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我试图开发一个android应用程序，它将连接到外部服务器以获取数据。我已经用我的.php文件成功读取了数据，但是我无法在我的应用程序中使用它，因为它会引发错误。解析DB JSON时出错
<?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));
?>
这就是我用来读取数据的php代码。这是我在Android Studio中
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;
}
使用代码这是JSONParser类：
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;
}
}
这是JSON的结果： 阵列（[0] =>数组（[测试] = > Name1）[1] => Array（[Name] => Name2））
错误信息： E/JSON解析器：解析数据时出错org.json.JSONException：Value java.lang.String类型的数组无法转换为JSONObject
我真的需要帮助。 预先感谢您。
上面的输出不是json。 JSON是'[{“Test”：“Name1”}，{“Name”：“Name2”}]'。试试'json_encode'PHP函数。 – mp911de
那么，我该怎么做呢？ $ var = json_encode（$ result），然后打印$ var？ – jk93
我试图对结果进行编码，但它没有写入“true”。 – jk93