Oracle SQL：以数字开头的搜索列

及其与此请求工作得很好：

WHERE trim(u_ods_val3.ods_itn_PHRSBMO.NO_ART_TECH_OI) IS NOT NULL 
    AND (SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1)='0' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='1' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='2 ' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='3' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='4' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='5' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='6' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='7' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='8' 
    OR SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) ='9')

但是实在是太长了。

谢谢你的帮助。

来源

2012-10-01 astrotouf

使用[LIKE]（http://docs.oracle.com/cd/B19306_01/server.102/b14200/conditions007.htm），它正是为了这个 – Yaroslav

尝试使用in：

WHERE trim(u_ods_val3.ods_itn_PHRSBMO.NO_ART_TECH_OI) IS NOT NULL 
    AND SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) in ('0','1','2','3','4','5','6','7','8','9')

来源

2012-10-01 09:20:37 Parado

Regexp_like将是方便和更短的

where regexp_like(trim(col_name), '^[0-9]')

或使用字符类

where regexp_like(trim(col_name), '^[[:digit:]]')

来源

2012-10-01 09:21:26

BETWEEN是你所需要的！（！NOT NULL在这种情况下隐含的）

WHERE SUBSTR(u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI,0,1) between '0' and '9'

如果你对此列的索引，并且不介意这种解决方案的小dirtyness，你甚至可以加速这一过程：

WHERE u_ods_val3.ODS_ITN_PHRSBMO.NO_ART_TECH_OI between '0' and '9~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~'

这假定NO_ART_TECH_OI不包含ASCII码大于126的字符。

来源

2012-10-01 09:28:29

Oracle SQL：以数字开头的搜索列

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