Question

我有两个表，一个用于下载，一个用于上传。它们几乎完全相同，但有一些其他列不同。我想为表格中的每个项目生成每个日期的统计信息列表。

我使用这两个查询但是在运行它们之后必须合并php中的数据。我想在一个查询中运行它们，它将从按日期分组的每一行中的两个查询中返回列。有时没有任何下载数据，只有上传数据，而且在以前的所有尝试中，如果它无法从两行中找到日志数据，它会跳过该行。

如何将这两个查询合并为一个，即使它在其中一个表中可用，它也会显示数据？

SELECT DATE(upload_date_added) as upload_date, SUM(upload_size) as upload_traffic, SUM(upload_files) as upload_files
FROM packages_uploads
WHERE upload_date_added BETWEEN '2011-10-26' AND '2011-11-16'
GROUP BY upload_date
ORDER BY upload_date DESC


SELECT DATE(download_date_added) as download_date, SUM(download_size) as download_traffic, SUM(download_files) as download_files
FROM packages_downloads
WHERE download_date_added BETWEEN '2011-10-26' AND '2011-11-16'
GROUP BY download_date
ORDER BY download_date DESC

我想获得这样的结果行：

date，upload_traffic，upload_files，download_traffic，download_files

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Answer 1

您可以执行两个查询，然后将其与UNION子句以及一个额外字段结合使用，以在不同的行上标识上载和下载：

SELECT
    'Uploads' TransmissionType,
    DATE(upload_date_added) as TransmissionDate,
    SUM(upload_size) as TransmissionTraffic,
    SUM(upload_files) as TransmittedFileCount
FROM
    packages_uploads 
WHERE upload_date_added BETWEEN '2011-10-26' AND '2011-11-16' 
GROUP BY upload_date 
ORDER BY upload_date DESC 
UNION 
SELECT
    'Downloads',
    DATE(download_date_added),
    SUM(download_size),
    SUM(download_files) 
FROM packages_downloads 
WHERE download_date_added BETWEEN '2011-10-26' AND '2011-11-16' 
GROUP BY download_date 
ORDER BY download_date DESC;

试一试!!!

Answer 2

您所询问的内容仅适用于具有相同的上传和下载添加日期的行。在这种情况下，我认为这个SQL应该有效：

SELECT 
    DATE(u.upload_date_added) as date,
    SUM(u.upload_size) as upload_traffic,
    SUM(u.upload_files) as upload_files,
    SUM(d.download_size) as download_traffic,
    SUM(d.download_files) as download_files
FROM 
    packages_uploads u, packages_downloads d
WHERE u.upload_date_added = d.download_date_added
AND u.upload_date_added BETWEEN '2011-10-26' AND '2011-11-16'
GROUP BY date
ORDER BY date DESC

Answer 3

如果不知道架构很难给出确切的答案，请将以下内容视为概念而非直接答案。

你可以尝试左连接，我不确定表包是否存在，但以下可能是值得深思的

SELECT 
  p.id,  
  up.date as upload_date
  dwn.date as download_date
FROM 
package p
LEFT JOIN package_uploads up ON 
    ( up.package_id = p.id WHERE up.upload_date = 'etc' )
LEFT JOIN package_downloads dwn ON 
    ( dwn.package_id = p.id WHERE up.upload_date = 'etc' )

以上将选择所有包并尝试加入，并且值不加入的位置将返回null。

Answer 4

有很多方法可以做到这一点。您可以使用主键和外键加入。如果表之间没有关系，

您可以使用，

LEFT JOIN / LEFT OUTER JOIN

返回左表中匹配的所有记录右表中的记录。结果是来自NULL 当没有比赛时右侧。

RIGHT JOIN / RIGHT OUTER JOIN

返回右表和匹配的所有记录左表中的记录。结果是左侧NULL 当没有比赛的时候。

FULL OUTER JOIN

当左表或右表记录中存在匹配时，返回所有记录。

UNION

用于组合两个或多个SELECT语句的结果集。

SELECT中的每个UNION语句必须具有相同的数量，   columns列还必须具有相似的数据类型列中，   每个SELECT语句也必须采用相同的顺序。

INNER JOIN

选择两个表中具有匹配值的记录。 - 这对你的情况有好处。

INTERSECT

不支持MySQL。

NATURAL JOIN

应匹配所有列名称。

由于您不需要更新这些，您可以从连接表创建view，然后您可以在PHP中使用较少的查询。但观点无法更新。你没有提到表之间的关系。因此，我必须使用UNION。

像这样，

CREATE VIEW checkStatus
AS
SELECT 
    DATE(upload_date_added) as upload_date,
    SUM(upload_size) as upload_traffic,
    SUM(upload_files) as upload_files
FROM packages_uploads
WHERE upload_date_added BETWEEN '2011-10-26' AND '2011-11-16'
GROUP BY upload_date
ORDER BY upload_date DESC

UNION

SELECT 
    DATE(download_date_added) as download_date,
    SUM(download_size) as download_traffic,
    SUM(download_files) as download_files
FROM packages_downloads
WHERE download_date_added BETWEEN '2011-10-26' AND '2011-11-16'
GROUP BY download_date
ORDER BY download_date DESC

然后，您想要选择的任何地方只需要一行：

SELECT * FROM checkStatus

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如何将这两个查询合并为一个？

4 个答案: