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我正在检索 JSON 数据数组并尝试显示和访问它。但是当我尝试显示它时，它只显示 JSON 数据数组中的最后一行。当我尝试执行类似 json1.get(2) 的操作时（这里的 json1 是包含 JSONObject 的 JSONArray）我得到数组越界错误。请告诉我我做错了什么以及如何检索 JSON 下面是这个问题的相关代码。

谢谢你。

我从 Freebies.java 类调用 UserFunctions.java 类的 getAllFreebies 函数

UserFunctions uf = new UserFunctions();
JSONObject json = uf.getAllFreebies();

下面是来自 UserFunctions.java 类的 getAllFreebies() 函数的代码：

public JSONObject getAllFreebies(){
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", getAllFreebies_tag));
    JSONObject json  = jsonParser.getJSONFromUrl(getAllFreebiesURL,params);
    JSONArray json1 = new JSONArray();
    json1.put(json);
    try {
        System.out.println(json1.get(2));
    } catch (JSONException e) {
        // TODO Auto-generated catch block
        e.printStackTrace();
    }
    return json;
}

下面是 index.php 类的代码，其中我调用 DB_Functions.php 类的 getFreebies() 函数

else if($tag = 'getAllFreebies'){
                $username = "";
                $catagory = "";
                $subcatagory = "";
                $title = "";
                $condition = "";
                $description = "";
                $address = "";
                $city = "";
                $state = "";
                $country = "";
                $zipcode = "";
                $posted_on= "";
                    $getAllFreebies = $db->getFreebies($username,$catagory,$subcatagory,$title,$condition,$description,$address,$city,$state,$country,$zipcode,$posted_on);
            $data = array();
            while($row = mysql_fetch_assoc($getAllFreebies)) {
            $response["success"] = 1;
            $data[] = array(
            $response["getAllFreebies"]["username"] = $row["username"],
            $response["getAllFreebies"]["catagory"] = $row["catagory"],
            $response["getAllFreebies"]["subcatagory"] = $row["subcatagory"],
            $response["getAllFreebies"]["title"] = $row["title"],
            $response["getAllFreebies"]["item_condition"] = $row["item_condition"],
            $response["getAllFreebies"]["description"] = $row["description"],
            $response["getAllFreebies"]["address"] = $row["address"],
            $response["getAllFreebies"]["city"] =  $row["city"],
            $response["getAllFreebies"]["state"] = $row["state"],
            $response["getAllFreebies"]["country"] = $row["country"],
            $response["getAllFreebies"]["zipcode"] = $row["zipcode"],
            $response["getAllFreebies"]["posted_on"] = $row["posted_on"]);}
            echo json_encode($response);
                    }// end of getAllFreebies tag

下面是 DB_Functions.php 的 getFreebies() 函数的代码

public function getFreebies(){
$result = mysql_query("SELECT * FROM freebie") or die(mysql_error());
return($result);
}

更新：我在我的 JSONParser 类中添加了以下内容以解析 JSONArray public JSONArray getJSONArrayFromUrl(String url, List params) {

    // Making HTTP request
    try {
        // defaultHttpClient
        DefaultHttpClient httpClient = new DefaultHttpClient();
        HttpPost httpPost = new HttpPost(url);
        httpPost.setEntity(new UrlEncodedFormEntity(params));
        HttpResponse httpResponse = httpClient.execute(httpPost);
        HttpEntity httpEntity = httpResponse.getEntity();
        is = httpEntity.getContent();

    } catch (UnsupportedEncodingException e) {
        e.printStackTrace();
    } catch (ClientProtocolException e) {
        e.printStackTrace();
    } catch (IOException e) {
        e.printStackTrace();
    }

    try {
        BufferedReader reader = new BufferedReader(new InputStreamReader(
                is, "iso-8859-1"), 8);
        StringBuilder sb = new StringBuilder();
        String line = null;
        while ((line = reader.readLine()) != null) {
            sb.append(line + "n");
        }
        is.close();
        json = sb.toString();

        Log.e("JSON", json);
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }

    // try parse the string to a JSON object
    try {
        jObj = new JSONObject(json);
        jarray.put(jObj);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    }

    // return JSON String
    return jarray;

}

我更改了 getAllFreebie() 以检索 JSONArray

public JSONArray getAllFreebies() {
    List<NameValuePair> params = new ArrayList<NameValuePair>();
    params.add(new BasicNameValuePair("tag", getAllFreebies_tag));
    JSONArray json = jsonParser.getJSONArrayFromUrl(getAllFreebiesURL,
            params);
    return json;
}

我仍然只得到单行而不是所有行。谁能告诉我我做错了什么。谢谢:(

4

2 回答 2

1

检查以下代码：

JSONObject json  = jsonParser.getJSONFromUrl(getAllFreebiesURL,params);
JSONArray json1 = new JSONArray();
json1.put(json);
try {
    System.out.println(json1.get(2));
} catch (JSONException e) {
    // TODO Auto-generated catch block
    e.printStackTrace();
}

您创建一个新的empty JSONArray，向其中添加一个元素，然后尝试访问它的第二个元素。

getJSONFromUrl应该解析一个对象还是一个数组？我相信您需要它来返回 a JSONArray。

于 2012-05-29T22:48:37.697 回答
1

将您的代码更新为：

// Initialize the array with your JSON string.
JSONArray json1 = new JSONArray(json);

// This line adds your String as the first element to
// the JSON array and is incorrect. You can safely remove it.
//json1.put(json);

编辑：试试你发布的代码的这个更新版本：

// Making HTTP request
try {
    // defaultHttpClient
    DefaultHttpClient httpClient = new DefaultHttpClient();
    HttpPost httpPost = new HttpPost(url);
    httpPost.setEntity(new UrlEncodedFormEntity(params));
    HttpResponse httpResponse = httpClient.execute(httpPost);

    try {
        // Get our response as a String.
        String jsonString = EntityUtils.toString(httpResponse.getEntity());
        Log.d(TAG, "JSON: "+jsonString);

        // Parse the JSON String into a JSONArray object.
        return new JSONArray(jsonString);
    } catch (JSONException e) {
        Log.e("JSON Parser", "Error parsing data " + e.toString());
    } catch (Exception e) {
        Log.e("Buffer Error", "Error converting result " + e.toString());
    }
} catch (UnsupportedEncodingException e) {
    e.printStackTrace();
} catch (ClientProtocolException e) {
    e.printStackTrace();
} catch (IOException e) {
    e.printStackTrace();
}
return null;
于 2012-05-29T22:51:30.867 回答