在一个查询中计算多个记录计数

发布于 2024-12-13 05:56:30 字数 387 浏览 0 评论 0原文

    SELECT li.listing_id, li.title, p.name, c.comment, COUNT( * ) AS pic_count
    FROM listings li
    LEFT JOIN photos p ON li.listing_id = p.listing_id
    LEFT JOIN comments c ON li.listing_id = c.listing_id
    WHERE li.listing_id =1
    GROUP BY li.listing_id
    LIMIT 1;

上面的查询为我提供了给定listing_id的照片数，同样我希望在同一查询中计算该listing_id的评论数，这可能吗？一次查询中有 2 个计数..

原文

    SELECT li.listing_id, li.title, p.name, c.comment, COUNT( * ) AS pic_count
    FROM listings li
    LEFT JOIN photos p ON li.listing_id = p.listing_id
    LEFT JOIN comments c ON li.listing_id = c.listing_id
    WHERE li.listing_id =1
    GROUP BY li.listing_id
    LIMIT 1;

above query gives me the count of photos for given listing_id, same way i want the count of comments for that listing_id in same query, is that possible ? 2 counts in one query ..

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小糖芽 2024-12-20 05:56:30

你的数错了。如果您想按列表分组，p.name 不应出现在字段列表中。但即使没有它，由于 * 的计数，没有照片的列表您也会得到 1。

如果您的照片有唯一的 ID，您可以执行以下操作：

SELECT li.listing_id, li.title,
  COUNT(DISTINCT p.photo_id) as photo_count,
  COUNT(DISTINCT c.comment_id) as comment_count
FROM listings li
  LEFT JOIN photos p ON li.listing_id = p.listing_id
  LEFT JOIN comments c ON li.listing_id = c.listing_id
WHERE li.listing_id =1
GROUP BY li.listing_id
LIMIT 1;

Your count is wrong. p.name should not be in the field list if you want to group by listing. But even without it, you get 1 for listings without a photo, because of the * in count.

If your foto's got a unique id, you can do this:

SELECT li.listing_id, li.title,
  COUNT(DISTINCT p.photo_id) as photo_count,
  COUNT(DISTINCT c.comment_id) as comment_count
FROM listings li
  LEFT JOIN photos p ON li.listing_id = p.listing_id
  LEFT JOIN comments c ON li.listing_id = c.listing_id
WHERE li.listing_id =1
GROUP BY li.listing_id
LIMIT 1;

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