Thanks for the great solution. This is really simple and powerful.
Nick
On Sep 30, 3:49 pm, Michael Moore <michaeljmo...@gmail.com> wrote:
> Here is a better solution created by my friend Shaun Batterton.
>
> SQL> with one as ( select 'This is it' x from dual),
>
> 2 two as ( select 'ThiS is not' x from dual)
>
> 3 select sum(decode(substr(one.x, level, 1) , substr(two.x, level, 1), 0,
> 1)) cnt_diff
>
> 4 from one,two
>
> 5 connect by level<= greatest(length(one.x), length(two.x));
>
> On Thu, Sep 29, 2011 at 3:47 PM, Michael Moore <michaeljmo...@gmail.com>wrote:
>
>
>
> > You could do this if you really hated the world ...
>
> > SELECT COUNT(*)
> > FROM (SELECT *
> > FROM (SELECT ROWNUM a_rnum, SUBSTR(nums, 1, INSTR(nums, ',') - 1)
> > a_num
> > FROM (SELECT n, SUBSTR(val, 1 + INSTR(val, ',', 1, n)) nums
> > FROM (SELECT ROWNUM AS n, list.val
> > FROM (SELECT
> > REGEXP_REPLACE('don...@5starcoxerage.com',
> > '(.)', ',\1')
> > || ','
> > val
> > FROM DUAL) list
> > CONNECT BY LEVEL <
> > LENGTH(list.val) -
> > LENGTH(REPLACE(list.val,
> > ',', ''))))) a
> > FULL OUTER JOIN
> > (SELECT ROWNUM b_rnum, SUBSTR(nums, 1, INSTR(nums, ',') - 1)
> > b_num
> > FROM (SELECT n, SUBSTR(val, 1 + INSTR(val, ',', 1, n)) nums
> > FROM (SELECT ROWNUM AS n, list.val
> > FROM (SELECT REGEXP_REPLACE('
> > doa...@5starcoverage.co',
> > '(.)', ',\1')
> > || ','
> > val
> > FROM DUAL) list
> > CONNECT BY LEVEL <
> > LENGTH(list.val) -
> > LENGTH(REPLACE(list.val,
> > ',', ''))))) b
> > ON a_rnum = b_rnum)
> > WHERE a_num != b_num OR a_num IS NULL OR b_num IS NULL;
>
> > On Thu, Sep 29, 2011 at 2:40 PM, ddf <orat...@msn.com> wrote:
>
> >> On Sep 28, 9:34 pm, Naveen <naveen.kuramse...@gmail.com> wrote:
> >> > string compare
>
> >> > On Wed, Sep 28, 2011 at 8:18 AM, Ninja Li <nickli2...@gmail.com> wrote:
> >> > > Hi,
>
> >> > > I have the following two email addresses:
>
> >> > > don...@5starcoverage.com
> >> > > doa...@5starcoverage.com
>
> >> > > The third and fourth characters of the two email addresses are not
> >> > > matching.
>
> >> > > Is there any way to use SQL to get the number of the characters not
> >> > > matching ( 2 ) in the the two string, e.g. using regexp functions? or
> >> > > I have to use PL/SQL.
>
> >> > > Thanks in advance.
>
> >> > > Nick
>
> >> > > --
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>
> >> > --
> >> > naveen- Hide quoted text -
>
> >> > - Show quoted text -
>
> >> That is a rather brief response, with no examples provided. Please
> >> provide an example so we can all see what this response means.
>
> >> David Fitzjarrell
>
> >> --
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>
> - Show quoted text -
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