我正在开发一个连接到远程MYSQL数据库的Android应用程序，并通过JSON数据检索信息。我遇到问题的应用程序部分涉及在应用程序中按userID进行搜索，并从数据库返回相关的用户信息，例如ID，名字，公司和位置。
问题是解析应用程序中的JSON数据。我得到的错误消息是System.err: org.json.JSONException: No value for id。
php页面显示JSON数据。输出两个JSON对象：success对象和result对象。应用程序正确解析success对象，并通过if语句告诉应用程序该做什么。所以if success == 1，应用程序执行一个代码块，该代码块应解析result对象，并将数组的每个元素分配给应用程序中的String值。 php页面的输出是：
{"success":1,"message":"UserID found!"}{"result":[{"id":"1100011","firstname":"Kevin","company":"company","position":"bartender"}]}
问题是应用程序没有解析results对象的值。这是php页面：
<?php
define('HOST','localhost');
define('USER','********');
define('PASS','**********');
define('DB','**********');
if (!empty($_POST)){
if (empty($_POST['userID'])){
$response["success"] = 0;
$response["message"] = "Please enter a User ID";
die(json_encode($response));
}
$userID = mysql_escape_string($_POST['userID']);
$con = mysqli_connect(HOST,USER,PASS,DB);
$sql = "SELECT * FROM users WHERE id = $userID";
$res = mysqli_query($con,$sql);
$result = array();
while($row = mysqli_fetch_array($res)){
array_push($result, array(
'id'=>$row[0],
'firstname'=>$row[4],
'company'=>$row[6],
'position'=>$row[7],
)
);
}
if($result){
$response["success"] = 1;
$response["message"] = "UserID found!";
echo json_encode($response); // if I comment out this line, the result array gets parsed properly by the app.
echo json_encode(array("result"=>$result));
}else{
$response["success"] = 0;
$response["message"] = "UserID not found. Please try again.";
die(json_encode($response));
}
mysqli_close($con);
} else {
?>
<h1>Search by User ID:</h1>
<form action="searchbyuserid.php" method="post">
Enter the UserID of the receipient:<br />
<input type="text" name="userID" placeholder="User ID" />
<br /><br />
<input type="submit" value="Submit" />
</form>
<a href="register.php">Register</a>
<?php
}
?>
如果我注释掉上面提到的那一行，那么应用程序中的日志会显示应用程序为results数组解析的正确数据：
D/UserID Lookup:: {"result":[{"id":"1100011","firstname":"Kevin","company":"company","position":"bartender"}]}，但后来W/System.err: org.json.JSONException: No value for success收到错误，因为success未被发送（显然）。
这是我的安卓代码：
import android.app.ProgressDialog; ...
public class SearchByUserID extends ActionBarActivity implements View.OnClickListener {
// Buttons
private Button mSubmitButton, mBackButton;
// EditText Field
EditText enterUserID;
// Progress Dialog
private ProgressDialog pDialog;
// JSON parser class
JSONParser jsonParser = new JSONParser();
// Variable for holding URL:
private static final String LOGIN_URL = "http://www.***********/webservice/searchbyuserid.php";
//JSON element ids from response of php script:
private static final String TAG_SUCCESS = "success";
private static final String TAG_MESSAGE = "message";
private static final String TAG_USERID = "id";
private static final String TAG_FIRSTNAME = "firstname";
private static final String TAG_COMPANY = "company";
private static final String TAG_POSITION = "POSITION";
@Override
protected void onCreate(Bundle savedInstanceState) {
super.onCreate(savedInstanceState);
this.supportRequestWindowFeature(Window.FEATURE_NO_TITLE);
setContentView(R.layout.search_by_user_id_layout);
mSubmitButton = (Button)findViewById(R.id.submit);
mBackButton = (Button)findViewById(R.id.back);
mSubmitButton.setOnClickListener(this);
mBackButton.setOnClickListener(this);
enterUserID = (EditText)findViewById(R.id.enterUserIdNumber);
}
@Override
public void onClick(View v) {
switch (v.getId()) {
case R.id.submit:
new SearchUserId().execute();
break;
case R.id.back:
finish();
break;
default:
break;
}
}
class SearchUserId extends AsyncTask<String, String, String> {
// Show progress dialog
boolean failure = false;
@Override
protected void onPreExecute() {
super.onPreExecute();
pDialog = new ProgressDialog(SearchByUserID.this);
pDialog.setMessage("Searching User ID...");
pDialog.setIndeterminate(false);
pDialog.setCancelable(true);
pDialog.show();
}
@Override
protected String doInBackground(String... args) {
// Check for success tag
int success;
String userID = enterUserID.getText().toString();
try {
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("userID", userID));
Log.d("UserID:", userID);
Log.d("request!", "starting");
JSONObject json = jsonParser.makeHttpRequest(LOGIN_URL, "POST", params);
// check your log for json response
Log.d("UserID Lookup:", json.toString());
// json success tag
success = json.getInt(TAG_SUCCESS);
if (success == 1) {
JSONObject json2 = jsonParser.makeHttpRequest(LOGIN_URL, "POST", params);
String userid = json2.getString(TAG_USERID);
String firstName = json2.getString(TAG_FIRSTNAME);
String company = json2.getString(TAG_COMPANY);
String position = json2.getString(TAG_POSITION);
Log.d("User ID Found!", json.toString());
Log.d("userid:", userid);
Log.d("firstName:", firstName);
Log.d("company:", company);
Log.d("position:", position);
Intent i = new Intent(SearchByUserID.this, HomeActivity.class);
finish();
startActivity(i);
return json.getString(TAG_MESSAGE);
}else{
Log.d("User ID not found.", json.getString(TAG_MESSAGE));
return json.getString(TAG_MESSAGE);
}
} catch (JSONException e) {
e.printStackTrace();
}
return null;
}
/**
* After completing background task Dismiss the progress dialog
* **/
protected void onPostExecute(String file_url) {
// dismiss the dialog once product deleted
pDialog.dismiss();
if (file_url != null){
Toast.makeText(SearchByUserID.this, file_url, Toast.LENGTH_LONG).show();
}
}
}
}
所以基本上我能够正确解析success对象或results对象，但不能同时解析两者。如果我尝试解析两者，我会收到no value for id的JSON错误。
答案 0 :(得分：0)
JSONObject json2 = jsonParser.makeHttpRequest(LOGIN_URL, "POST", params);
JSONArray jsArray = json2.getJSONArray("result");
String usrid = jsArray.getJSONObject("id");
String firstName = jsArray.getJSONObject("firstname");
String company = jsArray.getJSONObject("company");
String position = jsArray.getJSONObject("position");
尝试将代码转换为此内容。