Need help with regular Expression (RegEx)

Try to wrap your head around a regular Expression for the following format example: 0022-C-4452 OR 0022-C-4452-C

* The 4 digits are always numbers

* The last 5 digits are alpha numeric

* Last (if used) digit is always 'C' (in reference to the second structure)

Hold the dashes for "auto fill" if possible? This would be in the Custom Format? Sequence of keys? Or Validation? I appreciate any help!

I still think that you did not correctly describe your problem.

1 figures of 4 characters.

Optional separator.

Then 6 alphanumeric characters and ' - '.

1 figures of 4 characters.

Optional separator.

Then 6 alphanumeric characters and ' - '.

Optional separator.

Character 'C '.

Note that the "-" is not a number and not an alphabetic character. It is a white space character.

Try:

function {MyRe (cString)
var cFormatted = "";
var RE_MyCode0 = /^(\d{4})[-.]) {0,1} ([A-Za-a0 - 9-] {6}) $/;
var RE_MyCode1 = /^(\d{4})[-.]) {0,1} ([A-Za-a0 - 9-] {6}) [-.] {0,1} ([C]) $/;
If (RE_MyCode0.test (CString) == true) {}
cFormatted = RegExp. $1 + '-' + RegExp. $2;
}
If (RE_MyCode1.test (CString) == true) {}
cFormatted = RegExp. $1 + '-' + RegExp. $2 + '-' + RegExp. $3;
}
Return cFormatted;
} / / end of MyRe function;

some tests;
var MyString = "0022-C-4452; good channel;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));

var MyString = "0022-C-4452-C; string of Goo;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));

var MyString = "A022-C-4452" / / bad string;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));

var MyString = '0022-4452-CZ' / / bad string;
Console.println ("Input:" + MyString);
Console.println ("result:" + MyRe (MyString));

Tags: Acrobat

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Need help with regular expressions

Hi all

I need your help, because I have no ideas more...

I have the following problem: in the column of the database table, I have the string with the names of files already uploaded to the database. For example: + File1_V01.txt +, + File1_v02.txt +, + File1_01_v01.txt +, etc. The string _vxx * + or _Vxx * + (non-case sensitive) represents the version of the file.

Now the problem: I'll upload the file with the name + File1_v02.txt + (already exists in the table).
If the file name already exists in the table the pl/sql function should get the name of the file with the following version number. In my case it takes + File1_v03.txt +.

Is it possible to do this using SELECT with regular expressions?

Best regards and thanks!

with t as (
  select 'File_V05.txt' fn from dual union all
  select 'File_V04.txt' fn from dual union all
  select 'File2_v03.doc' fn from dual union all
  select 'File2_v115.doc' fn from dual union all
  select 'File2_v15.doc' fn from dual union all
  select 'File1_v03.doc' fn from dual union all
  select 'File1_v115.doc' fn from dual union all
  select 'File1_v999.doc' fn from dual union all
  select 'File2.doc' fn from dual union all
  select 'File2_v05.doc' fn from dual union all
  select 'File1_v01.txt' fn from dual union all
  select 'File1_v02.txt' fn from dual union all
  select 'File1_v1.txt' fn from dual union all
  select 'File1_v1.doc' fn from dual union all
  select 'File1_v2.txt' fn from dual union all
  select 'File2_v01.doc' fn from dual union all
  select 'File2_v02.doc' fn from dual union all
  select 'File1_ABC_v01_DEF.docx' fn from dual union all
  select 'File1_ABC_V02_ABC.docx' fn from dual union all
  select 'File1_ABC_v01_12_04_17.docx' fn from dual union all
  select 'ABC_V1_QWERT.pdf' fn from dual
)

select fn,
case when fn!=fn_new then
 last_value(fn_new)
 over(partition by regexp_replace(upper(fn),'V[[:digit:]]+','') --(.*?V0*)([1-9]+)(\..*?)$
 order by nv
 rows between unbounded preceding and unbounded following
 )
else fn
end fn_new
from (
    select
    case when v-1 <= 0 then fn
    else
           regexp_replace (fn,
                    '(_v|_V)(\d*)',
                    case
                    when length(substr(fn,v+1,p-v-1)+1) > (p-v-1)
                    then '\1'||to_char(substr(fn,v+1,p-v-1)+1)
                    else '\1'||lpad(substr(fn,v+1,p-v-1)+1,p- v-1,0)
                    end
           )
    end fn_new
    ,fn
    ,case when v-1 <= 0 then -1
     else
        substr(fn,v+1,p- v-1)+1
     end nv
    from (
        select fn, regexp_instr(upper(fn),'_V[[:digit:]]+',1,1,1) p, instr(upper(fn),'_V')+1 v from t
    )
)
order by fn

FN     FN_NEW
ABC_V1_QWERT.pdf     ABC_V2_QWERT.pdf
File_V04.txt     File_V06.txt
File_V05.txt     File_V06.txt
File1_ABC_v01_DEF.docx     File1_ABC_v02_DEF.docx
File1_ABC_v01_12_04_17.docx     File1_ABC_v02_12_04_17.docx
File1_ABC_V02_ABC.docx     File1_ABC_V03_ABC.docx
File1_v01.txt     File1_v3.txt
File1_v02.txt     File1_v3.txt
File1_v03.doc     File1_v1000.doc
File1_v1.doc     File1_v1000.doc
File1_v1.txt     File1_v3.txt
File1_v115.doc     File1_v1000.doc
File1_v2.txt     File1_v3.txt
File1_v999.doc     File1_v1000.doc
File2.doc     File2.doc
File2_v01.doc     File2_v116.doc
File2_v02.doc     File2_v116.doc
File2_v03.doc     File2_v116.doc
File2_v05.doc     File2_v116.doc
File2_v115.doc     File2_v116.doc
File2_v15.doc     File2_v116.doc

Need help with regular expression
I have a string like: separation by a comma as G, H, L

the table has values like this

Col1 Col2
ROW1 G
ROW2 F
ROW3 L

What is trying to achieve is to find out if G or H or L in the string of separated by commas are in col2. is it possible by using regular expressions or not we do split and the loops?

Thanks in advance.
Hello

ora1001 wrote:
I have a string like: separation by a comma as G, H, L

the table has values like this

Col1 Col2
ROW1 G
ROW2 F
ROW3 L

What is trying to achieve is to find out if G or H or L in the string of separated by commas are in col2. is it possible by using regular expressions or not we do split and the loops?

Thanks in advance.

You don't even need regular expressions
```
INSTR ( ',' || col1 || ','
      , ',' || col2 || ','
      )
```
will be greater than 0 if (and only if) co12 is one of the elements of col1.

If you're curious, a way of using regular expressions is
```
REGEXP_LIKE ( col1
         , '(^|,)' || col_2 || '($|,)'
         )
```
but it will be less effective than Instr.

Published by: Frank Kulash, November 5, 2010 11:38

HELP WITH REGULAR EXPRESSIONS
Hello
I'm just experimenting with regular expressions

My information in table are as shown below
```
TABLE NAME: -  EMP_TEST

STRUCTURE WITH DATA

  X_NAME
---------------------
12456com
ab1245com
AXM4554.com
```
Now, I pulled the following query to find the rows that have 1
```
select *
from emp_test
where regexp_like (x_name,'[1]')

RESULT AS EXPECTED : -


  X_NAME
---------------------
12456com
ab1245com
```
Now, I want to do a query that will give me tose lines that do not have 1

Initially I have run the following query
```
select *
from emp_test
where regexp_like (x_name,'[^1]')

RESULT NOT AS EXPECTED : -

  X_NAME
---------------------
12456com
ab1245com
AXM4554.com
```
After a bit of searching on Google I found an application that works as expectd
```
select *
from emp_test
where regexp_like (x_name,'^[^1]*$')

RESULT AS EXPECTED : -

  X_NAME
---------------------
AXM4554.com
```
Where my basic question is that all seeking a 1 in a row we do not put the beginning ^ end $ and so on and the
Works of query.
So why do [^ 1] does not work as shown, while excluding an amd why is ^ [^ 1] * $ necessary while excluding a
Hello

Elessar wrote:
So why do [^ 1] does not work as described

Because regexp_like(str,'[^1]') means: search str who have at least one character that is not 1.
When regexp_like(str,'^[^1]*$') means: search str done characters that are not 1 start (^) at the end ($)

have dell dimension 4700 xp need help with outlook express I have error message and can not send email

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I have

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How to add your Hotmail e-mail account to Microsoft Outlook Express
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If you don't want to use OE to access Hotmail and want to just use IE to access Hotmail, then just copy the link and paste it into a new message from Hotmail.

Need help with GREP expressions

I know - KNOW - there is a way to do it, but keep my GREP expressions.
I want to add an em dash before any sentence that is in a paragraph style. So I obviously want the em dash to be in the 'start of the paragraph' location. Lack of keep all my GREP searches. Can anyone help?

Adobe of the 'start of the subsection' implementation is just a little off the coast of standard GREP (which, in itself, is less a set of hard and fast rules but rather as "guidelines"). Yes - you can Search for beginning of paragraph using plain ^, but you cannot use it with a replace operation.

The trick is to give something to look for in InDesign. Search
```
^.
```
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```
~_$0
```
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Don't forget to put your paragraph style in the box of formatting to find.

Of workaround works around the failure of implementation, because indeed, you want to add something to an existing text. Unfortunately, another pretty standard GREP to search for empty paragraphs-^ $-does not work, because then you have no room for a wildcard... (and Yes, you pouvez use a hard return instead of code \r is first or last, but then it will not work at the beginning or at the end of the story). It's one of those things that I wish to see corrected with CS5.

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Please help with regular expression

Hello

With the help of my previous answers of the detachment, Re: procedure to extract several substring of a string , I've updated the query. But I don't get the answer you want in all cases. Could you please help me? My query is based on the previous announcement. Any other way to do this?
I'd really appreciate it.

select 
       ltrim ( regexp_substr(txt, '\[(\w+)', 1, level), '[')      as id, /* id is number */
       ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '\w+-*\d*', 1, 1), ':')  as qid, /* Qid could be char/number/space any combination except ':' */
       ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '\w+', 1, 2), ':')      as num,
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from  (
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                        from     dual
         )
        connect by level <= length(regexp_replace(txt, '[^[]'));

I should get:

ID             QID          NUM         EFFDATE
10946     M100     N     
10947     Q1222     N     
38198     PPP-2     N     
13935     PPP-6     N     
38244     QQQ-4     Y     01-JAN-10

But, be

ID             QID          NUM          EFFDATE
10946     M100     N     
10947     Q1222     N     
38198     PPP-2     2     
13935     PPP-6     6     
38244     QQQ-4     4     01-JAN-10

Thank you

Hello

If the column number is wrong, isn't it?
Describe what should be the num column. For example "num is the 3rd part of :-delimited list placed in square brackets.

If this is what you want, and then change the definition of number of

...                     ltrim ( regexp_substr(ltrim ( regexp_substr(txt, ':[^]]+', 1, level), ':'), '\w+', 1, 2), ':')      as num,

...                      REGEXP_SUBSTR  ( REGEXP_SUBSTR ( txt
                                      , '[^]]+'
                                            , 1
                                   , LEVEL
                                   )
                         , '[^:]+'
                         , 1
                         , 3
                         )       AS num,

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Delete all rules for Firefox in the list of permissions in the firewall and leave your firewall again ask permission to get full unlimited access to the internet for Firefox and the plugin-container and the update process.

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[click on images to enlarge]

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Then once approved, I need help to determine which component is causing the installation to fail and perhaps this element can be omitted from the installation:

If this element is not required, then maybe FCP6 can be installed successfully without it. And if that omitted element is necessary, perhaps a manual method to install it can be determined by pacifists.

It is my goal to help those who need to install and use FCP6 on their new Macs running El Capitan.

Here is the post on MacRumors with pre-installed Setup log:

http://forums.MacRumors.com/threads/i-need-help-with-an-installation-failure-to-interpret-and-troubleshoot-an-Installer-log.1954786/#post-22541389

Need help with regular Expression (RegEx)

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