Question

我有这张桌子

------------------------------------------
     id     |       Machines               
------------------------------------------
     1.     |       1 Truck              
------------------------------------------ 
     2.     |       1 Bobcat             
------------------------------------------
     3.     |       2 Platform          
------------------------------------------   
     4.     |       Telehender            
------------------------------------------
     5.     |       Teodolit, 3 Platform 
------------------------------------------
     6.     |       2 Tractor             
------------------------------------------
     7.     |         NULL             
------------------------------------------ 
  Result:   |         11                 
------------------------------------------

我想先对整数（1 + 1 + 2 + 3 + 2）求和，然后对不带整数的值求和（Telehender和Teodolit = 2），然后跳过NULL值。该表的结果必须为11。

我正在使用此查询

 SELECT Sum((Char_length(machines) - Char_length(Replace(machines, ',','')) + 1)) AS ukupno
FROM   izvestaji
WHERE  projekatid='8'
AND    datum='2019-10-03'

但是我得到的结果是9。

有没有办法进行查询？

Answer 1

错误的数据库设计...：/ 但是，在这里它是一个解决方案，其中我假设您不超过3个元素进入机器领域

SELECT 
        sum(case
                when length(machines)-length(replace(machines,',','')) = 0
                then if( CAST(machines AS UNSIGNED) = 0, 1, cast(machines as unsigned))

                when length(machines)-length(replace(machines,',','')) = 1
                  then ( if( CAST(SUBSTRING_INDEX(machines, ',', 1) AS UNSIGNED) = 0, 1, cast(SUBSTRING_INDEX(machines, ',', 1) as unsigned))) +  
                       (if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 1),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 1),','),'') as unsigned)))

                when length(machines)-length(replace(machines,',','')) = 2
                  then ( if( CAST(SUBSTRING_INDEX(machines, ',', 1) AS UNSIGNED) = 0, 1, cast(SUBSTRING_INDEX(machines, ',', 1) as unsigned))) +  
                       (if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 1),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 1),','),'') as unsigned))) +
                       (if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 2),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 2),','),'') as unsigned)))


             end) AS ukupno
FROM   izvestaji
#WHERE  projekatid='8'
#AND    datum='2019-10-03'

您可以获得2到M的其他情况，包括类似情况

when length(machines)-length(replace(machines,',','')) = M
                      then ( if( CAST(SUBSTRING_INDEX(machines, ',', 1) AS UNSIGNED) = 0, 1, cast(SUBSTRING_INDEX(machines, ',', 1) as unsigned))) +  
                           (if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 1),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 1),','),'') as unsigned))) +
                           (if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 2),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', 2),','),'') as unsigned))) +

...

(if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', j),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', j),','),'') as unsigned))) +


...

(if( CAST(replace(machines , concat(SUBSTRING_INDEX(machines, ',', M),','),'') AS UNSIGNED) = 0, 1, cast(replace(machines , concat(SUBSTRING_INDEX(machines, ',', M),','),'') as unsigned)))

如果您看到案例1或案例2，我认为您可以理解。

Answer 2

您可以使用sign()，abs()，substr()，instr()和length()函数的组合，并可以通过{{1} }公式：

string + 0

Demo

有没有办法只对行中的整数求和，然后对没有整数的字符串求和

2 个答案: