oracle - 在 Dash 之前删除字符

Question

我要从破折号 (-) 前后的字符串中提取字符。到目前为止，我能够使用下面的代码返回破折号之前的所有数字：

SELECT 
INSTR('100-7', '-'),
SUBSTR('100-7', 1, INSTR('100-7', '-')-1)
FROM dual;

现在我正在尝试找到一种方法来返回破折号后的所有字符。所以如果我有一个字符串：

20-150

查询应该只返回 150。无论如何我可以做到这一点吗？顺便说一句，我正在使用 Oracle。多谢你们 ：）

Answer 1

这将在破折号后返回所有内容

SELECT SUBSTR(value, INSTR(value, '-')+1) invalue
  FROM (SELECT '20-150' value FROM dual) t;

输出：

| INVALUE |
-----------
|     150 |

sqlfiddle

并将两者放在一起

SELECT SUBSTR(value, 1, INSTR(value, '-') - 1) lvalue, 
       SUBSTR(value, INSTR(value, '-') + 1) rvalue
  FROM (SELECT '20-150' value FROM dual) t;

输出：

| LVALUE | RVALUE |
-------------------
|     20 |    150 |

sqlfiddle

Answer 2

使用REGEXP_REPLACE：

SELECT REGEXP_REPLACE('20-150','(.*)-','') 
FROM dual;

Answer 3

使用 REGEXP_SUBSTR ：

SELECT REGEXP_SUBSTR ('20-15055555555', '[^-]+', 1, 2) rep
  FROM DUAL;

你可以试试这个

Answer 4

查询应该是这样的。

select substr('P_A_VA1234_1_EE',(INSTR('P_A_VA1234_1_EE','_',1,2)+1),(INSTR('P_A_VA1234_1_EE','_',1,4)-(INSTR('P_A_VA1234_1_EE','_',1,2)+1)))original from dual

select substr(shipment_gid,(INSTR(shipment_gid,'_',1,2)+1),(INSTR(shipment_gid,'_',1,4)-(INSTR(shipment_gid,'_',1,2)+1)))original from dual

SUBSTR(S.SHIPMENT_GID,8,8) LOADID,

select substr(shipment_gid,(INSTR(shipment_gid,'_',1,2)+1),(INSTR(shipment_gid,'_',1,4)-(INSTR(shipment_gid,'_',1,2)+1)))load,shipment_gid from shipment where shipment_gid like '%DHL/CLS.CLS_CV_%';

SUBSTR(S.SHIPMENT_GID,(INSTR(S.SHIPMENT_GID,'_',1,2)+1),(INSTR(S.SHIPMENT_GID,'_',1,4)-(INSTR(S.SHIPMENT_GID,'_',1,2)+1)))