無法檢索每個記錄額外查詢的值？

Question

我正在從數據庫中挑選用戶數據的表上工作。

問題是，即時通訊不能在我的條件，因爲我從不同的表和行獲取數據兩個語句。

@CODE

<?php 
$con = mysql_connect("localhost", "username", "password") or die(mysql_error()); 
mysql_select_db("database_name") or die(mysql_error()); 

$result = mysql_query("select * FROM users"); 

$space_used = mysql_query("SELECT SUM(fileSize) FROM file WHERE file.userId=users.id AND file.statusId=1"); 

$total_files = mysql_query("SELECT COUNT(id) FROM file WHERE file.userId=users.id AND file.statusId=1"); 

echo "<table class='table table-striped table-hover table-bordered' id='sample_editable_1'> 
<tr> 
    <th>Username</th> 
    <th>Type</th>         
    <th>Email</th> 
    <th>Last Login</th> 
    <th>Last IP</th> 
    <th>space_used</th> 
    <th>total_files</th> 
    <th>status</th> 
</tr>"; 

while($row = mysql_fetch_array($result)) 
{ 
    echo "<tr>"; 
     echo "<td>" . $row['username'] . "</td>"; 
     echo "<td>" . $row['level'] . "</td>"; 
     echo "<td>" . $row['email'] . "</td>"; 
     echo "<td>" . $row['lastlogindate'] . "</td>"; 
     echo "<td>" . $row['lastloginip'] . "</td>"; 
     echo "<td>" . $space_used['space_used'] . "</td>"; 
     echo "<td>" . $total_files['total_files'] . "</td>"; 
     echo "<td>" . $row['status'] . "</td>"; 
    echo "</tr>"; 
} 

echo "</table>"; 
mysql_close($con); 
?> 

什麼即時得到的用戶名，級別，電子郵件，lastlogindate的所有值，lastloginip &狀態。在我的桌子上沒有顯示的兩個是我想要實現的第二個條件：space_used和total_files。

我在做什麼錯了？我不是一個PHP的專家，事實上，我有一個混合的教程和腳本即時鏈接來獲得這個結果有點複雜。

謝謝

Answer 1

您需要重新修改您的查詢，您可以使用JOIN恢復您3個查詢到一個查詢：

SELECT u.*, 
     SUM(f.fileSize) AS space_used, 
     COUNT(f.id) AS total_files 
    FROM users u 
    JOIN file f 
     ON f.userId = u.id AND f.statusId = 1 
GROUP BY u.id 

然後，你可以這樣閱讀：

<?php 
// your database info 
$db_host = 'your MySQL server host'; 
$db_user = 'your username'; 
$db_pass = 'your password'; 
$db_name = 'your database name'; 
$con = new mysqli($db_host, $db_user, $db_pass, $db_name); 
if($con->connect_error) 
{ 
     die('Connect Error (' . mysqli_connect_errno() . ') '. mysqli_connect_error()); 
} 

if (!$result = $con->query("SELECT u.*, 
            SUM(f.fileSize) AS space_used, 
            COUNT(f.id) AS total_files 
           FROM users u 
           JOIN file f 
           ON f.userId = u.id AND f.statusId = 1 
          GROUP BY u.id")) 
{ 
    die('Select query error: ' . $con->error); 
} 
?> 
<table class='table table-striped table-hover table-bordered' id='sample_editable_1'> 
    <tr> 
     <th>Username</th> 
     <th>Type</th>         
     <th>Email</th> 
     <th>Last Login</th> 
     <th>Last IP</th> 
     <th>space_used</th> 
     <th>total_files</th> 
     <th>status</th> 
    </tr> 
<?php 
while ($row = $result->fetch_array()) 
{ 
?> 
    <tr> 
     <td><?php echo $row['username']; ?></td> 
     <td><?php echo $row['level']; ?></td> 
     <td><?php echo $row['email']; ?></td> 
     <td><?php echo $row['lastlogindate']; ?></td> 
     <td><?php echo $row['lastloginip']; ?></td> 
     <td><?php echo $row['space_used']; ?></td> 
     <td><?php echo $row['total_files']; ?></td> 
     <td><?php echo $row['status']; ?></td> 
    </tr> 
<?php 
} 
$con->close(); 
?> 
</table> 

Answer 2

首先查詢數據庫並將結果保存到變量中。

$result = mysql_query("select * FROM users"); 
$space_used = mysql_query("SELECT SUM(fileSize) FROM file WHERE file.userId=users.id AND file.statusId=1"); 
$total_files = mysql_query("SELECT COUNT(id) FROM file WHERE file.userId=users.id AND file.statusId=1"); 

然後你通過你的第一個結果的迭代與

while($row = mysql_fetch_array($result)) 

通知您的來電mysql_fetch_array。另外兩個結果你永遠不會撥打mysql_fetch_array。

查看PHP documentation on mysql_query瞭解更多詳情。具體而言，返回類型是資源。

返回的結果資源應該傳遞給mysql_fetch_array（）和其他用於處理結果表的函數來訪問返回的數據。

然而，

SELECT SUM(fileSize) FROM file WHERE file.userId=users.id AND file.statusId=1 

甚至不是你想要的。您檢查file.userId=users.id，但只能從file表中讀取。您缺少用戶源表。

Answer 3

您想要聚合每個userId。

拉這兩個查詢進入while循環。使用檢索到的用戶的id進行彙總查詢。

while($row = mysql_fetch_array($result)) 
{ 
    $space_used = mysql_query("SELECT SUM(fileSize) FROM file WHERE file.userId={$row['id']} AND file.statusId=1"); 
    $total_files = mysql_query("SELECT COUNT(id) FROM file WHERE file.userId={$row['id']} AND file.statusId=1"); 
    ... 
} 

結合他們到一個查詢。

$agragateDataResponce = mysql_query("SELECT SUM(fileSize) as size, 
              COUNT(id) as count 
             FROM file 
             WHERE file.userId={$row['id']} AND 
              file.statusId=1"); 
$agragateData = mysql_fetch_array($agragateDataResponce); 

通過$agragateData['size']和$agragateData['count']訪問;

然後查看如何將剩餘的兩個SQL合併爲一個連接:)。SO Question on topic

Answer 4

嘗試使用

$space_used = mysql_query("SELECT SUM(fileSize) 
          FROM file 
          WHERE file.userId=users.id AND 
            file.statusId=1"); 

$total_files = mysql_query("SELECT COUNT(id) 
           FROM file 
          WHERE file.userId=users.id AND 
            file.statusId=1"); 

以上內線While並使用user.id驗證

我是新來的stackoverflow，所以不要我的。我不知道如何讓代碼看起來像每個人一樣