Android:数据库处理程序SQLiteException

Question

大家好我在数据库处理程序类中有这个方法,这个类的作用是从product表中返回产品的ID.但是,我收到这个sqliteexception,我不知道为什么.请指教谢谢.

    private static final String TABLE_PRODUCT = "product"
    private static final String KEY_PRODUCTNAME = "productname";

public String getProductId(String productName) {
   String selectQuery = "SELECT productid FROM " + TABLE_PRODUCT+ " WHERE " +KEY_PRODUCTNAME +" = " + productName;

    SQLiteDatabase db = this.getReadableDatabase();
    Cursor cursor = db.rawQuery(selectQuery, null);
    String productid = cursor.toString();
    cursor.close();
    db.close();

    return productid;
}

错误:

E/AndroidRuntime(1884): FATAL EXCEPTION: main
E/AndroidRuntime(1884): android.database.sqlite.SQLiteException: unrecognized token: "Bluedress34.50" (code 1): , while compiling: SELECT productid FROM product WHERE productname = "Bluedress34.50"

我的活动课:

add.setOnClickListener(new View.OnClickListener() {

            public void onClick(View view) {

    DatabaseHandler db = new DatabaseHandler(getApplicationContext());

                    String productname = pname.getText().toString();
                    String productQTY = pqty.getText().toString();

                    String productnameid = db.getUProductId(productname);


                    JSONObject json = userFunction.addSales(productnameid, productQty);

}

}

我的用户功能类:

public JSONObject addSales(productnameid, productQty){
    // Building Parameters

    List<NameValuePair> paramsfile = new ArrayList();

    paramsfile.add(new BasicNameValuePair("productnameid", productnameid));
    paramsfile.add(new BasicNameValuePair("productQty", productQty));


    JSONObject jsonfileName  =  jsonParser.getJSONFromUrl(addFileURL, paramsfile);

   Log.e("JSON", jsonfileName.toString());
    return jsonfileName;
}

Answer 1

由于产品名称是字符串格式,因此'在查询中使用,如下所示,

String selectQuery = "SELECT productid FROM " + TABLE_PRODUCT 
      + " WHERE " + KEY_PRODUCTNAME +" ='" + productName  +"'";