[英]Trying to display User info from a database in my Android application
我已經為此工作了一段時間。 我嘗試的第一件事是將登錄的用戶存儲在會話中,然后稍后嘗試使用該變量,例如:
Login.php
<?php
session_start();
if($_SERVER['REQUEST_METHOD'] == 'POST'){
$sessionid = session_id();
require_once('connect.inc.php');
$sql = "SELECT username, password FROM USER WHERE username = ?";
$stmt = $conn->prepare($sql);
$username = $_POST["username"];
$password = $_POST["password"];
$stmt->bind_param("s", $username);
$stmt->execute();
$stmt->bind_result($user, $pass);
while($stmt->fetch()){
$verify = password_verify($password, $pass);
}
if($verify){
$_SESSION["username"] = $username;
echo 'connected';
echo $sessionid;
}else{
echo 'check details';
}
mysqli_close($conn);
}
?>
然后,我獲取登錄消息的響應並將其分為兩個變量。 登錄響應和會話ID。 我獲取會話ID並存儲在共享首選項中。 我試圖將會話ID存儲在我的java方法中,以便可以訪問會話用戶。 這是我的Java代碼,用於嘗試獲取用戶:
GetUserData Java方法
private void getUserData() {
SharedPreferences sharedPreferences = getSharedPreferences(Config.sharedPref, Context.MODE_PRIVATE);
String sessionId = sharedPreferences.getString(Config.SID, "SessionID");
StringRequest stringRequest = new StringRequest(Request.Method.GET, Config.SERVER_ADDRESS + "GetUserData.php?PHPSESSID=" + sessionId,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
JSONObject jsonObject = null;
try {
//json string to jsonobject
jsonObject = new JSONObject(response);
//get json sstring created in php and store to JSON Array
result = jsonObject.getJSONArray(Config.json_array);
//get username from json array
getUserInfo(result);
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
});
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
private void getUserInfo(JSONArray jsonArray){
for(int i = 0; i < jsonArray.length(); i++) {
try {
JSONObject json = jsonArray.getJSONObject(i);
userInfo.add(json.getString(Config.getUsername));
} catch (JSONException e) {
}
}
}
這是java方法嘗試調用的php文件:
GetUserData.php
<?php
session_start();
if($_SERVER['REQUEST_METHOD'] == 'GET'){
$username = $_SESSION['username'];
$sql = "SELECT * FROM USER WHERE username = '$username'";
require_once('connect.inc.php');
$run = mysqli_query($conn, $sql);
$result = array();
while($row = mysqli_fetch_array($run)){
array_push($result, array(
'id' => $row['id'],
'fname' => $row['fname'],
'lname' => $row['lname'],
'username' => $row['username'],
'email' => $row['email'],
));
}
echo json_encode(array('result'=>$result));
mysqli_close($conn);
}
?>
調試時,“結果”數組為空,因此由於某種原因,
$sql = "SELECT * FROM USER WHERE username = '$username'";
不管用。 我知道這與會話有關,但是我不確定問題出在哪里。
我的下一個嘗試將嘗試僅將登錄的用戶存儲在共享首選項中,然后從php文件中調用該變量並運行查詢以顯示帶有該變量的用戶信息。 我該怎么做?
謝謝。
您將要做的是,當用戶在此輸入用戶名和密碼時,向服務器發送請求。 注意我修改了您的代碼。
輸入的字符串用戶名=“用戶名”; 輸入的字符串密碼=“ xxxxxx”;
String uri = String.format("http://somesite.com/some_endpoint.php?param1=%1$s¶m2=%2$s", enteredUsername, enteredPassword);
StringRequest stringRequest = new StringRequest(Request.Method.GET, uri,
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
JSONObject jsonObject = null;
try {
// parse the response object and store user id and data in sharedpreference.
} catch (JSONException e) {
e.printStackTrace();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
});
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
檢查用戶尚未注冊,將用戶添加到數據庫中,並返回用戶數據庫的唯一ID以及所需的其他數據。
然后將用戶ID和用戶名保存到SharedPreference
SharedPreferences sharedPreferences = getSharedPreferences(Config.sharedPref, Context.MODE_PRIVATE);
sharedPreferences.Editor edit = prefs.edit();
edit.putStringSet("Personal Information", set);
edit.commit();
您應該首先檢查是否優先存儲用戶ID,以確定該用戶是否是注冊用戶。 如果用戶未注冊,則顯示登錄表單,否則將用戶重定向到配置文件活動頁面。
我已經有一個登錄類,可以確保用戶已注冊。 正如我所說的,我只是想從數據庫中獲取用戶信息。 這是我可能應該提供的登錄方法:
private void login(){
final String username = txtUsrnm.getText().toString().trim();
final String password = txtPswrd.getText().toString().trim();
//create string request
StringRequest stringRequest = new StringRequest(Request.Method.POST, Config.SERVER_ADDRESS + "Login.php",
new Response.Listener<String>() {
@Override
public void onResponse(String response) {
String responseOne = response.substring(0,9);
String responseTwo = response.substring(9);
if(responseOne.equalsIgnoreCase(Config.logInMessage)){
//create shared pref
SharedPreferences sharedPreferences = Login.this.getSharedPreferences(Config.sharedPref, Context.MODE_PRIVATE);
//editor stores values to the shared pref
SharedPreferences.Editor editor = sharedPreferences.edit();
//add values
editor.putBoolean(Config.sharedPrefBool, true);
editor.putString(Config.username, username);
editor.putString(Config.password, password);
editor.putString(Config.SID, responseTwo);
editor.commit();
Intent intent = new Intent(Login.this, Home.class);
startActivity(intent);
}else{
//display error message
Toast.makeText(Login.this, "Wrong Username or Password", Toast.LENGTH_LONG).show();
}
}
},
new Response.ErrorListener() {
@Override
public void onErrorResponse(VolleyError error) {
}
}){
@Override
protected Map<String, String> getParams() throws AuthFailureError {
//from android.com: A Map is a data structure consisting of a set of keys and
// values in which each key is mapped to a single value. The class of the objects
// used as keys is declared when the Map is declared, as is the class of the
// corresponding values.
Map<String,String> hashMap = new HashMap<>();
//maps specified string key, username and password, to specified string value
hashMap.put(Config.username, username);
hashMap.put(Config.password, password);
return hashMap;
}
};
//add string request to queue
RequestQueue requestQueue = Volley.newRequestQueue(this);
requestQueue.add(stringRequest);
}
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