数据库中的 Select 行,其中一列中的值不同且最新限制为 5
Select rows from database where the value in one column is distinct and limit to 5 latest
我有一个图像数据库,图像行用查看它们的人的最后一个 IP 更新,并用当前时间戳更新 date_updated 列。我正在尝试查看最后 5 张图片,但只有每个不同的 IP 地址,我不希望有人淹没最后查看的结果。
Fiddle::
http://sqlfiddle.com/#!2/d5b05/16
想要的结果:
从该数据集 select 中得到的期望结果:
SELECT * FROM `image` ORDER BY `date_updated` DESC;
| IMAGE | WIDTH | HEIGHT | DATE_ADDED | DATE_UPDATED | UPDATED_BY_IP |
|---------|-------|--------|------------|--------------|---------------|
| 1x1XGY4 | 1920 | 1080 | 1417546414 | 1421712314 | 192.168.0.7 |
| 1x1XGY3 | 1920 | 1080 | 1417546413 | 1421712313 | 192.168.0.7 |
| 1x1XGY2 | 1920 | 1080 | 1417546412 | 1421712312 | 192.168.0.10 |
| 1x1XGY1 | 1920 | 1080 | 1417546411 | 1421712311 | 192.168.0.10 |
| 1oApS54 | 1920 | 1080 | 1417138874 | 1421685474 | 192.168.0.2 |
| 1oApS53 | 1920 | 1080 | 1417138873 | 1421685473 | 192.168.0.2 |
| 1oApS52 | 1920 | 1080 | 1417138872 | 1421685472 | 192.168.0.10 |
| 1oApS51 | 1920 | 1080 | 1417138871 | 1421685471 | 192.168.0.10 |
| 1ydhtQ4 | 1920 | 1080 | 1421460434 | 1421685154 | 192.168.0.6 |
| 1ydhtQ3 | 1920 | 1080 | 1421460433 | 1421685153 | 192.168.0.7 |
| 1ydhtQ2 | 1920 | 1080 | 1421460432 | 1421685152 | 192.168.0.10 |
| 1ydhtQ1 | 1920 | 1080 | 1421460431 | 1421685151 | 192.168.0.5 |
| 1WyQib4 | 1920 | 1080 | 1420869354 | 1421634384 | 192.168.0.8 |
| 1WyQib3 | 1920 | 1080 | 1420869353 | 1421634383 | 192.168.0.2 |
| 1WyQib2 | 1920 | 1080 | 1420869352 | 1421634382 | 192.168.0.3 |
| 1WyQib1 | 1920 | 1080 | 1420869351 | 1421634381 | 192.168.0.10 |
| 1izDqg4 | 1920 | 1080 | 1416948144 | 1421608564 | 192.168.0.2 |
| 1izDqg3 | 1920 | 1080 | 1416948143 | 1421608563 | 192.168.0.2 |
| 1izDqg2 | 1920 | 1080 | 1416948142 | 1421608562 | 192.168.0.5 |
| 1izDqg1 | 1920 | 1080 | 1416948141 | 1421608561 | 192.168.0.10 |
使用伪select语句:
SELECT * FROM image WHERE updated_by_ip IS DISTINCT ORDER BY
date_updated DESC LIMIT 5
| IMAGE | WIDTH | HEIGHT | DATE_ADDED | DATE_UPDATED | UPDATED_BY_IP |
|---------|-------|--------|------------|--------------|---------------|
| 1x1XGY4 | 1920 | 1080 | 1417546414 | 1421712314 | 192.168.0.7 |
| 1x1XGY2 | 1920 | 1080 | 1417546412 | 1421712312 | 192.168.0.10 |
| 1oApS54 | 1920 | 1080 | 1417138874 | 1421685474 | 192.168.0.2 |
| 1ydhtQ4 | 1920 | 1080 | 1421460434 | 1421685154 | 192.168.0.6 |
| 1ydhtQ1 | 1920 | 1080 | 1421460431 | 1421685151 | 192.168.0.5 |
壁橱结果:
我能想到的最好的是:
SELECT DISTINCT updated_by_ip, MAX(date_updated) AS date_updated
FROM `image` GROUP BY updated_by_ip ORDER BY date_updated DESC LIMIT 5;
这给了我:
| UPDATED_BY_IP | DATE_UPDATED |
|---------------|--------------|
| 192.168.0.7 | 1421712314 |
| 192.168.0.10 | 1421712312 |
| 192.168.0.2 | 1421685474 |
| 192.168.0.6 | 1421685154 |
| 192.168.0.5 | 1421685151 |
其中我可以做一个
while (SELECT DISTINCT updated_by_ip ...)
{
$result_rows[] = SELECT * FROM image
WHERE updated_by_ip = query[updated_by_ip]
AND date_updated = query[date_updated] LIMIT 1
}
但是,我希望找到一种方法来做到这一点,而不必进行一堆 post 处理和额外的查询,以及 select 通过 updated_by_ip 和 date_updated好像不太稳定
谢谢。
这不是最漂亮的查询(根据 SQL 标准不正确)但它适用于 MySQL:
SELECT * FROM `image`
GROUP BY updated_by_ip
ORDER BY `date_updated` DESC
在 Postgres 中,您可以使用 DISTINCT ON(...)
,但 MySQL 不支持,因此最简单的解决方法是按要区分的列进行分组。另一种方法是使用子查询,但它的性能不太理想。
一种方法是使用变量来枚举行:
SELECT i.*
FROM (SELECT i.*,
(@rn := if(@uip = updated_by_ip, @rn + 1,
if(@uip := updated_by_ip, 1, 1)
)
)
FROM image i CROSS JOIN
(SELECT @uip := '', @rn := 0) vars
WHERE updated_by_ip
ORDER BY updated_by_ip, date_updated DESC
) i
WHERE seqnum <= 5;
要在没有 MySQL GROUP BY 扩展名的情况下执行此操作,您可以尝试以下操作:
首先,用这个子查询从五个不同的IP号中获取最近的更新时间。
SELECT updated_by_ip, MAX(date_updated) as date_updated
FROM image
GROUP BY updated_by_ip
ORDER BY 2 DESC
LIMIT 5
如果您的 table 很大,(updated_by_ip, date_updated)
上的索引将有助于提高性能。
然后,将其连接到该子查询的主查询以获得结果。
SELECT i.*
FROM image i
JOIN (
SELECT updated_by_ip, MAX(date_updated) as date_updated
FROM image
GROUP BY updated_by_ip
ORDER BY 2 DESC
LIMIT 5
) m USING(updated_by_ip, date_updated)
ORDER BY i.date_updated DESC
我有一个图像数据库,图像行用查看它们的人的最后一个 IP 更新,并用当前时间戳更新 date_updated 列。我正在尝试查看最后 5 张图片,但只有每个不同的 IP 地址,我不希望有人淹没最后查看的结果。
Fiddle:: http://sqlfiddle.com/#!2/d5b05/16
想要的结果: 从该数据集 select 中得到的期望结果:
SELECT * FROM `image` ORDER BY `date_updated` DESC;
| IMAGE | WIDTH | HEIGHT | DATE_ADDED | DATE_UPDATED | UPDATED_BY_IP |
|---------|-------|--------|------------|--------------|---------------|
| 1x1XGY4 | 1920 | 1080 | 1417546414 | 1421712314 | 192.168.0.7 |
| 1x1XGY3 | 1920 | 1080 | 1417546413 | 1421712313 | 192.168.0.7 |
| 1x1XGY2 | 1920 | 1080 | 1417546412 | 1421712312 | 192.168.0.10 |
| 1x1XGY1 | 1920 | 1080 | 1417546411 | 1421712311 | 192.168.0.10 |
| 1oApS54 | 1920 | 1080 | 1417138874 | 1421685474 | 192.168.0.2 |
| 1oApS53 | 1920 | 1080 | 1417138873 | 1421685473 | 192.168.0.2 |
| 1oApS52 | 1920 | 1080 | 1417138872 | 1421685472 | 192.168.0.10 |
| 1oApS51 | 1920 | 1080 | 1417138871 | 1421685471 | 192.168.0.10 |
| 1ydhtQ4 | 1920 | 1080 | 1421460434 | 1421685154 | 192.168.0.6 |
| 1ydhtQ3 | 1920 | 1080 | 1421460433 | 1421685153 | 192.168.0.7 |
| 1ydhtQ2 | 1920 | 1080 | 1421460432 | 1421685152 | 192.168.0.10 |
| 1ydhtQ1 | 1920 | 1080 | 1421460431 | 1421685151 | 192.168.0.5 |
| 1WyQib4 | 1920 | 1080 | 1420869354 | 1421634384 | 192.168.0.8 |
| 1WyQib3 | 1920 | 1080 | 1420869353 | 1421634383 | 192.168.0.2 |
| 1WyQib2 | 1920 | 1080 | 1420869352 | 1421634382 | 192.168.0.3 |
| 1WyQib1 | 1920 | 1080 | 1420869351 | 1421634381 | 192.168.0.10 |
| 1izDqg4 | 1920 | 1080 | 1416948144 | 1421608564 | 192.168.0.2 |
| 1izDqg3 | 1920 | 1080 | 1416948143 | 1421608563 | 192.168.0.2 |
| 1izDqg2 | 1920 | 1080 | 1416948142 | 1421608562 | 192.168.0.5 |
| 1izDqg1 | 1920 | 1080 | 1416948141 | 1421608561 | 192.168.0.10 |
使用伪select语句:
SELECT * FROM image WHERE updated_by_ip IS DISTINCT ORDER BY date_updated DESC LIMIT 5
| IMAGE | WIDTH | HEIGHT | DATE_ADDED | DATE_UPDATED | UPDATED_BY_IP |
|---------|-------|--------|------------|--------------|---------------|
| 1x1XGY4 | 1920 | 1080 | 1417546414 | 1421712314 | 192.168.0.7 |
| 1x1XGY2 | 1920 | 1080 | 1417546412 | 1421712312 | 192.168.0.10 |
| 1oApS54 | 1920 | 1080 | 1417138874 | 1421685474 | 192.168.0.2 |
| 1ydhtQ4 | 1920 | 1080 | 1421460434 | 1421685154 | 192.168.0.6 |
| 1ydhtQ1 | 1920 | 1080 | 1421460431 | 1421685151 | 192.168.0.5 |
壁橱结果:
我能想到的最好的是:
SELECT DISTINCT updated_by_ip, MAX(date_updated) AS date_updated
FROM `image` GROUP BY updated_by_ip ORDER BY date_updated DESC LIMIT 5;
这给了我:
| UPDATED_BY_IP | DATE_UPDATED |
|---------------|--------------|
| 192.168.0.7 | 1421712314 |
| 192.168.0.10 | 1421712312 |
| 192.168.0.2 | 1421685474 |
| 192.168.0.6 | 1421685154 |
| 192.168.0.5 | 1421685151 |
其中我可以做一个
while (SELECT DISTINCT updated_by_ip ...)
{
$result_rows[] = SELECT * FROM image
WHERE updated_by_ip = query[updated_by_ip]
AND date_updated = query[date_updated] LIMIT 1
}
但是,我希望找到一种方法来做到这一点,而不必进行一堆 post 处理和额外的查询,以及 select 通过 updated_by_ip 和 date_updated好像不太稳定
谢谢。
这不是最漂亮的查询(根据 SQL 标准不正确)但它适用于 MySQL:
SELECT * FROM `image`
GROUP BY updated_by_ip
ORDER BY `date_updated` DESC
在 Postgres 中,您可以使用 DISTINCT ON(...)
,但 MySQL 不支持,因此最简单的解决方法是按要区分的列进行分组。另一种方法是使用子查询,但它的性能不太理想。
一种方法是使用变量来枚举行:
SELECT i.*
FROM (SELECT i.*,
(@rn := if(@uip = updated_by_ip, @rn + 1,
if(@uip := updated_by_ip, 1, 1)
)
)
FROM image i CROSS JOIN
(SELECT @uip := '', @rn := 0) vars
WHERE updated_by_ip
ORDER BY updated_by_ip, date_updated DESC
) i
WHERE seqnum <= 5;
要在没有 MySQL GROUP BY 扩展名的情况下执行此操作,您可以尝试以下操作:
首先,用这个子查询从五个不同的IP号中获取最近的更新时间。
SELECT updated_by_ip, MAX(date_updated) as date_updated
FROM image
GROUP BY updated_by_ip
ORDER BY 2 DESC
LIMIT 5
如果您的 table 很大,(updated_by_ip, date_updated)
上的索引将有助于提高性能。
然后,将其连接到该子查询的主查询以获得结果。
SELECT i.*
FROM image i
JOIN (
SELECT updated_by_ip, MAX(date_updated) as date_updated
FROM image
GROUP BY updated_by_ip
ORDER BY 2 DESC
LIMIT 5
) m USING(updated_by_ip, date_updated)
ORDER BY i.date_updated DESC