Question

嗨，大家好，我需要你的帮助。我在我的应用程序中创建SQLite数据库，然后将数据插入其中。现在我想从中检索数据，但我只想插入一个数据并检索它然后将其显示到TextView中。

请帮助我们这是我第一次使用SQLite数据库。

public class Db_sqlit extends SQLiteOpenHelper{

    String TABLE_NAME = "BallsTable";

    public final static String name = "db_data";

    public Db_sqlit(Context context) {
        super(context, name, null, 1);
    }

    @Override
    public void onCreate(SQLiteDatabase db) {
        db.execSQL("create table "+TABLE_NAME+" (id INTEGER PRIMARY KEY AUTOINCREMENT, ball TEXT)");

    }

    @Override
    public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
        db.execSQL("DROP TABLE IF EXISTS "+TABLE_NAME);
        onCreate(db);
    }

    public boolean insertData(String balls){
      SQLiteDatabase db = this.getWritableDatabase();
      ContentValues contentValues = new ContentValues();

      contentValues.put("ball",balls);

      long result = db.insert(TABLE_NAME,null,contentValues);
      if(result == -1){
          return false;
      }
      else
          return true;
    }

    public void list_balls(TextView textView) {

        Cursor res = this.getReadableDatabase().rawQuery("select ball from "+TABLE_NAME+"",null);
        textView.setText("");
        while (res.moveToNext()){
            textView.append(res.getString(1));
        }  
    }
}

Answer 1

以下是我如何实现这一目标的一个例子。

在此示例中，我将store，retrieve，update和delete作为学生姓名和年龄。

首先创建一个类，我称之为

<强> DBManager.java

public class DBManager {
private Context context;
private SQLiteDatabase database;
private SQLiteHelper dbHelper;

public DBManager(Context c) {
    this.context = c;
}

public DBManager open() throws SQLException {
    this.dbHelper = new SQLiteHelper(this.context);
    this.database = this.dbHelper.getWritableDatabase();
    return this;
}

public void close() {
    this.dbHelper.close();
}

public void insert(String name, String desc) {
    ContentValues contentValue = new ContentValues();
    contentValue.put(SQLiteHelper.NAME, name);
    contentValue.put(SQLiteHelper.AGE, desc);
    this.database.insert(SQLiteHelper.TABLE_NAME_STUDENT, null, contentValue);
}


public Cursor fetch() {
    Cursor cursor = this.database.query(SQLiteHelper.TABLE_NAME_STUDENT, new String[]{SQLiteHelper._ID, SQLiteHelper.NAME, SQLiteHelper.AGE}, null, null, null, null, null);
    if (cursor != null) {
        cursor.moveToFirst();
    }
    return cursor;
}

public int update(long _id, String name, String desc) {
    ContentValues contentValues = new ContentValues();
    contentValues.put(SQLiteHelper.NAME, name);
    contentValues.put(SQLiteHelper.AGE, desc);
    return this.database.update(SQLiteHelper.TABLE_NAME_STUDENT, contentValues, "_id = " + _id, null);
}

public void delete(long _id) {
    this.database.delete(SQLiteHelper.TABLE_NAME_STUDENT, "_id=" + _id, null);
    }
}

然后创建一个名为我的SQLiteOpenHelper

<强> SQLiteHelper.java

public class SQLiteHelper extends SQLiteOpenHelper {
public static final String AGE = "age";
private static final String CREATE_TABLE_STUDENT = " create table STUDENTS ( _id INTEGER PRIMARY KEY AUTOINCREMENT, name TEXT NOT NULL , age TEXT );";
private static final String DB_NAME = "STUDENTS.DB";
private static final int DB_VERSION = 1;
public static final String NAME = "name";
public static final String TABLE_NAME_STUDENT = "STUDENTS";
public static final String _ID = "_id";

public SQLiteHelper(Context context) {
    super(context, DB_NAME, null, 1);
}

public void onCreate(SQLiteDatabase db) {
    Log.e("SQLITE", "table reated");
    db.execSQL(CREATE_TABLE_STUDENT);
}

public void onUpgrade(SQLiteDatabase db, int oldVersion, int newVersion) {
    db.execSQL("DROP TABLE IF EXISTS STUDENTS");
    onCreate(db);
}

}

添加：

在此示例中，我从EditText获取文本，当单击按钮时，我会检查EditText是否为空。如果它不是空的并且学生尚不存在，我将学生姓名和年龄插入数据库。我显示Toast，让用户知道状态：

btnAdd.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View view) {
            if (edtName.getText().toString().trim().length() == 0) {
                Toast.makeText(getApplicationContext(), "Please provide your students name", Toast.LENGTH_SHORT).show();

            } else
                try {
                    if (edtAge.getText().toString().trim().length() != 0) {

                        String name = edtName.getText().toString().trim();
                        String age = edtAge.getText().toString().trim();
                        String query = "Select * From STUDENTS where name = '"+name+"'";
                        if(dbManager.fetch().getCount()>0){
                            Toast.makeText(getApplicationContext(), "Already Exist!", Toast.LENGTH_SHORT).show();
                        }else{
                            dbManager.insert(name, age);

                            Toast.makeText(getApplicationContext(), "Added successfully!", Toast.LENGTH_SHORT).show();

                        }

                    } else {

                        Toast.makeText(getApplicationContext(), "please provide student age!", Toast.LENGTH_SHORT).show();


                    }
                } catch (Exception e) {
                    e.printStackTrace();
                }


        }
    });

更新：

在这里，我将文本放在EditText中，并在单击按钮时更新学生。您还可以将以下内容放在try/catch中，以确保其成功更新。

btnupdate.setOnClickListener(new View.OnClickListener() {

        @Override
        public void onClick(View view) {
            String name = nameText.getText().toString();
            String age = ageText.getText().toString();

            dbManager.update(_id, name, age);
            Toast.makeText(getApplicationContext(), "Updated successfully!", Toast.LENGTH_SHORT).show();


        }
    });

删除：

dbManager.delete(_id);
Toast.makeText(getApplicationContext(), "Deleted successfully!", Toast.LENGTH_SHORT).show();

获取：

在这里，我获得了学生的姓名并将其显示在TextView

中

DBManager dbManager = new DBManager(getActivity());
dbManager.open();

Cursor cursor = dbManager.fetch();
cursor.moveToFirst();
final TextView studentName = (TextView) getActivity().findViewById(R.id.nameOfStudent);
studentName.settext(cursor.getString(0));

Answer 2

然后我在主 Java 类中实现了代码，我想在其中使用 cursor.moveToNext() 显示

searchButton.setOnClickListener(new View.OnClickListener() {
        @Override
        public void onClick(View v) {
            Cursor result = databaseSQLite2.searchData(searchET.getText().toString());
            while (result.moveToNext()){
                searchresultTV.setText(result.getString(2));
            }
        }
    });

Answer 3

为了从 sqlite 获取数据，我在 DatabaseHelper 类中完成了这个方法

public Cursor searchData(String id){
    SQLiteDatabase sqLiteDatabase = this.getWritableDatabase();
    //String qry = "SELECT * FROM "+TABLE_NAME+" WHERE ID="+id;
    Cursor cursor = sqLiteDatabase.rawQuery("SELECT * FROM "+TABLE_NAME+" WHERE ID="+id,null);
    return cursor;
}

android studio从Sqlite数据库中检索数据并将其显示到textview中

3 个答案: