如何获取对 JSONObject 的 JsonArrayRequest JSONArray 响应答案

【问题标题】：How To Get JsonArrayRequest JSONArray Response To JSONObject如何获取对 JSONObject 的 JsonArrayRequest JSONArray 响应
【发布时间】：2016-06-02 13:10:34
【问题描述】：

从昨天开始我就一直遇到这个问题，从我的服务器获取我的数据。我在 POSTMAN 中测试我的 php 代码并将格式更改为 JSON，我什至在 JSONLint 上测试它并返回有效的 json。我注意到它返回 JSONObject 而不是 JSONArray （据我了解 JSONArray 以 [ 开头）。

这是我的 java 代码：orgFragment.java

public void parseJSONData(){
    RequestQueue requestQueue = Volley.newRequestQueue(getContext());
    JsonArrayRequest jsonArray = new JsonArrayRequest(ServerScripts.PHP_SCRIPT_PATH + ServerScripts.PHP_GET_FEEDS, new Response.Listener<JSONArray>() {
        @Override
        public void onResponse(JSONArray response) {
            if (response.length() > 0){
                try {
                    organizationDataList.clear();
                    for (int i = 0; i < response.length(); i++ ){
                        JSONObject jsonObject = response.getJSONObject(i);
                        OrganizationData organizationData = new OrganizationData();

                        //SQL TABLE NAME
                        if (!jsonObject.isNull("organizationName")){
                            organizationData.orgName = jsonObject.getString("organizationName");
                        }

                        if (!jsonObject.isNull("organizationType")){
                            organizationData.orgType = jsonObject.getString("organizationType");
                        }

                        if (!jsonObject.isNull("organizationDescription")){
                            organizationData.orgDesc = jsonObject.getString("organizationDescription");
                        }

                        if (!jsonObject.isNull("organizationCategory")){
                            organizationData.orgCategory = jsonObject.getString("organizationCategory");
                        }

                        if (!jsonObject.isNull("organizationCurrentMembers")){
                            organizationData.orgCurrentMembers = jsonObject.getInt("organizationCurrentMembers");
                        }

                        if (!jsonObject.isNull("organizationMaxMembersNo")){
                            organizationData.orgMaxMembers = jsonObject.getInt("organizationMaxMembersNo");
                        }
                        //ADD
                        organizationDataList.add(organizationData);
                    }
                        //NOTIFY
                        orgListAdapter.notifyDataSetChanged();

                }catch (Exception e){
                    e.printStackTrace();
                    Toast.makeText(getContext(),e.toString(),Toast.LENGTH_LONG).show();
                    Log.e("JSON Parser", "Error parsing data " + e.toString());
                }
            }
        }
    }, new Response.ErrorListener() {
        @Override
        public void onErrorResponse(VolleyError error) {
            String getErrorMsg = error.toString();
            Toast.makeText(getContext(),"Failed to Fetch Data" + getErrorMsg,Toast.LENGTH_LONG).show();
        }

    });
    requestQueue.add(jsonArray);

}

错误：我刚刚从 Logcat 截取了 json 错误响应 org.json.JSONObject 类型的“成功”} 无法转换为 JSONArray

【问题讨论】：

标签： java php json jsonobject jsonresult

【解决方案1】：

我只是修改我的 php 代码，返回一个 JSONArray。

//if($_SERVER['REQUEST_METHOD'] == ''){


     require('dbConnect.php');

     $SQLi_ORG_FEEDS = "SELECT organizationName,organizationDescription,organizationCategory,
                        organizationCurrentMembers,organizationMaxMembersNo,organizationType FROM org_information";

     $query = mysqli_query($dbConnect,$SQLi_ORG_FEEDS) or die("Error".mysqli_error($dbConnect));

     $checkRow = mysqli_num_rows($query);

     $response = array();

     if($checkRow > 0){
         while ($getRecord = mysqli_fetch_array($query)) {
            $response[] = $getRecord;

        /*array_push($response,array(
            "organizationName"=>$getRecord['organizationName'],
            "organizationDescription"=>$getRecord['organizationDescription'],
            "organizationCategory"=>$getRecord['organizationCategory'],
            "organizationCurrentMembers"=>$getRecord['organizationCurrentMembers'],
            "organizationMaxMembersNo"=>$getRecord['organizationMaxMembersNo'],
            "organizationType"=>$getRecord['organizationType']));*/
         }

            //$response['success'] = 'success';
            echo json_encode($response);
            //echo json_encode(array("response"=>$response));
     }
     else {
            $response['failed'] = 'failed';
            echo json_encode($response);
     }

【讨论】：