使用IN關鍵字的android系統源碼

Question

我有兩個表：

1) Employee -- _id, employee_name. 
2) Salary -- _id, amount, emp_id. 

的樣本數據：

Employee: 
1 John 
2 Rocky 
3 Marry 

Salary: 

1 500 1 //salary for John 
2 400 1 //salary for John 
3 600 2 //salary for Rocky 
4 700 2 //salary for Rocky 
5 350 3 //salary for Marry 

現在，我想在工資表中進行搜索，查看我所支付的工資。讓我說如果我在工資表中搜索'John'，它應該返回John的第1行和第2行。

這裏就是我想：

String where = " emp_id in (select _id from Employee where employee_name like ?)"; 

String[] whereArgs = new String[] {"'%" + mFilter + "%'" }; 

Cursor c = getDB(mContext).query("Salary", null, where, whereArgs, 
       null, null, null); 

但它總是返回任何結果。請幫忙。

更新

我調試代碼，發現下面的查詢在光標被執行：

SELECT * FROM Salary WHERE emp_id in (select _id from Employee where employee_name like ?); 

Answer 1

選擇ARGS將自動用作字符串。在那裏

String[] whereArgs = new String[] {"%" + mFilter + "%" }; (removed ' from your strings) 

有了額外'它取消了文本，併成爲''%John%''建設者自動處理'的選擇ARGS：更改此：

String[] whereArgs = new String[] {"'%" + mFilter + "%'" }; 

了這一點。

編輯
更改您的查詢也是這樣：

String sql = "SELECT * FROM Salary WHERE emp_id in (select _id from Employee where employee_name like ?)"; 
Cursor c = getDB(mContext).rawQuery(sql, whereArgs); 

EDIT 2

我重新創建您的設置與下面的類和我的代碼跑就好了。我從用戶約翰拉出並得到了兩個結果。我相信問題出在你的數據庫創建中，或者你的數據庫中沒有數據。使用DDMS來提取數據庫並用SQLite Browser打開它。檢查數據庫中是否有任何數據。如果確實如此，那麼你的表創建類型不匹配選擇語句。當我調用GetMyValues（）時，我得到了從遊標返回的2條記錄。

public class DataBaseHandler extends SQLiteOpenHelper { 

    private static final String TAG = "DBHandler"; 

    //Database VERSION 
    private static final int DATABASE_VERSION = 2; 

    //DATABASE NAME 
    private static final String DATABASE_NAME = "test"; 

    //DATABASE TABLES 
    private static final String TABLE_SALARY = "Salary"; 
    private static final String TABLE_EMP = "Employee"; 

    //DATABASE FIELDS 
    private static final String SalaryID= "_id"; 
    private static final String SalaryEmpName = "employee_name"; 
    private static final String EmpID= "_id"; 
    private static final String EmpAmt = "amount"; 
    private static final String EmpSalID = "emp_id"; 

    //DATABASE TYPES 
    private static final String INTPK = "INTEGER PRIMARY KEY"; 
    private static final String INT = "INTEGER"; 
    private static final String TEXT = "TEXT"; 

    //CREATE TABLES 
    private static final String CREATE_SALARY_TABLE = "CREATE TABLE " + TABLE_SALARY + "(" 
       + EmpID + " " + INTPK + "," + EmpAmt + " " + INT + "," 
       + EmpSalID + " " + INT + ")"; 

     //CREATE TABLE Salary(_id INTEGER PRIMARY KEY,amount INTEGER,emp_id INTEGER) 

    private static final String CREATE_EMPLOYEE_TABLE = "CREATE TABLE " + TABLE_EMP + "(" 
       + SalaryID + " " + INTPK + "," + SalaryEmpName + " " + TEXT + ")"; 

     //CREATE TABLE Employee(_id INTEGER PRIMARY KEY, employee_name TEXT) 

    public DataBaseHandler(Context context){ 
     super(context, DATABASE_NAME, null, DATABASE_VERSION); 
    } 

    @Override 
    public void onCreate(SQLiteDatabase db) { 
     db.execSQL(CREATE_EMPLOYEE_TABLE); 
     db.execSQL(CREATE_SALARY_TABLE); 

     insertEmployeeValues(db); 
     insertSalaryValues(db);  
    } 

    private void insertEmployeeValues(SQLiteDatabase db){ 
     ContentValues values = new ContentValues(); 
     values.put(SalaryEmpName, "John"); 
     db.insert(TABLE_EMP, null, values); 
     values.clear(); 
     values.put(SalaryEmpName, "Rocky"); 
     db.insert(TABLE_EMP, null, values); 
     values.clear(); 
     values.put(SalaryEmpName, "Marry"); 
     db.insert(TABLE_EMP, null, values); 
     values.clear(); 
    } 

    private void insertSalaryValues(SQLiteDatabase db){ 
     ContentValues values = new ContentValues(); 
     values.put(EmpAmt, 500); 
     values.put(EmpSalID, 1); 
     db.insert(TABLE_SALARY, null, values); 
     values.clear(); 
     values.put(EmpAmt, 400); 
     values.put(EmpSalID, 1); 
     db.insert(TABLE_SALARY, null, values); 
     values.clear(); 
     values.put(EmpAmt, 600); 
     values.put(EmpSalID, 2); 
     db.insert(TABLE_SALARY, null, values); 
     values.clear(); 
     values.put(EmpAmt, 700); 
     values.put(EmpSalID, 2); 
     db.insert(TABLE_SALARY, null, values); 
     values.clear(); 
     values.put(EmpAmt, 350); 
     values.put(EmpSalID, 3); 
     db.insert(TABLE_SALARY, null, values); 
     values.clear(); 
    } 

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

    public int GetMyValues(){ 
     String mFilter = "John"; 
     String[] whereArgs = new String[]{"%" + mFilter + "%"}; 
     int count = 0; 
     SQLiteDatabase db = this.getWritableDatabase(); 
     String where = " emp_id in (select _id from Employee where employee_name like ?)"; 
     Cursor c = db.query("Salary",null, where, whereArgs,null,null,null); 
     count = c.getCount(); 
     c.close(); 
     return count; 
    } 
} 

Dev Reference:

您可能包括哪些內容？S IN的選擇，這將是由值 代替從selectionArgs兩個，以使他們出現在選擇。該 值將被綁定爲字符串

Answer 2

使用遊標是更好，我猜，

演示代碼：

 Cursor c = db.query(TABLE_CONTACTS, columns, KEY_MOBILE + " like '%" 
      + mobno + "'", null, null, null, order); 
return c; 

它將返回值，如果這個值實例是數據庫！ ！