在单个查询中检索结果集和其他列的计数答案

【问题标题】：Retrieve count of the result set and other columns in a single query在单个查询中检索结果集和其他列的计数
【发布时间】：2016-10-21 11:25:10
【问题描述】：

我有以下查询对contents 和tag 表进行内部联接。（使用mysql）

            SELECT
                contents.original_duration,
                tag.tagconfig,
                contents.id_contents
            FROM
                contents
                    inner JOIN
                tag
                    ON
                contents.id_contents = tag.id_contents
                AND contents.id_host = tag.id_host
                AND contents.id_vhost = tag.id_vhost

            WHERE
                tag.video_insert_status='idle'
                AND contents.type = 'video'
                AND contents.subtype = 'video'
                AND original_duration is not NULL;

id_contents、id_host 和 id_vhost 是机器人 tag 和 contents 表中的主键。以上查询结果260条。

如何在单个查询中选择此计数？

我试过了

SELECT
                contents.original_duration,
                tag.tagconfig,
                contents.id_contents,
                count(contents.original_duration)

但是，它似乎没有给出正确的结果。

【问题讨论】：

您希望将结果作为新列还是其他内容？

标签： mysql sql

【解决方案1】：

您可以使用的一个技巧是 GROUP BY 选择列表中的所有列，然后使用 COUNT(*) 获取记录数：

SELECT
    contents.original_duration,
    tag.tagconfig,
    contents.id_contents,
    COUNT(*) AS recordCount
FROM
    contents
        inner JOIN
    tag
        ON
    contents.id_contents = tag.id_contents
    AND contents.id_host = tag.id_host
    AND contents.id_vhost = tag.id_vhost
WHERE
    tag.video_insert_status='idle'
    AND contents.type = 'video'
    AND contents.subtype = 'video'
    AND original_duration is not NULL;
GROUP BY
    contents.original_duration,
    tag.tagconfig,
    contents.id_contents

【讨论】：

【解决方案2】：

将原始查询的 SELECT 子句替换为

SELECT COUNT(*)

将给出计数。

【讨论】：

【解决方案3】：

我认为您可以使用两个相同的查询并将它们加入以获得您想要的内容，例如：

SELECT T1.*, T2.CNT
FROM (
    SELECT
        contents.original_duration,
        tag.tagconfig,
        contents.id_contents
    FROM
        contents
            inner JOIN
        tag
            ON
        contents.id_contents = tag.id_contents
        AND contents.id_host = tag.id_host
        AND contents.id_vhost = tag.id_vhost

    WHERE
        tag.video_insert_status='idle'
        AND contents.type = 'video'
        AND contents.subtype = 'video'
        AND original_duration is not NULL) T1
INNER JOIN (
    SELECT
        COUNT(*) AS CNT
    FROM
        contents
            inner JOIN
        tag
            ON
        contents.id_contents = tag.id_contents
        AND contents.id_host = tag.id_host
        AND contents.id_vhost = tag.id_vhost

    WHERE
        tag.video_insert_status='idle'
        AND contents.type = 'video'
        AND contents.subtype = 'video'
        AND original_duration is not NULL
) T2

或者你可以执行下面的sql，你可以通过最后一条记录来计数：

SELECT
    contents.original_duration,
    tag.tagconfig,
    contents.id_contents,
    @CNT := @CNT + 1 AS CNT
FROM
    contents
        inner JOIN
    tag
        ON
    contents.id_contents = tag.id_contents
    AND contents.id_host = tag.id_host
    AND contents.id_vhost = tag.id_vhost
    CROSS JOIN (SELECT @CNT := 0) T
WHERE
    tag.video_insert_status='idle'
    AND contents.type = 'video'
    AND contents.subtype = 'video'
    AND original_duration is not NULL
ORDER BY CNT

【讨论】：