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我正在開發一個連接到遠程MYSQL數據庫並通過JSON數據檢索信息的Android應用程序。我遇到的部分應用涉及在應用中搜索userID，並從數據庫中返回關聯的用戶信息，例如ID，名字，公司和位置。需要幫助從MYSQL數據庫獲取JSON數據到Android應用程序
問題是解析應用程序內的JSON數據。我得到的錯誤消息是System.err: org.json.JSONException: No value for id。
php頁面顯示JSON數據。輸出兩個JSON對象：一個success對象和一個result對象。 success對象被應用程序正確解析並通過if語句告訴應用程序要執行什麼操作。因此，if success == 1，應用程序會執行一個代碼塊，該代碼塊應該分析result對象並將該數組的每個元素分配給應用程序中的String值。從PHP頁面的輸出是：
{"success":1,"message":"UserID found!"}{"result":[{"id":"1100011","firstname":"Kevin","company":"company","position":"bartender"}]}
的問題是從results對象中的值不被由應用解析。這裏是PHP頁面：
<?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
}
?>
如果我註釋掉線上面提到的，從應用程序日誌顯示了正確的數據由應用程序解析爲results陣列：
D/UserID Lookup:: {"result":[{"id":"1100011","firstname":"Kevin","company":"company","position":"bartender"}]}，後來我因爲success未被髮送（顯然），所以得到W/System.err: org.json.JSONException: No value for success的錯誤。
這裏是我的Android代碼：
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();
}
}
}
}
所以基本上我能正確解析無論是success對象或results對象，但不能同時使用。如果我嘗試解析兩者，則會出現no value for id的JSON錯誤。
這不是有效的JSON。您正在輸出兩個JSONObjects。只需將該數組添加到響應中，並確保只調用一次「echo json_encode（）」。結果如下所示：http://pastebin.com/4QxBvyZF作爲一種替代方案，您也可以將它封裝在帶有兩個JSONObjects的JSONArray中，結果如下所示：http://pastebin.com/0JHQYGqh –
**警告**：使用'mysqli'時，您應該使用[參數化查詢]（http://php.net/manual/en/mysqli.quickstart.prepared-statements.php）和['bind_param']（http：/ /php.net/manual/en/mysqli-stmt.bind-param.php）將用戶數據添加到您的查詢。 **不要**使用字符串插值或連接來完成此操作，因爲您創建了嚴重的[SQL注入漏洞]（http://bobby-tables.com/）。 **絕不**將'$ _POST'或'$ _GET'數據直接放入查詢中，如果有人試圖利用您的錯誤，這會非常有害。 – tadman
在代碼中間的'die'調用中阻塞也是不好的形式。如果您有錯誤，請設置響應代碼，渲染一些內容，然後執行常規清理。 – tadman