache.http.HttpEntity;
import org.apache.http.HttpResponse;
import org.apache.http.NameValuePair;
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.json.JSONObject;
import android.os.Bundle;
import android.app.Activity;
import android.util.Log;
import android.view.Menu;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;
import android.widget.Toast;

public class MainActivity extends Activity {

    String name;
    String id;
    InputStream is=null;
    String result=null;
    String line=null;
    int code;

    @Override
    public void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        final EditText e_id=(EditText) findViewById(R.id.editText1);
        final EditText e_name=(EditText) findViewById(R.id.editText2);
        Button insert=(Button) findViewById(R.id.button1);

        insert.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View v) {
            // TODO Auto-generated method stub

            id = e_id.getText().toString();
            name = e_name.getText().toString();

            insert();
        }
    });
    }

    public void insert()
    {
        ArrayList<NameValuePair> nameValuePairs = new ArrayList<NameValuePair>();

    nameValuePairs.add(new BasicNameValuePair("id",id));
    nameValuePairs.add(new BasicNameValuePair("name",name));

        try
        {
        HttpClient httpclient = new DefaultHttpClient();
            HttpPost httppost = new HttpPost("http://10.0.2.2/insert.php");
            httppost.setEntity(new UrlEncodedFormEntity(nameValuePairs));
            HttpResponse response = httpclient.execute(httppost); 
            HttpEntity entity = response.getEntity();
            is = entity.getContent();
            Log.e("pass 1", "connection success ");
    }
        catch(Exception e)
    {
            Log.e("Fail 1", e.toString());
            Toast.makeText(getApplicationContext(), "Invalid IP Address",
            Toast.LENGTH_LONG).show();
    }     

        try
        {
            BufferedReader reader = new BufferedReader
            (new InputStreamReader(is,"iso-8859-1"),8);
            StringBuilder sb = new StringBuilder();
            while ((line = reader.readLine()) != null)
        {
                sb.append(line + "\n");
            }
            is.close();
            result = sb.toString();
        Log.e("pass 2", "connection success ");
    }
        catch(Exception e)
    {
            Log.e("Fail 2", e.toString());
    }     

    try
    {
            JSONObject json_data = new JSONObject(result);
            code=(json_data.getInt("code"));

            if(code==1)
            {
        Toast.makeText(getBaseContext(), "Inserted Successfully",
            Toast.LENGTH_SHORT).show();
            }
            else
            {
         Toast.makeText(getBaseContext(), "Sorry, Try Again",
            Toast.LENGTH_LONG).show();
            }
    }
    catch(Exception e)
    {
            Log.e("Fail 3", e.toString());
    }
    }

    @Override
    public boolean onCreateOptionsMenu(Menu menu) {
        getMenuInflater().inflate(R.menu.activity_main, menu);
        return true;
    }    
}

有些基础知识的人能找到一点时间给我解释一下，或者给我一些建议如何解决我的问题？

Answer 1

您的菜单文件名是main.xml，因此您必须将代码更改为：

@Override
    public boolean onCreateOptionsMenu(Menu menu) {
        getMenuInflater().inflate(R.menu.main, menu);
        return true;
    }

使用PHP在Android上连接MYSQL

1 个答案: