MySQL查询，其中JOIN依赖于CASE [英] MySQL query where JOIN depends on CASE

问题描述

我现在知道，CASE只能在WHERE上下文中使用.虽然，我需要根据column值使用不同的表.我尝试过的内容如下:

As I am now aware, CASE can be used only in WHERE context. Though, I need to use different table depending on column value. What I've tried looks like this:

SELECT
    `ft1`.`task`,
    COUNT(`ft1`.`id`) `count`
FROM
    `feed_tasks` `ft1`
CASE
    `ft1`.`type`
WHEN
    1
THEN
    (INNER JOIN `pages` `p1` ON `p1`.`id` = `ft1`.`reference_id`)
WHEN
    2
THEN
    (INNER JOIN `urls` `u1` ON `u1`.`id` = `ft1`.`reference_id`)
WHERE
    `ft1`.`account_id` IS NOT NULL AND
    `a1`.`user_id` = {$db->quote($user['id'])}

现在我知道这是无效的语法，最接近的替代方法是什么?

Now that I know this is invalid syntax, what's the closest alternative?

推荐答案

可能需要进行调整以返回正确的结果，但我希望您能明白:

It probably needs tweaking to return the correct results but I hope you get the idea:

SELECT ft1.task, COUNT(ft1.id) AS count
FROM feed_tasks ft1
LEFT JOIN pages p1 ON ft1.type=1 AND p1.id = ft1.reference_id
LEFT JOIN urls u1 ON ft1.type=2 AND u1.id = ft1.reference_id
WHERE COALESCE(p1.id, u1.id) IS NOT NULL
AND ft1.account_id IS NOT NULL
AND a1.user_id = :user_id

关于CASE...END的一点说明.您的原始代码无法运行，因为与PHP或JavaScript不同，SQL CASE并不是一种流控制结构，它无法选择将运行代码的哪一部分.而是返回一个表达式.因此，您可以执行以下操作:

A little note about CASE...END. Your original code does not run because, unlike PHP or JavaScript, the SQL CASE is not a flow control structure that allows to choose which part of the code will run. Instead, it returns an expression. So you can do this:

SELECT CASE
    WHEN foo<0 THEN 'Yes'
    ELSE 'No'
END AS is_negative
FROM bar

...但不是这样:

-- Invalid
CASE 
    WHEN foo<0 THEN SELECT 'Yes' AS is_negative
    ELSE SELECT 'No' AS is_negative
END
FROM bar

这篇关于MySQL查询，其中JOIN依赖于CASE的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！