php - 将多个查询数据放入单个 HTML 表(PHP、MySQL)

Question

这是我当前的代码。我正在尝试将来自三个单独查询的数据显示到一个包含多个列的表中。我的 while 声明在这里错了吗？它正在打印 1 个表数据，然后是后面的数据，而不是在同一行中的旁边。

echo "<table border='1'>
<tr>
<TH COLSPAN=2>July 2010</TH>
<TH COLSPAN=2>August 2010</TH>
<TH COLSPAN=2>September 2010</TH>
</tr>
<tr>
<th>User</th>
<th>Posts</th>
<th>User</th>
<th>Posts</th>
<th>User</th>
<th>Posts</th>
</tr>";

while (($row = mysql_fetch_assoc($july)) || ($row2 = mysql_fetch_assoc($aug)) || ($row3 = mysql_fetch_assoc($sept))) {
echo "<tr>";
echo "<td>" . $row['cUsername'] . "</td>";
echo "<td>" . $row['postCount'] . "</td>";
echo "<td>" . $row2['cUsername'] . "</td>";
echo "<td>" . $row2['postCount'] . "</td>";
echo "<td>" . $row3['cUsername'] . "</td>";
echo "<td>" . $row3['postCount'] . "</td>";
echo "</tr>";
}

echo "</table>";

Answer 1

$data = array();

while($row = mysql_fetch_assoc($july)) {$data['row'][] = $row;}
while($row = mysql_fetch_assoc($aug))  {$data['row2'][] = $row;}
while($row = mysql_fetch_assoc($sept)) {$data['row3'][] = $row;}

$count = count($data['row']);

for($i=0;$i<=$count;$i++)
{
    echo '<tr>';
        if(($i % 3) == 1)
        {
            echo "<td>" . $data['row3'][$i]['cUsername'] . "</td>";
            echo "<td>" . $data['row3'][$i]['postCount'] . "</td>";
        }else if(($i % 2) == 1)
        {
            echo "<td>" . $data['row2'][$i]['cUsername'] . "</td>";
            echo "<td>" . $data['row2'][$i]['postCount'] . "</td>";
        }else /*Never try find remainder of 1 as theres always a multiple of 1*/
        {
            echo "<td>" . $data['row'][$i]['cUsername'] . "</td>";
            echo "<td>" . $data['row'][$i]['postCount'] . "</td>";
        }
    echo '</tr>';
}

通过将结果单独提取到本地数组中，而不是尝试同时提取 3 个不同的行，您应该单独执行它们并将它们存储在局部变量中，如果它是一个大数组，只需在 words 之后取消设置变量。

我的代码未经测试。