Question

我正在尝试将db文件复制到assets文件夹，Toats通知：复制成功，但是当我实现显示表信息时，＆＃34; Android SQLiteException没有这样的表＆＃34;显示错误。请帮我。非常感谢你。

主：包com.example.huyvu.home_sqlite;

import android.database.Cursor;
import android.database.sqlite.SQLiteDatabase;
import android.support.v7.app.AppCompatActivity;
import android.os.Bundle;
import android.util.Log;
import android.widget.ArrayAdapter;
import android.widget.ListView;
import android.widget.Toast;

import java.io.File;
import java.io.FileOutputStream;
import java.io.InputStream;
import java.io.OutputStream;

import model.Contact;

public class MainActivity extends AppCompatActivity {
String DATABASE_NAME = "dbContact.sqlite";
String DB_PATH_SUFFIX = "/databases/";
SQLiteDatabase database = null;
ListView lvContact;
ArrayAdapter<Contact> adapter;

@Override
protected void onCreate(Bundle savedInstanceState) {
    super.onCreate(savedInstanceState);
    setContentView(R.layout.activity_main);
    processCopy();
    addControls();
    hienThiToanBoSP();
}

private void hienThiToanBoSP() {
    database = openOrCreateDatabase(DATABASE_NAME,MODE_PRIVATE,null);
    Cursor cursor = database.rawQuery("select * from Contact",null);
    adapter.clear();
    while (cursor.moveToNext()){
        int ma = cursor.getInt(0);
        String ten = cursor.getString(1);
        String phone = cursor.getString(2);
        Contact contact = new Contact(ma,ten,phone);
        adapter.add(contact);
    }
    cursor.close();
}

private void addControls() {
    lvContact = (ListView) findViewById(R.id.lvContact);
    adapter = new ArrayAdapter<Contact>(MainActivity.this, 
android.R.layout.simple_list_item_1);
    lvContact.setAdapter(adapter);
}

private void processCopy(){
    try {
        File dbFile = getDatabasePath(DATABASE_NAME);
        if(!dbFile.exists()){
            copyDatabaseFromAsset();
            Toast.makeText(MainActivity.this,"Copy DB to device success",Toast.LENGTH_SHORT).show();
        }
    }
    catch (Exception ex){

Toast.makeText(MainActivity.this,ex.toString(),Toast.LENGTH_SHORT).show();
        Log.e("Error",ex.toString());
    }
}
private String getDatabasePath(){
    return getApplicationInfo().dataDir+DB_PATH_SUFFIX+DATABASE_NAME;
}
private void copyDatabaseFromAsset() {
    try {
        InputStream myInput = getAssets().open(DATABASE_NAME);
        String outFileName = getDatabasePath();
        File f = new File(getApplicationInfo().dataDir+DB_PATH_SUFFIX);
        if(!f.exists())
            f.mkdir();//create named databases folder
        OutputStream myOutput = new FileOutputStream(outFileName);
        byte[] buffer = new byte[1024];
        int length;
        while ((length=myInput.read(buffer))>0){
            myOutput.write(length);
            myOutput.flush();
            myOutput.close();
            myInput.close();
        }
    }
    catch (Exception ex){
        Log.e("Error",ex.toString());
    }
}
}

这是我的联系班级：包装模型; / *

* /

public class Contact {
private int ma;
private String ten;
private String phone;

public Contact() {
}

public Contact(int ma, String ten, String phone) {
    this.ma = ma;
    this.ten = ten;
    this.phone = phone;
}

public int getMa() {
    return ma;
}

public void setMa(int ma) {
    this.ma = ma;
}

public String getTen() {
    return ten;
}

public void setTen(String ten) {
    this.ten = ten;
}

public String getPhone() {
    return phone;
}

public void setPhone(String phone) {
    this.phone = phone;
}
public String toString(){
    return this.ma + "-" + this.ten + "\n" + this.phone;
}
}

Answer 1

你永远不会创建表，你应该从异常中理解它，因为它包含“没有这样的表”。在代码中，您永远不会创建数据库，您只能尝试从中读取数据。解决方案非常简单：打开/创建数据库后添加SQL查询：

CREATE TABLE IF NOT EXISTS Contacts (Rows here)

Android：将db文件复制到Access文件夹，SQLiteException：没有这样的表

1 个答案: