[英]Android Studio parsing JSON
我正在嘗試開發一個Android應用程序,該應用程序將連接到外部服務器以獲取數據。 我已經成功讀取了.php文件中的數據,但是由於拋出錯誤,因此無法在我的應用中使用它。
<?php
/**
* A class file to connect to database
*/
// connect to database
try {
$pdo = new PDO('mysql:host=localhost;dbname=myDBname',' user', 'pass');
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$pdo->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
} catch(PDOException $err) {
die($err->getMessage());
}
$stmt = $pdo->prepare("select * from Test");
$result = $stmt->execute();
print_r($stmt->fetchAll(PDO::FETCH_ASSOC));
?>
那就是我用來讀取數據的php代碼。 這是我在Android Studio中使用的代碼
protected String doInBackground(String... args) {
// Building Parameters
List params = new ArrayList();
// getting JSON string from URL
JSONObject json = jParser.makeHttpRequest(server_php_url, "GET", params);
// Check your log cat for JSON reponse
Log.d("All Products: ", json.toString());
try {
// Checking for SUCCESS TAG
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
// products found
// Getting Array of Products
products = json.getJSONArray(TAG_PRODUCTS);
// looping through All Products
//Log.i("ramiro", "produtos.length" + products.length());
for (int i = 0; i < products.length(); i++) {
JSONObject c = products.getJSONObject(i);
// Storing each json item in variable
String name = c.getString(TAG_NAME);
// creating new HashMap
HashMap map = new HashMap();
// adding each child node to HashMap key => value
map.put(TAG_NAME, name);
empresaList.add(map);
}
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
這是JSONParser類:
public class JSONParser {
static InputStream is = null;
static JSONObject jObj = null;
static String json = "";
// constructor
public JSONParser() {
}
// function get json from url
// by making HTTP POST or GET mehtod
public JSONObject makeHttpRequest(String url, String method,
List params) {
// Making HTTP request
try {
// check for request method
if(method == "POST"){
// request method is POST
// 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();
}else if(method == "GET"){
// request method is GET
DefaultHttpClient httpClient = new DefaultHttpClient();
String paramString = URLEncodedUtils.format(params, "utf-8");
url += "?" + paramString;
HttpGet httpGet = new HttpGet(url);
HttpResponse httpResponse = httpClient.execute(httpGet);
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();
} 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);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
// return JSON String
return jObj;
}
}
這是JSON的結果:
Array ( [0] => Array ( [Test] => Name1 ) [1] => Array ( [Name] => Name2 ) )
錯誤是:
E/JSON Parser﹕ Error parsing data org.json.JSONException: Value Array of type java.lang.String cannot be converted to JSONObject
我真的需要幫助 先感謝您。
在返回變量中使用json_decode(),您將獲得json格式的輸出,如下所示:
return json_decode(jObj);
在后台方法中重做json對象
protected String doInBackground(String... args) {
// Building Parameters
List params = new ArrayList();
// getting JSON string from URL
JSONObject json = jParser.makeHttpRequest(server_php_url, "GET", params);
// Check your log cat for JSON reponse
Log.d("All Products: ", json.toString());
try {
// Checking for SUCCESS TAG
int success = json.getInt(TAG_SUCCESS);
if (success == 1) {
// products found
// Getting Array of Products
products = json.getJSONArray(TAG_PRODUCTS);
// looping through All Products
//Log.i("ramiro", "produtos.length" + products.length());
for (int i = 0; i < products.length(); i++) {
JSONObject c = products.getJSONObject(i);
// Storing each json item in variable
String name = c.getString(TAG_NAME);
// creating new HashMap
HashMap map = new HashMap();
// adding each child node to HashMap key => value
map.put(TAG_NAME, name);
empresaList.add(map);
}
}
} catch (JSONException e) {
e.printStackTrace();
}
return json;
據我所知,該錯誤是在JSONParser類中最具體地拋出:
try {
jObj = new JSONObject(json);
} catch (JSONException e) {
Log.e("JSON Parser", "Error parsing data " + e.toString());
}
我終於將輸出編碼為[{“ Test”:“ Name1”}],現在我的錯誤已更改為解析數據org.json.JSONException的錯誤:org類型的值[{“ Test”:“ Name1”}]]。 json.JSONArray無法轉換為JSONObject
如果您可以使用mysql發布相同的.php代碼,我將不勝感激,因為我嘗試編寫它,但是它說“ mysql已過時”,而且我不知道如何使用mysqli。
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