如果用戶的登錄用戶名與Drupal的另一個表中的用戶名相匹配，我該如何獲取用戶詳細信息？

Question

我已經用下面的代碼attemped它：

<?php 
session_start(); 

// connects to the database 
$mysqli = new mysqli("localhost", "root", ""); 

$query = "SELECT First_Name, Last_Name, Balance FROM customers WHERE username = '".$_SESSION['name']."'"; 
if($result = $mysqli->query($query)) 
{ 
while($row = $result->fetch_assoc()) 
{ 
echo "<div align=\"center\">"; 
echo "<br />Your <b><i>Profile</i></b> is as follows:<br />"; 
echo "<b>First name:</b> ". $row['First_Name']; 
echo "<br /><b>Last name:</b> ".$row['Last_Name']; 
echo "<br /><b>Balance:</b> ".$row['balance']; 
echo "</div>"; 
} 
$result->free(); 
} 
else 
{ 
echo "No results found"; 
} 

我得到以下錯誤：

Notice: A session had already been started - ignoring session_start() in include_once() (line 2 of  C:\wamp\www\drupalNew\sites\all\modules\New_Balance\balance_module.module). 
Notice: Undefined index: name in include_once() (line 7 of C:\wamp\www\drupalNew\sites\all\modules\New_Balance\balance_module.module). 

我想這給關於誰是Drupal的7個網站登錄的用戶的詳細信息。

Answer 1

只需使用：

function your_function() { 
    global $user; 

    $username = $user->name; // Instead of $_SESSION['name'] 

    // ... your query 
}