Android setText在TextView上不起作用

Question

我試圖改變一個TextView一些資料我查詢解析為的價值，但由於某些原因， .setText未服用影響。

我嘗試了谷歌搜索，但.getString也不起作用。 我對解析任何建議還很陌生？

我的代碼如下：

import android.os.Bundle;
import android.support.v4.app.Fragment;
import android.util.Log;
import android.view.LayoutInflater;
import android.view.View;
import android.view.ViewGroup;
import android.widget.TextView;

import com.parse.FindCallback;
import com.parse.GetCallback;
import com.parse.ParseException;
import com.parse.ParseObject;
import com.parse.ParseQuery;

import org.w3c.dom.Text;

import java.util.List;


/**
* A simple {@link Fragment} subclass.
 */
public class productDetail extends Fragment {


public productDetail() {
    // Required empty public constructor
}






@Override
public View onCreateView(LayoutInflater inflater, ViewGroup container,
                         Bundle savedInstanceState) {

    final View theView = inflater.inflate(R.layout.fragment_product_detail, container, false);

    final TextView productNameLabel = (TextView) theView.findViewById(R.id.productNameTextView);
    final TextView productDetailsLabel = (TextView) theView.findViewById(R.id.productDetails);
    final TextView shippingFeeLabel = (TextView) theView.findViewById(R.id.shippingFee);
    final TextView productPriceLabel = (TextView) theView.findViewById(R.id.productPrice);
    final TextView productNumberOfReviewsLabel = (TextView) theView.findViewById(R.id.productNumberOfReviews);

    final String chosenProduct = getArguments().getString("productName");
    Log.i("AppInfo", chosenProduct);

    ParseQuery<ParseObject> query = new ParseQuery<ParseObject>("Inventory");
    query.whereEqualTo("productName", chosenProduct);
    query.getFirstInBackground(new GetCallback<ParseObject>() {
        @Override
        public void done(ParseObject parseObject, ParseException e) {
            if (parseObject == null) {

                Integer fee = parseObject.getInt("ShippingFee");
                Integer review = parseObject.getInt("numberOfReviews");
                String information = parseObject.get("information").toString();
                Integer price = parseObject.getInt("price");


                productNameLabel.setText(chosenProduct);
                productDetailsLabel.setText(String.valueOf(information));
                shippingFeeLabel.setText(String.valueOf(fee));
                productPriceLabel.setText(String.valueOf(price));
                productNumberOfReviewsLabel.setText(String.valueOf(entries));

            }

        }
    });




    // Inflate the layout for this fragment
    return theView;
}

}

Answer 1

如果輸入if語句中的所有內容，它將拋出NullPointerException :)

將if語句更改為此

if (parseObject != null)

也許你是這個意思？

if (e == null)