Question

我一直在努力争取这个;我会尽可能简单地在这里解释一下。

考虑这个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       |
+----------+-----------+---------+--------+

session_id和pilot_id都是外键，引用另一个表中的主键。相同的pilot_id可以与不同的session_id相关联，但每个pilot_id - session_id组合都是唯一的。

我需要创建一个HTML表格（在PHP中），它将显示如下数据：

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

因此，行为pilot_id，列为session_id。当pilot_id - session_id组合的present值为1时，应检查相应的单元格。（即，当行组合为零或MySQL表中不存在组合时，HTML表中不应出现任何内容）

呼。

有什么想法吗？

谢谢！

我已经尝试过erisco提出的答案，但我很困惑。（评论字段对于我的解释来说太小了，因此我的问题更新了。）

这是我正在使用的实际数据：

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

我使用$rows = mysqli_fetch_array($result);。我已经确认查询正在返回正确的数据。

但是，当我使用ericso提出的答案时，我看起来似乎是任意数据。这是生成的HTML表：

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

此外，'X'位置与present值无关。

为什么会发生这种情况？

谢谢！

Answer 1

幸运的是，您只需要一个查询。假设$ rows是从数据库中提取的数据格式：

<?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

你可以使用这样的算法：

$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>";

来自多个MySQL表的单个HTML表

2 个答案: