PHP-MYSQL RAND()无法生成正确的列表

我在选择正确的结果时遇到问题.表wp_rdp_category_images具有3列：id(唯一)-category_name-category_image.

在下面的选择中,作为结果,我需要最多4个category_name,并且每个category_name中需要1个category_image.

有人可以帮助我更正此选择吗？会很好.提前致谢.

function show_game_category_images_func() {
    $content = '';
    $i = 1;
    $record = $GLOBALS['wpdb']->get_results('
    SELECT * FROM wp_rdp_category_images
    ORDER BY `category_name`, RAND()
    LIMIT 4 
    ');

    if($record){
        foreach($record as $key => $rec){                   
            // if($key == 'picture' && $rec != null && $rec != '')     
            $rec = '<img src="'.$upload_url.$record->category_image.'" width="150" height="150">';

            if ($i == 1) {
                echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span>';  // even
            }
            if ($i == 2) {
                echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span><br>';  // even
            }
            else {
                echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span>';   // odd
            }
            $i++;
        }
    }   
    return $content;
}
add_shortcode('show_game_category_images', 'show_game_category_images_func');

我的sql：

SELECT * FROM `wp_rdp_category_images`    
ORDER BY `category_name`, RAND() LIMIT 4

这是一个测试表：

CREATE TABLE IF NOT EXISTS `wp_rdp_category_images` (
  `id` int(12) NOT NULL AUTO_INCREMENT,
  `category_name` varchar(32) NOT NULL,
  `category_image` varchar(64) NOT NULL,
   PRIMARY KEY (`id`)
) ENGINE=InnoDB  DEFAULT CHARSET=latin1 AUTO_INCREMENT=22 ;

INSERT INTO `wp_rdp_category_images` (`id`, `category_name`, `category_image`) VALUES
(1, 'natuur', 'natuur1.jpg'),
(2, 'natuur', 'natuur4.jpg'),
(4, 'aarde', 'aarde3.jpg'),
(5, 'sport', 'sport2.jpg'),
(6, 'aarde', 'aarde3.jpg'),
(7, 'auto', 'autogrijs.jpg'),
(8, 'auto', 'autowit.jpg'),
(9, 'auto', 'autoblack.jpg'),
(10, 'auto', 'autoroodzwart.jpg'),
(11, 'aarde', 'aarde1.jpg'),
(12, 'aarde', 'aarde2.jpg'),
(13, 'sport', 'sport4.jpg'),
(14, 'sport', 'sport3.jpg'),
(15, 'sport', 'sport1.jpg'),
(16, 'natuur', 'natuur2.jpg'),
(17, 'natuur', 'natuur3.jpg'),
(18, 'people', 'people1.jpg'),
(19, 'people', 'people2.jpg'),
(20, 'people', 'people3.jpg'),
(21, 'people', 'people4.jpg');

解决方法:

您可以使用以下查询：

select     category_name,
           (select   category_image
            from     wp_rdp_category_images
            where    category_name = main.category_name
            order by rand()
            limit    1) category_image
from       wp_rdp_category_images main
group by   category_name
order by   rand()
limit      4

SQL Fiddle

如果需要结果集中的所有列,请使用以下变体：

select     main.*
from       (
            select     category_name,
                       (select   id
                        from     wp_rdp_category_images
                        where    category_name = main.category_name
                        order by rand()
                        limit    1) id
             from       wp_rdp_category_images main
            ) sub
inner join  wp_rdp_category_images main
        on  main.id = sub.id
group by    main.category_name
order by   rand()
limit      4

SQL Fiddle

PHP 代码

您的PHP有问题,因此重复了第一行.请注意,当$i为1时,您将在第一个if中执行该部分,而且还将在第二个if中执行该部分.

如果出现以下情况,则需要在第二个之前放置else：

if ($i == 1) {
    echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span>';  // even
} else if ($i == 2) { // !!!! ADDED "else" here!
    echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span><br>';  // even
}
else {
    echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span>';   // odd
}

但是您的意见似乎表明您想创建两列.在这种情况下,最好这样写：

echo '<span><img src="'.$upload_url.$rec->category_image.'" ></span>';  // all
if ($i % 2 == 0) echo '<br>';  // even only

PHP-MYSQL RAND()无法生成正确的列表

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