來自多個MySQL表格的單個HTML表格

Question

我一直在努力爭取一段時間，我會盡可能簡單地在這裏解釋它。

考慮這個MySQL表：

+----------+-----------+---------+--------+ 
|status_id |session_id |pilot_id |present | 
+----------+-----------+---------+--------+ 
|1   |61   |901  |1  | 
|2   |63   |901  |1  | 
|3   |62   |901  |0  | 
|4   |62   |902  |1  | 
|5   |63   |903  |1  | 
+----------+-----------+---------+--------+ 

兩個session_id和pilot_id外鍵使得參照另一個表的主鍵。同一pilot_id可以與不同的session_id相關聯，但每個pilot_id - session_id組合是唯一的。

我需要一個HTML表（PHP）會顯示這樣的數據：

+----------+---------+---------+---------+ 
|   |61  |62  |63  | 
+----------+---------+---------+---------+ 
|901  |X  |   |X  | 
|902  |   |X  |   | 
|903  |   |   |X  | 
+----------+---------+---------+---------+ 

因此，行是pilot_id和列session_id。當pilot_id - session_id組合的值爲1時，應檢查相應的單元格。 （即，當行組合爲零或組合不存在於MySQL表格中時，HTML表格中不應該出現任何內容）

Phew。

任何想法？

謝謝！

---

我試過erisco提出的答案，但我很困惑。 （評論字段太小，我的解釋，因此這個更新我的問題）。

這是我與工作實際數據：

+----------+-----------+---------+--------+ 
|status_id |session_id |pilot_id |present | 
+----------+-----------+---------+--------+ 
|7   |65   |33  |1  | 
|8   |66   |33  |1  | 
|9   |65   |17  |0  | 
|10  |66   |16  |1  | 
+----------+-----------+---------+--------+ 

我用$rows = mysqli_fetch_array($result);。我已經確認查詢正在返回正確的數據。

但是，當我使用ericso提出的答案時，我看起來似乎是任意的數據。下面是生成的HTML表格：

+----------+---------+---------+---------+---------+ 
|   |1  |3  |6  |7  | 
+----------+---------+---------+---------+---------+ 
|1   |X  |   |   |   | 
|3   |   |   |   |   | 
|6   |   |   |   |   | 
|7   |   |   |   |   | 
+----------+---------+---------+---------+---------+ 

此外，「X」的位置保持他毫無來由present值相同。

任何想法爲什麼會發生這種情況？

謝謝！

Answer 1

幸運的是，你只需要一個查詢。假定$行是您從數據庫中提取的數據的格式：

<?php 

$rows = array(
    array(
    'status_id' => 1, 
    'session_id' => 61, 
    'pilot_id' => 901, 
    'present' => 1, 
), 
    array(
    'status_id' => 2, 
    'session_id' => 63, 
    'pilot_id' => 901, 
    'present' => 1, 
), 
    array(
    'status_id' => 3, 
    'session_id' => 62, 
    'pilot_id' => 901, 
    'present' => 0, 
), 
    array(
    'status_id' => 4, 
    'session_id' => 62, 
    'pilot_id' => 902, 
    'present' => 1, 
), 
    array(
    'status_id' => 5, 
    'session_id' => 63, 
    'pilot_id' => 903, 
    'present' => 1, 
) 
); 

$session_ids = array(); 
$pilot_ids = array(); 
$crosses = array(); 

foreach ($rows as $row) { 
    $session_ids[$row['session_id']] = $row['session_id']; 
    $pilot_ids[$row['pilot_id']] = $row['pilot_id']; 
    if ($row['present'] == 1) { 
    $cross_index = $row['session_id'].'.'.$row['pilot_id']; 
    $crosses[$cross_index] = $cross_index; 
    } 
} 

sort($session_ids); 
sort($pilot_ids); 

?> 

<table> 
    <tr> 
    <th></th> 
    <?php foreach ($session_ids as $sess_id): ?> 
    <th><?php echo $sess_id; ?></th> 
    <?php endforeach; ?> 
    </tr> 
    <?php foreach ($pilot_ids as $pilot_id): ?> 
    <tr> 
    <th><?php echo $pilot_id; ?></th> 
    <?php foreach ($session_ids as $sess_id): ?> 
    <?php if (isset($crosses[$sess_id.'.'.$pilot_id])): ?> 
    <td>X</td> 
    <?php else: ?> 
    <td></td> 
    <?php endif; ?> 
    <?php endforeach; ?> 
    </tr> 
    <?php endforeach; ?> 
</table> 

Answer 2

可以使用的算法是這樣的：

$sql = "SELECT DISTINCT session_id AS sid FROM pilot_session ORDER BY 1 ASC"; 
$rs = mysql_query($sql, $conn); 
$sessions = array(); 
while(false !== ($r = mysql_fetch_array($rs))){ 
    $sessions[] = $r['sid']; 
} 

$sql = "SELECT DISTINCT pilot_id AS pid FROM pilot_session ORDER BY 1 ASC"; 
$rs = mysql_query($sql, $conn); 
$pilots = array(); 
while(false !== ($r = mysql_fetch_array($rs))){ 
    $pilots[] = $r['pid']; 
} 

$pilot_presence = array(); 
$sql = "SELECT session_id, pilot_id, present FROM pilot_session"; 
$rs = mysql_query($sql, $conn); 
while(false !== ($r = mysql_fetch_array($rs))){ 
    $s_presence[$r['pilot_id']][$r['session_id']] = $r['present']; 
} 

echo "<table><tr><td>&nbsp</td>"; 
foreach($sessions as $s){ 
    echo "<td>$s</td>"; 
} 
echo "</tr>"; 
foreach($pilots as $p){ 
    echo "<tr><td>$p</td>"; 
    foreach($sessions as $s){ 
     $tp = ''; 
     if(isset($s_presence[$p][$s])){ 
      if($s_presence[$p][$s] == '1'){ 
       $tp = 'X'; 
      } 
     } 
     echo "<td>".$tp."</td>"; 
    }; 
    echo "</tr>"; 
} 
echo "</table>";