Question

这是我的JSonArray

{  
   "vendor":[  
      {  
         "vendor_name":"Tapan Moharana",
         "vendor_description":"",
         "vendor_slug":"tapan",
         "vendor_logo":null,
         "contact_number":null
      }
   ],
   "products":[  
      {  
         "name":"Massage",
         "price":"5000.0000",
         "image":"http:\/\/carrottech.com\/lcart\/media\/catalog\/product\/cache\/1\/image\/150x\/9df78eab33525d08d6e5fb8d27136e95\/2\/9\/29660571-beauty-spa-woman-portrait-beautiful-girl-touching-her-face.jpg"
      },
      {  
         "name":"Chicken Chilly",
         "price":"234.0000",
         "image":"http:\/\/carrottech.com\/lcart\/media\/catalog\/product\/cache\/1\/image\/150x\/9df78eab33525d08d6e5fb8d27136e95\/c\/h\/cheicken.jpg"
      },
      {  
         "name":"Chicken Biryani",
         "price":"500.0000",
         "image":"http:\/\/carrottech.com\/lcart\/media\/catalog\/product\/cache\/1\/image\/150x\/9df78eab33525d08d6e5fb8d27136e95\/placeholder\/default\/image_1.jpg"
      }
   ]
}

这是我的java代码：

    JSONObject jsono = new JSONObject(response);
    JSONArray children = jsono.getJSONArray("vendor");
    JSONArray childrenProducts = jsono.getJSONArray("products");
    for (int i = 0; i <children.length(); i++) {
        JSONObject jsonData = children.getJSONObject(i);
        System.out.print(jsonData.getString("vendor_name") + "<----");
        //  String vendorThumbNailURL=jsonData.getString("")
        //jvendorImageURL.setImageUrl(local, mImageLoader);
        vendorLogo=vendorLogo+jsonData.getString("vendor_logo").trim();
        jvendorImageURL.setImageUrl(vendorLogo, mImageLoader);
        jvendorName.setText(jsonData.getString("vendor_name"));
        jvendorAbout.setText(jsonData.getString("vendor_description"));
        jvendorContact.setText(jsonData.getString("contact_number"));
        System.out.print(jsonData.getString("products") + "<----");
    }
    for(int i=0;i<childrenProducts.length();i++){
        JSONObject jsonData = childrenProducts.getJSONObject(i);
        System.out.println("inside products");
        System.out.print(jsonData.getString("name") + "<----dd");
    }

第一个for循环工作正常，但第二个forloop不是..如果我尝试在第二个for循环中执行那些print语句，我什么也得不到..请帮助我!!

Answer 1

为什么不使用Gson来简单地解析JSON字符串？

您需要首先声明类以匹配JSON响应，如下所示。

public class Vendor {

    private String vendor_name;
    private String vendor_description;
    private String vendor_slug;
    private String vendor_logo;
    private String contact_number;

    public Vendor() {
    }

    public String getVendor_name() {
        return vendor_name;
    }

    public String getVendor_description() {
        return vendor_description;
    }

    public String getVendor_slug() {
        return vendor_slug;
    }

    public String getVendor_logo() {
        return vendor_logo;
    }

    public String getContact_number() {
        return contact_number;
    }
}

...

public class Product {

    private String name;
    private String price;
    private String image;

    public Product() {
    }

    public String getName() {
        return name;
    }

    public String getPrice() {
        return price;
    }

    public String getImage() {
        return image;
    }
}

现在声明像这样的Response类

public class Response {

    private List<Vendor> vendor;
    private List<Product> products;

    public Response() {
    }

    public List<Vendor> getVendor() {
        return vendor;
    }

    public List<Product> getProducts() {
        return products;
    }
}

现在，一旦你拥有了JSON字符串，就可以轻松地使用GSON将数据反弹到这样的Response类中。

Gson gson = new Gson();
Response mResponse = gson.fromJson(jsonString, Response.class);

简单！

Answer 2

这不是你的第二个for循环，因为你的第一个for循环中有SomeException。

您的执行将被抛出到任何catch()子句，并且将不会执行进一步的执行，包括您的第二个for循环。

试试这样：

    JSONObject jsono = new JSONObject(response);
    JSONArray children = jsono.getJSONArray("vendor");
    JSONArray childrenProducts = jsono.getJSONArray("products");

    //this will be executed now..!!

    for(int i=0;i<childrenProducts.length();i++){
        JSONObject jsonData = childrenProducts.getJSONObject(i);
        System.out.println("inside products");
        System.out.print(jsonData.getString("name") + "<----dd");
    }

    for (int i = 0; i <children.length(); i++) {
        JSONObject jsonData = children.getJSONObject(i);
        System.out.print(jsonData.getString("vendor_name") + "<----");
        //  String vendorThumbNailURL=jsonData.getString("")
        //jvendorImageURL.setImageUrl(local, mImageLoader);
        vendorLogo=vendorLogo+jsonData.getString("vendor_logo").trim();
        jvendorImageURL.setImageUrl(vendorLogo, mImageLoader);
        jvendorName.setText(jsonData.getString("vendor_name"));
        jvendorAbout.setText(jsonData.getString("vendor_description"));
        jvendorContact.setText(jsonData.getString("contact_number"));
        System.out.print(jsonData.getString("products") + "<----");
    }

Answer 3

这就是我解决它的方法：

   try {
                    JSONObject jsono = new JSONObject(response);
                    JSONArray children = jsono.getJSONArray("vendor");
                    for (int i = 0; i <children.length(); i++) {
                        JSONObject jsonData = children.getJSONObject(i);
                        System.out.print(jsonData.getString("vendor_name") + "<----");
                      //  String vendorThumbNailURL=jsonData.getString("")
                        //jvendorImageURL.setImageUrl(local, mImageLoader);
                        vendorLogo=vendorLogo+jsonData.getString("vendor_logo").trim();
                        jvendorImageURL.setImageUrl(vendorLogo, mImageLoader);
                        jvendorName.setText(jsonData.getString("vendor_name"));
                        jvendorAbout.setText(jsonData.getString("vendor_description"));
                        jvendorContact.setText(jsonData.getString("contact_number"));
                        System.out.print(jsonData.getString("products") + "<----");
                    }

                } catch (JSONException e) {
                    e.printStackTrace();
                }

                try {
                    JSONObject jsono = new JSONObject(response);
                    JSONArray childrenProducts = jsono.getJSONArray("products");
                    System.out.println(childrenProducts.length()+"LENGTH");
                    for(int i=0; i<childrenProducts.length(); i++){
                        JSONObject jsonData1 = childrenProducts.getJSONObject(i);
                        System.out.println(childrenProducts.length() + "LENGTH");
                        System.out.print(jsonData1.getString("name") + "<----dd");
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }
            }

只需要采取两个单独的试块...有人可以告诉为什么它不能在一个试块中工作？上面的代码工作

使用android volley迭代来自url的JSonArray响应

3 个答案: