1. 有哪些是目前質檢體系發現不了的問題(發生過,只是 是沒有事先察覺)
2. 哪些是管理層需要了解的但現有質檢體系滿足不了的
3. 現有的質檢體系介紹,哪些規
select * from (
select t.platform as platform,
t.version_id as version_id,
t.version as oversion,
regexp_substr(t.version, '^[[:digit:]\.]') as bigver,
decode(regexp_substr(version, '^[[:digit:]\.]'),4,regexp_substr(version, '^[[:digit:]](\.[[:digit:]]+)+00$'),version) as version
from dim_version t
where regexp_substr(t.version, '^[[:digit:]](\.[[:digit:]]+)+$') is not null
and t.version != 'all'
) where version is not null
select platform,pn,regexp_substr(t.pn, '^[[:alnum:]](\_)*([[:alnum:]]+)+$') ,pn_id from dim_pn t
select platform,
pn,pn_id
from dim_pn t
where regexp_substr(t.pn, '^[:alnum:]*_*[[:alnum:]]*$') is not null and t.pn !='all'
select
distinct platform,regexp_extract(version,'^([1-4])\.',1), version
from
rc_pt
where
1=1and accessday >= '20130704'
and accessday <= '20130704'
and datatype='first'
and product='online'
and platform in('anphone','win','anpad','')
and length(deviceid)=40
and regexp_extract(version,'^[1-4](\.)',1) != '.'
and version not rlike '(^[1-3]\.[0-9]+(\.[0-9]+)$)|(^4\.[0-9]\.[0-9]\.[1-9]0?00$)';
select
distinct platform,regexp_extract(version,'^([1-4])\.',1), version
from
rc_pt
where
1=1and accessday >= '20130704'
and accessday <= '20130704'
and datatype='first'
and product='online'
and platform in('win','mac','iphone','ipad','anpad','anphone')
and regexp_extract(version,'^[1-4](\.)',1) = '.'
and version rlike '(^[1-3]\.[0-9]+(\.[0-9]+)$)|(^4\.[0-9]\.[0-9](\.[1-9]0?00)?$)';
and length(deviceid)=40;
select
distinct platform,regexp_extract(version,'^([1-4])\.',1), version
from
rc_pt
where
1=1and accessday >= '20130704'
and accessday <= '20130704'
and datatype='first'
and product='online'
and platform in('win','mac','iphone','ipad','anpad','anphone')
and version rlike '(^4\.[0-9]\.[0-9])|(^4\.[0-9]\.[0-9]\.[1-9]0?00$)'
and length(deviceid)=40;
select
distinct platform,regexp_extract(version,'^([1-4])\.',1), version
from
rc_pt
where
1=1and accessday >= '20130704'
and accessday <= '20130704'
and datatype='first'
and product='online'
and platform in('win','mac','iphone','ipad','anpad','anphone')
and version regexp '^4(\.[0-9])(\.[1-9]0?00$)?'
and length(deviceid)=40;
(^[1-3]\.[0-9]+(\.[0-9]+)$)|(^4\.[0-9]\.[0-9](\.[1-9]0?00$)?)
select
distinct platform,regexp_extract(version,'^([1-4])\.',1), version
from
rc_pt
where
1=1and accessday >= '20130704'
and accessday <= '20130704'
and datatype='first'
and product='online'
and platform in('anphone')
and length(deviceid)=40
and version rlike '^[1-3](\.[0-9]+)+$';
(?(exp)yes|no)
^[1-9a-z]*_*[1-9a-z]*$
select
distinct platform,(case when platform='anphone' and (substr(pn,1,1)='1' or substr(pn,1,1)='2') then 'user-defined' else pn end) as pn
from
rc_pt
where
1=1and accessday >= '20130704'
and accessday <= '20130704'
and datatype='oc'
and product='online'
and platform in('win','mac','iphone','ipad','anpad','anphone')
and length(deviceid)=40
and pn rlike '^[a-za-z0-9_]+$';
oracle 正規表示式
with temp as select br.777 1 a from dual union all select bs.6 175 a from dual union all select d.140 supplement 1 a from dual union all select e.191 ...
oracle正規表示式
在資料庫可直接執行下面語句,但在程式中不能執行,regexp like t.ipaddress,25 0 5 2 0 4 0 9 01 0 9 0 9 25 0 5 2 0 4 0 9 01 0 9 0 9 報錯有特殊字元 在句末再加乙個 改為 regexp like t.ipaddress,25 ...
oracle正規表示式
oracle 10g支援正規表示式的四個新函式分別是 regexp like regexp instr regexp substr 和regexp replace。它們使用posix 正規表示式代替了老的百分號 和萬用字元 字元。特殊字元 匹配輸入字串的開始位置,在方括號表示式中使用,此時它表示不接...