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如何获取json数组的每个值？我使用来自 textview 的追加来读取所有值。但我想要的是如何获取每个值来设置每个文本视图。

public void JSONparse() {
    String url = "https://api.myjson.com/bins/mat2d";
    JsonObjectRequest request = new JsonObjectRequest(Request.Method.GET, url, null, new Response.Listener<JSONObject>() {
        @Override public void onResponse(JSONObject response) {
            try {
                JSONArray JA = response.getJSONArray("products");
                for (int i=0; i<JA.length();i++){
                    Log.d("Result",JA.getString(i));
                    t1.append(JA.getString(i)+"\n");
                }
            } catch (JSONException e) {
                e.printStackTrace();
            }
        }
    }, new Response.ErrorListener() {

        @Override public void onErrorResponse(VolleyError error) {
            error.printStackTrace();
        }
    });
    rq.add(request);
}

我的日志猫

02-28 09:29:50.132 3101-3101/com.example.coorsdev.testing D/Result: Alfonso_Brandy 1000mL Imported Brandy
02-28 09:29:50.132 3101-3101/com.example.coorsdev.testing D/Result: Gran_Matador 350mL Regular Brandy
4

1 回答 1

0

用于Stringutils.split()按白人步数分割字符串。例如StringUtils.split("Hello World")返回"Hello"和"World"; 或者您可以使用whitespace regex：

 String str = "Hello I'm your String";
    String[] splited = str.trim().split("\\s+"); //use trim to avoid empty strings

结果将是：

splited [0] == "Hello";
splited [1] == "I'm";
splited [2] == "Your";
splited [3] == "String";

  public void JSONparse() {
        String url = "https://api.myjson.com/bins/mat2d";
        JsonObjectRequest request = new JsonObjectRequest(Request.Method.GET, url, null, new Response.Listener<JSONObject>() {
            @Override public void onResponse(JSONObject response) {
                try {
                    JSONArray JA = response.getJSONArray("products");
                    for (int i=0; i<JA.length();i++){

                        StringUtils.split(JA.getString(i))

                             // or

                    String[] splited = JA.getString(i).trim().split("\\s+");

                        Log.d("Result",JA.getString(i));
                        t1.append(JA.getString(i)+"\n");
                    }
                } catch (JSONException e) {
                    e.printStackTrace();
                }
            }
        }, new Response.ErrorListener() {

            @Override public void onErrorResponse(VolleyError error) {
                error.printStackTrace();
            }
        });
        rq.add(request);
    }
于 2018-02-28T06:08:00.500 回答