[英]Mysql count and sum from two different tables
i have a problem with some querys in php and mysql: I have 2 different tables with one field in common: 我在php和mysql中有一些查询问题:我有2个不同的表,其中一个字段相同:
table 1 表格1
id | id | hits |
点击 num_g |
num_g | cats |
猫| usr_id |active
usr_id |活动
1 | 1 | 10 |
10 | 11 |
11 | 1 |
1 | 53 |
53 | 1
1个
2 | 2 | 13 |
13 | 16 |
16 | 3 |
3 | 53 |
53 | 1
1个
1 | 1 | 10 |
10 | 22 |
22 | 1 |
1 | 22 |
22 | 1
1个
1 | 1 | 10 |
10 | 21 |
21 | 3 |
3 | 22 |
22 | 1
1个
1 | 1 | 2 |
2 | 6 |
6 | 2 |
2 | 11 |
11 | 1
1个
1 | 1 | 11 |
11 | 1 |
1 | 1 |
1 | 11 |
11 | 1
1个
table 2 表2
id | id | usr_id |
usr_id | points
点数
1 | 1 | 53 |
53 | 300
300
Now i use this statement to sum just the total from the table 1 every id count + 1 too 现在,我使用此语句对每个id计数+ 1也只汇总表1的总数
SELECT usr_id, COUNT( id ) + SUM( num_g + hits ) AS tot_h FROM table1 WHERE usr_id!='0' GROUP BY usr_id ASC LIMIT 0 , 15
and i get the total for each usr_id 我得到每个usr_id的总数
usr_id| usr_id | tot_h |
tot_h |
53 | 53 | 50
50
22 | 22 | 63
63
11 | 11 | 20
20
until here all is ok, now i have a second table with extra points (table2) I try this: 直到这里一切都OK,现在我有了另一个带有加分的表(table2),我尝试这样做:
SELECT usr_id, COUNT( id ) + SUM( num_g + hits ) + (SELECT points FROM table2 WHERE usr_id != '0' ) AS tot_h FROM table1 WHERE usr_id != '0' GROUP BY usr_id ASC LIMIT 0 , 15
but it seems to sum the 300 extra points to all users: 但似乎给所有用户加了300分:
usr_id| usr_id | tot_h |
tot_h |
53 | 53 | 350
350
22 | 22 | 363
363
11 | 11 | 320
320
Now how i can get the total like the first try but + the secon table in one statement? 现在,我如何才能像第一次尝试一样获得总计,但在一条语句中加上secon表? because now i have just one entry in the second table but i can be more there.
因为现在我在第二张表中只有一个条目,但是我可以在其中更多。 thanks for all the help.
感谢所有的帮助。
hi thomas thanks for your reply, i think is in the right direction, but i'm getting weirds results, like 嗨,托马斯,谢谢您的答复,我认为方向是正确的,但是我得到的结果很奇怪,例如
usr_id | usr_id | tot_h
tot_h
22 | 22 | NULL <== i think the null its because that usr_id as no value in the table2
NULL <==我认为它是null,因为usr_id作为table2中的无值
53 | 53 | 1033
1033
Its like the second user is getting all the the values. 就像第二个用户正在获取所有值。 then i try this one:
然后我尝试这个:
SELECT table1.usr_id, COUNT( table1.id ) + SUM( table1.num_g + table1.hits + table2.points ) AS tot_h
FROM table1
LEFT JOIN table2 ON table2.usr_id = table1.usr_id
WHERE table1.usr_id != '0'
AND table2.usr_id = table1.usr_id
GROUP BY table1.usr_id ASC
Same result i just get the sum of all values and not by each user, i need something like this result: 同样的结果,我只是得到所有值的总和,而不是每个用户,我需要像这样的结果:
usr_id | usr_id | tot_h
tot_h
53 | 53 | 53 <==== plus 300 points on table1
53 <====加table1上的300点
22 | 22 | 56 <==== plus 100 points on table2
56 <====加table2上的100分
/////////the result i need //////////// //////////我需要的结果/////////////
usr_id | usr_id | tot_h
tot_h
53 | 53 | 353 <==== plus 300 points on table2
353 <====在table2上加300点
22 | 22 | 156 <==== plus 100 points on table2
156 <====加上table2的100点
I think the structure need to be something like this Pseudo statements ;) 我认为结构需要像这样的伪语句;)
from table1 count all id to get the number of record where the usr_id are then sum hits + num_g and from table2 select the extra points where the usr_id are the same as table1 and get the result: 从table1计数所有id以获取usr_id为总和+ num_g的记录数,并从table2选择usr_id与table1相同的加点,并得到结果:
usr_id | usr_id | tot_h
tot_h
53 | 53 | 353
353
22 | 22 | 156
156
There is nothing in your subquery which calculates extra points to correlate it to the outer Table1. 子查询中没有任何东西可以计算额外的点以将其与外部Table1相关联。 So, one solution is to add that correlation:
因此,一种解决方案是添加相关性:
SELECT usr_id
, COUNT( id ) + SUM( num_g + hits )
+ (SELECT points
FROM table2
WHERE table2.usr_id = table1.usr_id ) AS tot_h
FROM table1
WHERE usr_id != '0'
GROUP BY usr_id ASC
LIMIT 0 , 15
Another solution would be to simply join to it directly: 另一个解决方案是直接将其直接加入:
SELECT table1.usr_id
, COUNT( table1.id )
+ SUM( table1.num_g + table1.hits + table2.points )
AS tot_h
FROM table1
Left Join table2
On table2.usr_id = table1.usr_id
WHERE table1.usr_id != '0'
GROUP BY table1.usr_id ASC
LIMIT 0 , 15
i think get the solution, i dont know if its the best but it works for me, if you know a way to optimize this i really apreciate it. 我想得到解决方案,我不知道它是否是最好的,但是对我有用,如果您知道一种优化方法,我真的很感激。
SELECT usr_id , COUNT( id ) + SUM( num_g + hits )as sumtot ,
(SELECT points FROM table2 WHERE usr_id = table1.usr_id ) AS tot_h
FROM table1
WHERE usr_id != '0'
GROUP BY usr_id ASC
with this i get something like this 有了这个我得到这样的东西
usr_id |sumtot | usr_id | sumtot | tot_h
tot_h
5 |557 | 5 | 557 | NULL
空值
53 |2217 | 53 | 2217 | 300
300
So then i just sum the result and show it in a while loop. 因此,然后我只是对结果求和并在while循环中显示它。
<?php
//some mysql here
//then the while loop
// and then the final sum
$final_result=$r_rank['tot_h']+$r_rank['sumtot'];
?>
Thanks a lot for your help thomas :) 非常感谢您对托马斯的帮助:)
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