Question

我有这样的数据我见过函数和子串和LEFT，右也但它不符合我的目的

declare @t table (val varchar(50))
INSERT INTO @t(val)values ('E-001GHDEM120ENDORSEMENT'),
 ('E-001GHDEM120Renewal'),
  ('E-001GHDEM120Adjustment'),
   ('E-001GHDEM120ENDORSEMENT')
select * from @t

输出

ENDORSEMENT
Renewal
Adjustment
ENDORSEMENT

我需要在where条件中使用该语句来过滤记录

Answer 1

试试这个。从你的例子来看，这就是我的理解。

select right(val,patindex('%[0-9]%', reverse(val))-1) 
 from @t

Answer 2

尝试此选择

library(dplyr)

prod <- c('P1','P2','P3')
level <- c('L1','L2','L3')
part <- c('p1','p2','p3','p4','p5')

axis_x <- list(L1 = list('Ordering' = 'id'),
               L2 = list('Ordering' = 'id', 'Part name' = 'part'),
               L3 = list('Ordering' = 'id', 'Part name' = 'part'))

# Data for module 1
set.seed(123)
module1_df <- data.frame(prod = sample(prod,300, replace = T), 
                        level = sample(level, 300, replace = T), 
                        part = sample(part, 300, replace = T),
                        value = rnorm(300))

module1_df <- module1_df %>% 
        group_by(prod) %>% 
        mutate(id = 1:n()) %>% 
        arrange(prod, id)

# Data for module 2
set.seed(321)
module2_df <- data.frame(prod = sample(prod,300, replace = T), 
                         level = sample(level, 300, replace = T), 
                         part = sample(part, 300, replace = T),
                         value = rnorm(300))

module2_df <- module2_df %>% 
        group_by(prod) %>% 
        mutate(id = 1:n()) %>% 
        arrange(prod, id)

如何选择最后一个字符串

2 个答案: