如何決議volley庫中的JSON資料？

我想決議 JSON 回應。我無法決議回應。它顯示org.json.JSONException: No value for String Response此錯誤。

這是我的代碼`public void onResponse(JSONObject response) {

                        Log.d("TAG", "Details:"   response);
                        responseTV.setText("String Response : "   response.toString());

                        try {
                            JSONObject jsonObject = response.getJSONObject("String Response" response);
                            strcode = jsonObject.getString("responseCode");
                            strtext = jsonObject.getString("responseText");
                            strname = jsonObject.getString("personName");
                            Log.i("TAG","parseData:" strname);
                            response_code.setText(""  strcode);
                            response_text.setText(""  strtext);
                            person_name.setText(""  strname);

                        } catch (JSONException e) {
                            Log.d("TAG", "profile: "   e);
                        }

                }`

uj5u.com熱心網友回復：

我猜您使用 Volley JsonObjectRequest，因此您需要將回應轉換為 String 并將其決議為 a JSONObject，如下所示：

jsonObjectRequest = new JsonObjectRequest(Request.Method.GET, url, null,
                response -> {

                    try {

                        JSONObject jsonObject = new JSONObject(response.toString());
                        strcode = jsonObject.getString("responseCode");
                        strtext = jsonObject.getString("responseText");
                        strname = jsonObject.getString("personName");
                        Log.i("TAG","parseData:" strname);
                        response_code.setText(""  strcode);
                        response_text.setText(""  strtext);
                        person_name.setText(""  strname);

                    } catch (JSONException e) {
                        Log.d("TAG", "profile: "   e);
                    }


                }, error -> {}
        );

uj5u.com熱心網友回復：

 try {
                        JSONObject jsonObject = response.getJSONObject(response);
                        strcode = jsonObject.getString("responseCode");
                        strtext = jsonObject.getString("responseText");
                        strname = jsonObject.getString("personName");
                        Log.i("TAG","parseData:" strname);
                        response_code.setText(""  strcode);
                        response_text.setText(""  strtext);
                        person_name.setText(""  strname);

                    } catch (JSONException e) {
                        Log.d("TAG", "profile: "   e);
                    }

在進行決議之前，您將通過 API exp 檢查所有接收值：- responseCode、responseText、personName ( https://jsonlint.com/ )，如果您接收到所有值，請檢查其格式是否正確，然后您將決議資料

你也用這個，它會處理 JsonException

jsonObject.optString("responseText");

uj5u.com熱心網友回復：

用這段代碼解決了這個問題

public void onResponse(JSONObject response) {
    Log.d("TAG", "Details:"   response);
    responseTV.setText("String Response : "   response.toString());

    try {
        // JSONObject jsonObject = response.getJSONObject(response.toString());
        strcode = response.getString("responseCode");
        strtext = response.getString("responseText");
        strname = response.getString("personName");
        Log.i("TAGParser","parseData:" strname);
        response_code.setText(""  strcode);
        response_text.setText(""  strtext);
        person_name.setText(""  strname);
    } catch (JSONException e) {
        Log.d("TAG", "profile: "   e);
    }
}

標籤：安卓 json json解析器

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