2015-08-23 62 views
-4

當我嘗試登錄時顯示不正確的用戶詳細信息。儘管數據正在被插入但沒有被抓取。 以下是我的兩個PHP文件。無法獲取名稱,用戶名和密碼

Register.php

<?php 

$con=mysqli_connect("mysql.hostinger.in","name","password","u721174658_swap"); 


$name = $_POST["name"]; 

$username = $_POST["username"]; 
$password = $_POST["password"]; 

$statement = mysqli_prepare($con, "INSERT INTO `user` (name,username, password) VALUES (?, ?, ?)"); 
mysqli_stmt_bind_param($statement, "sss", $name,$username, $password); 
mysqli_stmt_execute($statement); 



mysqli_stmt_close($statement); 

mysqli_close($con); 
?> 

FetchUserData.php

<?php 
$con=mysqli_connect("mysql.hostinger.in","name","password","u721174658_swap"); 



$username = $_POST["username"]; 
$password = $_POST["password"]; 

$statement = mysqli_prepare($con, "SELECT * FROM `user` WHERE username = ? AND password = ?"); 
mysqli_stmt_bind_param($statement, "ss", $username, $password); 
mysqli_stmt_execute($statement); 

mysqli_stmt_store_result($statement); 
mysqli_stmt_bind_result($statement, $userID, $name,$username, $password); 

$user = array(); 

while(mysqli_stmt_fetch($statement)){ 
    $user["name"] = $name; 

    $user["username"] = $username; 
    $user["password"] = $password; 
} 

echo json_encode($user); 
mysqli_close($con); 
?> 
+2

**不要將密碼存放在純文本** 。 – SLaks

+1

另外 - 你不應該在你的while循環中分配'$ user'嗎? 'while($ user = mysqli_stmt_fetch($ statement)){' – andrewsi

+0

@ andrewsi-試過這樣做..仍然問題依然存在 –

回答

0

使用,以獲取以下方式和會話開始

while($row= mysql_fetch_array($statement){ 
    $name = $row['name']; 
} 
+0

用戶正在使用'mysqli_' - 'mysql_fetch_array'不起作用。 – andrewsi