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我有一個問題... 我試圖將設備的udid存儲到我的數據庫當用戶註冊... 我得到這個錯誤logcat：JSON + PHP + Java的 - Android登錄到mysql
10-18 17:17:08.141: E/JSON(1712): <br />
10-18 17:17:08.141: E/JSON(1712): <b>Warning</b>: Missing argument 4 for DB_Functions::storeUser(), called in /home/matbest1/public_html/android_login_api/index.php on line 64 and defined in <b>/home/matbest1/public_html/android_login_api/include/DB_Functions.php</b> on line <b>25</b><br />
10-18 17:17:08.141: E/JSON(1712): {"tag":"register","success":0,"error":1,"error_msg":"Error occured in Registartion"}
10-18 17:17:08.151: E/JSON Parser(1712): Error parsing data org.json.JSONException: Value <br of type java.lang.String cannot be converted to JSONObject
10-18 17:17:08.381: D/dalvikvm(1712): GC_CONCURRENT freed 296K, 5% free 9394K/9799K, paused 2ms+4ms
我試圖存儲在數據庫中的UDID ... 繼承人的registeractivity：
public JSONObject registerUser(String name, String email, String password, String uid){
// Building Parameters
List<NameValuePair> params = new ArrayList<NameValuePair>();
params.add(new BasicNameValuePair("tag", register_tag));
params.add(new BasicNameValuePair("name", name));
params.add(new BasicNameValuePair("email", email));
params.add(new BasicNameValuePair("password", password));
params.add(new BasicNameValuePair("uid", uid));
// getting JSON Object
JSONObject json = jsonParser.getJSONFromUrl(registerURL, params);
// return json
return json;
}
註冊：
UserFunctions userFunction = new UserFunctions();
JSONObject json = userFunction.registerUser(name, email, password, uid);
PHP代碼：
public function storeUser($name, $email, $password, $uid) {
$uuid = $uid;
$hash = $this->hashSSHA($password);
$encrypted_password = $hash["encrypted"]; // encrypted password
$salt = $hash["salt"]; // salt
$result = mysql_query("INSERT INTO users(unique_id, name, email, encrypted_password, salt, created_at) VALUES('$uuid', '$name', '$email', '$encrypted_password', '$salt', NOW())");
// check for successful store
if ($result) {
// get user details
$uid = mysql_insert_id(); // last inserted id
$result = mysql_query("SELECT * FROM users WHERE uid = $uid");
// return user details
return mysql_fetch_array($result);
} else {
return false;
}
}
&
if ($tag == 'register') {
// Request type is Register new user
$name = $_POST['name'];
$email = $_POST['email'];
$password = $_POST['password'];
$uid = $_POST['uid'];
// check if user is already existed
if ($db->isUserExisted($email)) {
// user is already existed - error response
$response["error"] = 2;
$response["error_msg"] = "User already existed";
echo json_encode($response);
} else {
// store user
$user = $db->storeUser($name, $email, $password);
if ($user) {
// user stored successfully
$response["success"] = 1;
$response["uid"] = $user["unique_id"];
$response["user"]["name"] = $user["name"];
$response["user"]["email"] = $user["email"];
$response["user"]["created_at"] = $user["created_at"];
$response["user"]["updated_at"] = $user["updated_at"];
echo json_encode($response);
} else {
// user failed to store
$response["error"] = 1;
$response["error_msg"] = "Error occured in Registartion";
echo json_encode($response);
}
}
} else {
echo "Invalid Request";
}
} else {
echo "Access Denied";
}
有人能幫助我嗎？我對此很陌生...非常感謝。
'java.lang.String類型的值不能轉換爲JSONObject' - 看起來好像傳遞給jsonParser的JSON格式錯誤。 –
所以當我解析udid時有一些錯誤？ – matteo
閱讀錯誤消息。它會拋出字符串''標籤的開頭。 '
'標籤在JSON中甚至不合法，所以有些東西肯定是錯誤的。看看JSON。 –