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mysql - 根据特定的 meta_key 和 meta_value 过滤帖子元

转载 作者:行者123 更新时间:2023-11-29 00:32:11 25 4
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我有以下 SQL 语句来计算基于 3 个 Wordpress 表的几个字段。我需要的是仅包含 meta_key='odd' 和 meta_value >= '1.5' 的结果。我在 WHERE 语句中尝试过,但它不起作用,因为此查询中的大多数结果显示为 Null。

非常感谢任何帮助!

SELECT (display_name) AS 'user',
SUM(Case When meta_key = 'status' Then meta_value = 'won' else Null End) AS 'Won',
SUM(Case When meta_key = 'status' Then meta_value = 'lost' else Null End) AS 'Lost',
COUNT(Case When meta_key = 'odd' Then meta_value else Null End) AS 'Total',
ROUND (AVG(Case When meta_key = 'odd' Then meta_value else Null End), 2) AS 'Avg odd',
ROUND (AVG(Case When meta_key = 'bet' Then meta_value else Null End), 2) AS 'Avg bet',
ROUND (SUM(Case When meta_key = 'balance' Then meta_value else Null End), 2) AS 'Balance'
FROM wp_postmeta pm
INNER JOIN wp_posts p ON pm.post_id = p.ID
INNER JOIN wp_users u ON p.post_author = u.ID
WHERE Month(post_date) = MONTH(CURRENT_DATE) AND p.post_status='publish' AND pm.meta_key='odd' AND pm.met.value >='1.5'
GROUP BY (display_name)
ORDER BY Balance DESC

最佳答案

您需要将其移至 having 子句:

SELECT (display_name) AS 'user',
SUM(Case When meta_key = 'status' Then meta_value = 'won' else Null End) AS 'Won',
SUM(Case When meta_key = 'status' Then meta_value = 'lost' else Null End) AS 'Lost',
COUNT(Case When meta_key = 'odd' Then meta_value else Null End) AS 'Total',
ROUND (AVG(Case When meta_key = 'odd' Then meta_value else Null End), 2) AS 'Avg odd',
ROUND (AVG(Case When meta_key = 'bet' Then meta_value else Null End), 2) AS 'Avg bet',
ROUND (SUM(Case When meta_key = 'balance' Then meta_value else Null End), 2) AS 'Balance'
FROM wp_postmeta pm
INNER JOIN wp_posts p ON pm.post_id = p.ID
INNER JOIN wp_users u ON p.post_author = u.ID
WHERE Month(post_date) = MONTH(CURRENT_DATE) AND p.post_status='publish'
GROUP BY (display_name)
having SUM(case when pm.meta_key='odd' AND pm.meta_value >='1.5' then 1 else 0 end) > 0
ORDER BY Balance DESC

这决定了对于每个 display_name 是否存在这样的一行。

关于mysql - 根据特定的 meta_key 和 meta_value 过滤帖子元,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16019828/

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