Question

我有JSON回复

{
"data":{
    "last_name":"abcdef",
    "email":"xyz@zzz.com",
    "data1":{
        "data2":{
            "membership":"senior"
        },
        "Skills":[
            {
                "C":"GOOD",
                "C++":"AVERAGE",
                "JAVA":"GOOD",
                "WEB":"GOOD"
            }
        ]
    }
}
}

现在我使用以下方法通过GSON

解析它

public class ABC {
    public String last_name;
    public String email;
    .
    .
    .
}

然后我做

让我们考虑JSON OBJECT是== responseObject

ABC obj = (ABC)responseObject  // TypeCasting

obj.last_name给出了姓氏的结果如何实现数组和其他data1字典？

Answer 1

您可以使用gson库，这样可以更轻松地解析，也很快。制作下面的类，它会将json转换为你的java对象

public class Details {

   public MyData data;

}

public class MyData{

   public String last_name;
   public String email;
   public MyData1 data1;

}

public class MyData1 {

   public MyData2 data2;
   public List<Skill> Skills;
}

public class Mydata2{

    public string membership;
}

public class Skill {

   public String C;
   public String C++;
   public String JAVA;
   public String WEB;
}

// call following statement where you want to parse
Details details= new Gson().fromJson(json, Details.class);

Answer 2

试试这个：

im = ax.imshow(data)
im.set_extent((0, new_data.shape[1], new_data.shape[0], 0))
im.set_data(new_data)

ABC.java

    try {
        JSONObject object = new JSONObject(jsonString);
        if (object.has("data")) {
            JSONObject dataObject = object.getJSONObject("data");
            String name = "", email = "", membership = "";
            if (dataObject.has("last_name")) {
                name = dataObject.getString("last_name");
            }
            if (dataObject.has("email")) {
                email = dataObject.getString("email");
            }
            ABC abc = new ABC();
            abc.setLast_name(name);
            abc.setEmail(email);
            abc.setMembership(membership);
            if (dataObject.has("data1")) {
                JSONObject data1JsonObject = dataObject.getJSONObject("data1");
                if (data1JsonObject.has("data2")) {
                    membership = data1JsonObject.getJSONObject("data2").getString("membership");
                }

                if (data1JsonObject.has("Skills")) {
                    JSONArray jsonArray = data1JsonObject.getJSONArray("Skills");
                    Skills skills = new Skills();
                    for (int i = 0; i < jsonArray.length(); i++) {
                        JSONObject skillsJsonObject = jsonArray.getJSONObject(i);
                        if (skillsJsonObject.has("C")) {
                            skills.setC(skillsJsonObject.getString("C"));
                        }
                        if (skillsJsonObject.has("C++")) {
                            skills.setcPlusPlus(skillsJsonObject.getString("C++"));
                        }
                        if (skillsJsonObject.has("JAVA")) {
                            skills.setJava(skillsJsonObject.getString("JAVA"));
                        }
                        if (skillsJsonObject.has("WEB")) {
                            skills.setWeb(skillsJsonObject.getString("WEB"));
                        }
                    }
                    abc.setSkills(skills);
                }
            }
        }
    } catch (JSONException e) {
        e.printStackTrace();
    }

Skills.java

public class ABC {
private String last_name;
private String email;
private String membership;
private Skills skills;

public String getLast_name() {
    return last_name;
}

public void setLast_name(String last_name) {
    this.last_name = last_name;
}

public String getEmail() {
    return email;
}

public void setEmail(String email) {
    this.email = email;
}

public String getMembership() {
    return membership;
}

public void setMembership(String membership) {
    this.membership = membership;
}

public Skills getSkills() {
    return skills;
}

public void setSkills(Skills skills) {
    this.skills = skills;
}
 }

如何使用GSON和java解析复杂的JSON

2 个答案: