Question

我的数据库中有一个表calle“cat”。这就是“game1”，“game2”，“game3”等。我想知道如何在一个列表中输出这个，每个类别的链接都会像“http://mypage.com/game1”，“{{3} “等等。

我尝试了一些mysql输出，但我只是没有让它工作：/

Answer 1

<?php
//Array with data from table.
$array = array('game1', 'game2', 'game3', 'game4', 'game5');
echo '<pre>';
print_r($array);
echo '</pre>';
//This shows
//Array
//(
//    [0] => game1
//    [0] => game2
//    [0] => game3
//    [0] => game4
//    [0] => game5
//)

//Now we gonna use the implode() function.
echo '<ul>';
echo '<li>' . implode("</li><li>", $array) . '</li>';
echo '</ul>';
//This shows
//   game1
//   game2
//   game3
//   game4
//   game5
?>

print_r()函数，您主要用于检查数组的填充内容。

implode()函数是在数组的每个元素之间推送一些东西。

Answer 2

<?php

# Replace $base_url = "http://mypage.com/" with your Website URL you need to set
$base_url = "http://mypage.com/";

# --- Remove this piece of code, it's for demonstration purpose only
# List of all categories from database "cat" table
foreach(range('Z', 'A') as $char){
    //$categories[$char] = array("$char-Category-1","$char-Category-2");
    $categories[] = "$char-Category-1";
    $categories[] = "$char-Category-2";
}
# --- Remove above code

# $categories will be your database cat table Categories
$category_by_alphabets = array();

foreach ($categories as $cat_name){
    $character = strtoupper(substr($cat_name, 0, 1));
    $category_by_alphabets[$character][] = $cat_name;
}
ksort($category_by_alphabets);
?>

<?php 
    if(!empty($category_by_alphabets) && is_array($category_by_alphabets)):
?>

    <ul>
        <?php foreach($category_by_alphabets as $alphabets => $categories): ?>

            <li> <?php echo $alphabets;?></li>
            <ul>
                <?php foreach($categories as $category_name): ?>
                <li> 
                    <a href="<?php echo $base_url. $category_name;?>"> 
                    <?php echo $category_name;?></a>
                </li>
                <?php endforeach; ?>
            </ul>
        <?php endforeach; ?>
    </ul>

<?php endif; ?>

希望，它会帮助你。

从类别输出一个很好的列表

2 个答案: