使用IN关键字的android系统源码

Question

我有两个表：

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

的样本数据：

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 

现在，我想在工资表中进行搜索，查看我所支付的工资。让我说如果我在工资表中搜索'John'，它应该返回John的第1行和第2行。

这里就是我想：

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); 

但它总是返回任何结果。请帮忙。

更新

我调试代码，发现下面的查询在光标被执行：

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

Answer 1

选择ARGS将自动用作字符串。在那里

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

有了额外'它取消了文本，并成为''%John%''建设者自动处理'的选择ARGS：更改此：

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

了这一点。

编辑
更改您的查询也是这样：

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

我重新创建您的设置与下面的类和我的代码跑就好了。我从用户约翰拉出并得到了两个结果。我相信问题出在你的数据库创建中，或者你的数据库中没有数据。使用DDMS来提取数据库并用SQLite Browser打开它。检查数据库中是否有任何数据。如果确实如此，那么你的表创建类型不匹配选择语句。当我调用GetMyValues（）时，我得到了从游标返回的2条记录。

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:

您可能包括哪些内容？S IN的选择，这将是由值 代替从selectionArgs两个，以使他们出现在选择。该 值将被绑定为字符串

Answer 2

使用游标是更好，我猜，

演示代码：

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

它将返回值，如果这个值实例是数据库！ ！