Question

更新：抱歉，我的问题第一次没有明确，希望这更有意义。我有一个查询列出了所有类别，在它迭代并将其放在段落标记中时。

$data = $conn->query('SELECT * FROM category ORDER BY category_name');

foreach($data as $row) {
echo '
<p><a href="category/'.$row['category_slug'].'">'.$row['category_name'].'</a></p>';
}

如何按字母顺序将所有名称放在列中，即以A开头的列放在一列中，b放在另一列中，依此类推。类似的东西：

一个--------------乙--------ç

动作-------- ----布拉赫卡通

冒险--- ----布拉赫漫画

我可以创建一个列，然后在每个列中执行查询，但这意味着查询字母表中的每个字母。我可以用一个查询来实现这一点。

Answer 1

我认为他想要的是这样的：

$wordArray = array('august', 'March', 'october', 'April', 'January', 'February', 'June', 'July');

foreach($wordArray as $word){

    // Get first letter uppercased
    $letter = ucfirst(substr($word, 0, 1));

    $ordered[$letter][] = $word;
}

echo '<pre>';
print_r($ordered);
echo '</pre>';

输出是：

    Array
(
    [A] => Array
        (
            [0] => august
            [1] => April
        )

    [M] => Array
        (
            [0] => March
        )

    [O] => Array
        (
            [0] => october
        )

    [J] => Array
        (
            [0] => January
            [1] => June
            [2] => July
        )

    [F] => Array
        (
            [0] => February
        )

)

然后你可以构建一个像@CodeGodie这样的表，然后你会完成它。）

Answer 2

当您抓取数据并显示数据时，您可能需要定义显示顺序的内容。

$query = "SELECT * FROM category ORDER BY category_name ORDER BY `category` ASC";

$row = mysql_query($query);
$row = mysql_fetch_object($row);

//Count number of pets
$count = mysql_num_rows($query);

//Figure out how many pets to put in each column
$columncount = ceil($count/3);
$lastcolumncount = $count - $columncount *2;
$middlecolumnstart = $count - $columncount - $lastcolumncount +1;
$lastcolumnstart = $count-$lastcolumncount +1;

//Put the pets into three columns
if ($row): 
    $count = 0;
    ?>
    <div class="column">
        <?php
        foreach ($rows as $cat):
            $count ++;

            if($count == $middlecolumnstart) {
                echo '</div><div class="column">';
            }

            if($count == $lastcolumnstart) {
                echo '</div><div class="column last-column">';
            }

           ?>

            <a href=""><?php echo $cat->category_name; ?></a>

        <?php 
        endforeach;
    endif; ?>
</div>
<div style="clear:both;"></div>

CSS：

.column {
    width:180px;
    float:left;
    margin-right:26px;
}

.olumn a {
    display:block;
    width:100%;
    margin-bottom:10px;

}

.column.last-column {
    margin-right:0;
}

Answer 3

首先，你排序就像这样：

$array = $data->fetch_assoc();
$arr = array();
foreach ($array as $n) {
    $f = substr($n, 0, 1);
    $arr[$f] = array();
    foreach ($array as $n2) {
        $f2 = substr($n2, 0, 1);
        if ($f2 === $f) {
           array_push($arr[$f], $n2);
        }
    }
}
?>

然后您可以将其迭代到构建您的列，如下所示：

$cols = "";
foreach ($arr as $letter => $names) {
    $column = "<div class='column'>";
    $column .= "<h3>$letter</h3>";
    foreach ($names as $name) {
        $column .= "<div>$name</div>";
    }
    $column .= "</div>";
    $cols .= $column;
}
?>

现在在 HTML ：

<body>
    <?= $cols ?>
</body>

<强> CSS ：

.column{
    float:left;
    padding:10px;
}

Answer 4

问题不明确。

快速解决方案：

 $data = $conn->query('SELECT * FROM category ORDER BY category_name');
 echo '<table><tr><td>';
 $curIndexLatter='';
 foreach($data as $row) {
    if ($curIndexLatter != '' && $curIndexLatter!=substr($row['category_name'],0,1) )
    echo '</td><td>';
    echo '<p><a href="category/'.$row['category_slug'].'">'.$row['category_name'].'</a></p>';

    $curIndexLatter=substr($row['category_name'],0,1); 
}
echo '</td</tr></table>';

按字母顺序显示列中的mysql结果

4 个答案: