Oracle SQL:以数字开头的搜索列

Question

我想找到列值以数字开头的所有行.

它适用于此请求:

    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')

但它太长了.

谢谢您的帮助.

Answer 1

Regexp_like会派上用场,而且会更短

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

或使用字符类

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

Answer 2

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')