Question

我希望将数据库（类别列表）中的值显示到包含2列和x行数的表中。

我希望我的网页显示如下：

Apes Cats
Apples Cherries
Bats Tigers
Berries Zebras

而不是

Apes Apples
Bats Bears
Cats Cherries
Tigers Zebras

到目前为止，这是我的代码：

<table border="0" bordercolor="#FFCC00" style="background-color:#FFFFCC" width="400" cellpadding="3" cellspacing="3">
    <?php
        $query = "SELECT * FROM category ORDER BY cat_name";
        $data = mysqli_query($dbc, $query);

        while ($category = mysqli_fetch_array($data)) {
    ?>
    <tr>
        <td><?php echo $category['cat_name'] ?></td>
    </tr>
    <?php } ?>
</table>

Answer 1

这是基本想法：

您可以通过num_rows获取数据的计数除以2。现在，您的除法结果将是行数。

输出行x和x + num行的循环回显值。例如，第1行的输出为：

<tr><td>$row[val1][data]</td><td>$row[val5][data]</td></tr>

所以，你的循环最终会输出：

val 1  |  val 5
val 2  |  val 6
val 3  |  val 7
val 4  |  val 8

当你的递增变量= num_rows时，循环应该结束。从那里应该非常简单。祝你好运。

Answer 2

你可以在循环中添加另一个变量，例如$ i，并按如下方式递增：

<table border="0" bordercolor="#FFCC00" style="background-color:#FFFFCC" width="400" cellpadding="3" cellspacing="3">
        <?php
        $query = "SELECT * FROM category ORDER BY cat_name";
        $data = mysqli_query($dbc, $query);

        $i=1;

        while ($category = $data->fetch_row()) {
        ?>
            <tr>
                <td><?php echo $i; ?></td>   
                <td><?php echo $category[1] ?></td>
            </tr>   
        <?php
        $i++;
        } ?>
    </table>

编辑：更新以获取每个结果的行数据，并假设cat_name是数组中的第二个项目，即$ category [1]。

Answer 3

如果它不一定是一个表，你的浏览器支持CSS 3，那么你可以使用CSS列：

http://jsfiddle.net/QKuDL/

否则你需要先对结果进行排序（未经测试 - PHP不是我的强项）：

<?php
$query = "SELECT * FROM category ORDER BY cat_name";
$data = mysqli_query($dbc, $query);

$col_count = 2;
$max_items_per_col = ceil( mysqli_num_rows ( $data ) / $col_count );
$cols = array(array());

$col = 0;

while ($category = mysqli_fetch_array($data)) {
   if (count($cols[$col]) >= $max_items_per_col) {
     $col++;
   }
   $cols[$col][] = $category['cat_name'];
}
?>
<table> <!-- all of those attributes should be CSS instead -->
 <?php for ($i = 0; $i < $max_items_per_col; $i++) { ?>
    <tr>
       <?php foreach ($cols as $col) { ?>
         <td><?php if(isset($col[$i])) echo $col[$i]; ?></td>
       <?php } ?>       
    </tr>
 <?php } ?>
 </table>

Answer 4

<?php
$query = "SELECT * FROM category ORDER BY cat_name";
$data = mysqli_query($dbc, $query);
$midpoint = ceil($data->num_rows / 2);
$i=0;

while ($category = $data->fetch_array()) {
  if ($i < $midpoint)
    $left[$i] = "<td>" . $category['cat_name'] . "</td>";
  else 
    $right[$i - $midpoint] = "<td>" . $category['cat_name'] . "</td>";
  $i++;
}

print "<table>";
for ($j=0; $j < $i; $j++)
{
    print "<tr>" . $left[$j];
    if (array_key_exists($j, $right)) print $right[$j];
    print "</tr>";     
}
print "</table>";

Answer 5

试试这个（虽然没有测试过）：

<table border="0" bordercolor="#FFCC00" style="background-color:#FFFFCC" width="400" cellpadding="3" cellspacing="3">
    <?php
    $query = "SELECT * FROM category ORDER BY cat_name";
    $data = mysqli_query($dbc, $query);

    # Calculate total rows and half rows, rounded up
    $full_row_count = mysqli_num_rows($data);
    $half_row_count = ceil($full_row_count / 2);

    $i = 0;
    while ($i <= $half_row_count) {
        # Set the result pointer for first column ...
        mysqli_data_seek($data, $i);
        $category_1 = mysqli_fetch_array($data);

        # ... then calculate the offset for the second column ...
        $col_2_offset = $i + $half_row_count;

        # .. and make sure it's not larger than total rows - 1
        if ($col_2_offset <= $full_row_count - 1) {
            mysqli_data_seek($data, $col_2_offset);
            $category_2 = mysqli_fetch_array($data);
        } else {
            $category_2 = array('cat_name' => '');
        }
    ?>
    <tr>    
        <td><?php echo $category_1['cat_name'] ?></td>
        <td><?php echo $category_2['cat_name'] ?></td>
    </tr>
    <?php
        $i++;
    }
    ?>
</table>

希望这有帮助！

将数据库中的值显示到html表中

5 个答案: