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
>> >
>> > > --
>> > > You received this message because you are subscribed to the Google
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>> > > To post to this group, send email to Oracle-PLSQL@googlegroups.com
>> > > To unsubscribe from this group, send email to
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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
>>
>> --
>> You received this message because you are subscribed to the Google
>> Groups "Oracle PL/SQL" group.
>> To post to this group, send email to Oracle-PLSQL@googlegroups.com
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>>
>
>

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