在将订单添加到数据库之前检查条件是否满足

Question

我试图确保用户已经选择了一个waiter id（第一个if语句），并且该id是有效的，并且与数据库中的记录相关联（else if语句）。

我正在从一个数字（Android中的GUI文本）中收集ID，并选择number值，如下所示

 EditText waiter_id;
 waiter_id = (EditText) findViewById(R.id.waiter_id);

textUtils.isEmpty(WAITER_ID.getText()检查以确保输入了一个值，并显示一条消息，如果没有，则此操作正常。

之后，我调用了在DatabaseHelper isEmployee（String id)类中创建的方法，该方法使用用户输入的id在数据库中搜索该记录。如果在数据库中找不到id，则会出现一条消息来警告用户。

Order onClick OrderActivity

order.setOnClickListener(new View.OnClickListener() {
            Order order;
            @Override
            public void onClick(View view) {
                if (TextUtils.isEmpty(waiter_id.getText())) {
                    Toast.makeText(OrderActivity.this, "Please select waiter id", Toast.LENGTH_SHORT).show();

                // This is where the method which returns the boolean is called using
                // the value in the Number field in xml file
                } else if (!dbHelp.isEmployee(String.valueOf(waiter_id))){
                    Toast.makeText(OrderActivity.this, "Id not valid", Toast.LENGTH_SHORT).show();
                }

              

        });

用于查询数据库DataBaseHelper类的方法

public Boolean isEmployee(String id){
    SQLiteDatabase db = this.getReadableDatabase();

    String findEmployeeUsingId = "SELECT * FROM " + EMP_TABLE +
            " WHERE " + ID_EMP + " = " + id;
    Cursor cursor = db.rawQuery(findEmployeeUsingId, null);


    if (cursor.getCount() == 0) {
        return false;
    }
    else
        return true;
}

我收到的错误消息是:

com.example.dropit E/SQLiteLog: (1) near ".": syntax error in "SELECT * FROM EMP_TABLE WHERE ID = androidx.appcompat.widget.AppCompatEditText{93d8cbb VFED..CL. .F...... 249,193-829,317 #7f08011f app:id/waiter_id aid=1073741824}"

android.database.sqlite.SQLiteException: near ".": syntax error (code 1 SQLITE_ERROR): , while compiling: SELECT * FROM EMP_TABLE WHERE ID = androidx.appcompat.widget.AppCompatEditText{93d8cbb VFED..CL. .F...... 249,193-829,317 #7f08011f app:id/waiter_id aid=1073741824}

    at com.example.dropit.DataBaseHelper.isEmployee(DataBaseHelper.java:283)
    at com.example.dropit.OrderActivity$1.onClick(OrderActivity.java:58)

Answer 1

WAITER_ID是EditText，而不是字符串。
调用IsEmployee()时，应传递其文本:

else if (!dbHelp.isEmployee(waiter_id.getText().toString()))

还要在rawquery()内使用？占位符，而不是串联参数:

String findEmployeeUsingId = "SELECT * FROM " + EMP_TABLE + " WHERE " + ID_EMP + " = ?";
Cursor cursor = db.rawQuery(findEmployeeUsingId, new String[] {id});