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我尝试使用db.insert(inputfname, inputlname, inputnickname) 插入；但它是错误的。它告诉我“FriendsDB insertActivity.java 类型的方法 insertData(EditText, EditText, EditText) 未定义”你能告诉我吗

这是我的代码（Mainactivity）包 com.example.sqlite；

import java.security.PublicKey;

import com.example.sqlite.db.FriendsDB;

import android.app.Activity;
import android.content.Context;
import android.content.Intent;
import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.EditText;

public class insertActivity extends Activity {
private Context context;
private FriendsDB db;
EditText inputfname;
EditText inputlname;
EditText inputnickname;
Button save;
Button cancle;

@Override
protected void onCreate(Bundle savedInstanceState) {
    // TODO Auto-generated method stub
    super.onCreate(savedInstanceState);
    setContentView(R.layout.insert_layout);

    //view matching
    inputfname = (EditText)findViewById(R.id.fname);
    inputlname = (EditText)findViewById(R.id.lname);
    inputnickname = (EditText)findViewById(R.id.nickname);
    save = (Button)findViewById(R.id.save);
    cancle = (Button)findViewById(R.id.cancle);

    //context
            context = this;
            db = new FriendsDB(context);

    //button event
    save.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View arg0) {
            Intent insertData = new Intent(insertActivity.this,FriendsListActivity.class);
            insertData.putExtra("fname", inputfname.getText().toString());
            insertData.putExtra("lname", inputlname.getText().toString());
            insertData.putExtra("nickname", inputnickname.getText().toString());
            if(insertData!=null){
                  db.insert(inputfname, inputlname, inputnickname); <<<<<< ERROR
            }
        }
    } );

这是我的 FriendsDB 代码

package com.example.sqlite.db;

import java.util.ArrayList;

import com.example.sqlite.entry.FriendEntry;

import android.content.ContentValues;
import android.content.Context;
import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;

//ใช้ในการ inert update delete
public class FriendsDB {
    private FriendsDBHelper helper;
private SQLiteDatabase db;

public FriendsDB(Context context){
    helper = new FriendsDBHelper(context);
    db = helper.getWritableDatabase();
}
//insert
    public long insert(String fname,String lname,String nickname){
        ContentValues values = new ContentValues();
        values.put(FriendsDBHelper.TABLE_KEY_FNAME, fname);
        values.put(FriendsDBHelper.TABLE_KEY_LNAME, lname);
        values.put(FriendsDBHelper.TABLE_KEY_NICKNAME, nickname);
        return db.insert(FriendsDBHelper.TABLE_NAME, null ,values);

    }
//select all
    public ArrayList<FriendEntry> selectAll(){
        ArrayList<FriendEntry> friends = new ArrayList<FriendEntry>();
        //   cursor คือ ชุดข้อมูล
        Cursor cursor = db.rawQuery("SELECT*FROM"+FriendsDBHelper.TABLE_NAME+"WHERE    id != ?",new String[]{Integer.toString(0)});
        cursor.moveToFirst();
        if(cursor.getCount()!=0){
            do{
                FriendEntry friend = new FriendEntry();
                  friend.setId(cursor.getInt(cursor.getColumnIndex(FriendsDBHelper.TABLE_KEY_ID)));
                friend.setFname(cursor.getString(cursor.getColumnIndex(FriendsDBHelper.TABLE_KEY_FNAME)));
                friend.setLname(cursor.getString(cursor.getColumnIndex(FriendsDBHelper.TABLE_KEY_LNAME)));
                friend.setNickname(cursor.getString(cursor.getColumnIndex(FriendsDBHelper.TABLE_KEY_NICKNAME)));
                friends.add(friend);

            }while(cursor.moveToNext());

        }

        return friends;
    }
4

1 回答 1

1

改变这个

db.insert(inputfname, inputlname, inputnickname);

到

db.insert(inputfname.getText().toString(), inputlname.getText().toString(), inputnickname.getText().toString());

您的数据库插入方法将字符串作为参数。

insert(String fname,String lname,String nickname)// arguments are strings
于 2013-08-19T17:24:09.513 回答