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Java 从JSON获取所有用户并将其放入TextView_Java_Android - Fatal编程技术网

Java 从JSON获取所有用户并将其放入TextView

Java 从JSON获取所有用户并将其放入TextView,java,android,Java,Android,我正在尝试在我的应用程序上解析这个JSON,并尝试将结果放在TextView上。从现在起,我将JSON放在字符串中,这是我的JSON [{ "id": 1, "name": "Leanne Graham", "username": "Bret", "email": "Sincere@april.biz", "address": { "street": "Kulas Light", "suite": "Apt. 556",

我正在尝试在我的应用程序上解析这个
JSON
,并尝试将结果放在
TextView
上。从现在起,我将
JSON
放在
字符串中,这是我的
JSON

[{
    "id": 1,
    "name": "Leanne Graham",
    "username": "Bret",
    "email": "Sincere@april.biz",
    "address": {
        "street": "Kulas Light",
        "suite": "Apt. 556",
        "city": "Gwenborough",
        "zipcode": "92998-3874",
        "geo": {
            "lat": "-37.3159",
            "lng": "81.1496"
        }
    },
    "phone": "1-770-736-8031 x56442",
    "website": "hildegard.org",
    "company": {
        "name": "Romaguera-Crona",
        "catchPhrase": "Multi-layered client-server neural-net",
        "bs": "harness real-time e-markets"
    }
},

我试过这个:

JSONArray users = new JSONArray(result);
tvNames.setText(users.get("name"));

但是它不起作用。

您正在尝试从
JSONArray
获取名称,您需要遍历它并找到带有键“name”的
JSONObject,如下所示:

private String getAllNames(String result){
        StringBuilder names = new StringBuilder();
        try{
            JSONArray users = new JSONArray(result);
            for(int i = 0; i<users.length();i++){
                JSONObject user = users.getJSONObject(i);
                names.append(user.getString("name")).append("\n");
          }
        }catch(JSONException ex){
            Log.d("Debug","Error parsing json " + ex.getMessage());
        }
        return names.toString();
    }
tvNames.setText()