MySQL：条件带走笔记

我有一个请求，它统计所有未移动的笔记和具有特定值的未移动笔记：

SELECT
    catalogs.id AS id,
    COUNT(DISTINCT t.id) AS total,
    COUNT(DISTINCT c.id) AS closed
FROM catalogs
LEFT JOIN links t ON catalogs.id = t.catalog
LEFT JOIN links c ON catalogs.id = c.catalog
WHERE catalogs.removed = 0
    AND ( t.removed = 0 OR t.removed is NULL )
    AND ( c.removed = 0 OR c.removed is NULL )
    AND ( c.is_open = 0 OR c.is_open is NULL )
GROUP BY catalogs.id
ORDER BY catalogs.id;

但在response中，我只能看到notes，其中total=0，或者至少存在一个c.is_open=0的notes

upd 0:我对sql不是很熟悉，但我意识到我试图解决问题的方法。。。我真丢脸：(

upd 1:我找到了另一种（答案中的第一个）使用SUM（）进行查询的方法，语法是

SUM(case when links.removed = 1 then 1 else 0 end) AS removed

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对我来说，这一项更容易。

尝试将其他列上的检查从WHERE子句移动到on子句：-

SELECT
    catalogs.id AS id,
    COUNT(DISTINCT t.id) AS total,
    COUNT(DISTINCT c.id) AS closed
FROM catalogs
LEFT JOIN links t ON catalogs.id = t.catalog AND t.removed = 0
LEFT JOIN links c ON catalogs.id = c.catalog AND c.removed = 0 AND ( c.is_open = 0 OR c.is_open is NULL )
WHERE catalogs.removed = 0
GROUP BY catalogs.id
ORDER BY catalogs.id;

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请提供实际和预期结果的样本数据