sql - ORACLE/SYBASE SQL 说明

Question

数据长度在特定字段中是随机的，如下所示：

我只想选择特殊字符之前的字符。

样品：

123456_3
454656484784_13
15051514745_7

预期结果如下：

123456
454656484784
15051514745

Answer 1

获得第一部分

select substr(your_column, 1, INSTR(your_column, '_') - 1) 
from your_table

得到最后一部分

select substr(your_column, INSTR(your_column, '_') + 1) 
from your_table

SQLFiddle 演示

Answer 2

一般示例：

substring(your_data, 0, instr(your_data, '_'))

Answer 3

快速而肮脏 - 您可以通过 Oracle/Sybase 程序或类似的程序一步一步弄清楚如何做到这一点......

-- Ignore 0 or negative numbers --
-- This gives you your end pos for substr --
SELECT distinct(Instr('123456_3 454656484784_13 15051514745_7', '_', LEVEL)-1) substr_end_pos
  FROM dual
  CONNECT BY LEVEL < length('123456_3 454656484784_13 15051514745_7') 
  /

 -- This is start pos for substr --
 SELECT Distinct(Instr('123456_3 454656484784_13 15051514745_7', ' ', LEVEL)+1)  substr_start_pos
   FROM dual
   CONNECT BY LEVEL < length('123456_3 454656484784_13 15051514745_7')
  /

 -- These are substr start/end numbers returned by above queries --
 Select Substr('123456_3 454656484784_13 15051514745_7',  1, 6) From dual -- substr_start_pos=1, substr_end_pos=6
 /

 Select Substr('123456_3 454656484784_13 15051514745_7',  10, 21-9) From dual -- substr_start_pos=10, substr_end_pos=21-substr_start_pos-1
 /

 Select Substr('123456_3 454656484784_13 15051514745_7',  26, 36-25) From dual -- substr_start_pos=26 substr_end_pos=36-substr_start_pos-1
/