WSA S370 CLI - Use regular expression with "tail"

Hello
it is possible to somehow filter output from "tail" access logs to match string pattern - similar to "Enter the regular expression" for grep? We have lot of users and sometime I wanted to see real on-the-fly traffic via access logs, but I wanted to point specific IP/username/URL ... but now I am not possible as using "tail" I am able only to specify log type.
> version
Current Version
===============
Product: Cisco IronPort S370 Web Security Appliance
Model: S370
Version: 7.5.0-833
thanks
michal

Hi Michal,
Let me try to rephrase.
On the WSA:
'tail' will tail all the access logs in real time.
'grep' can tail the access logs in real time while grepping for the RegEx you'd like. Be aware that when you issue the 'grep', there will be an option in which you must answer YES to (tail the output).
-Vance

Similar Messages

Rplacing space with &nbsb; in html using regular expressions

Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What number does this represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

sorry i missed some information in above,The modified information was in red color
Hi
I want to replace space with &nbsb; in HTML.
I used the below method to replace space in my html file.
var spacePattern11:RegExp =/(\s)/g;
str= str.replace(spacePattern," "
Here str varaible contains below html file.In this html file i want to replace space present between " What number does this represents" with &nbsb;
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADING="2"> What number does this Roman numeral represents MDCCCXVIII ?</TEXTFORMAT>
</body>
</html>
But by using the above regular expression i am getting like this.
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head><body>
<TEXTFORMAT LEADING="2"></TEXTFORMAT><TEXTFORMAT LEADIN G="2"> What number does this represents</TEXTFORMAT>
</body>
</html>
Here what happening means it was replacing space with &nbsb; in HTML tags also.But want to replace space with &nbsb; present in the outside of the HTML tags.I want like this using regular expressions in FLEX
<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>
<body>What&nbsb;number&nbsb;does&nbsb;this&nbsb;represents</body>
</html>
Hi,Please give me the solution to slove the above problem using regular expressions
Thanks in Advance to all
Regards
ssssssss

Regular expressions with boolean connectives (AND, OR, NOT) in Java?

I'd like to use regular expression patterns that are made up of simple regex patterns connected via AND, OR, or NOT operators, in order to do some keyword-style pattern matching.
A pattern could look like this:
(.*Is there.*) && (.*library.*) && !((.*badword.*) || (^$))
Is there any Java regex library that allows these operators?
I know that in principle these operators should be available, since Regular languages are closed under union, intersection, and complement.

AND is implicit,
xy -- means x AND yThat's not what I need, though, since this is just
concatenation of a regex.
Thus, /xy/ would not match the string "a y a x",
because y precedes x.So it has to contain both x and y, but they could be
in any order?
You can't do that easily or generally.
"x.*y|y.*x" wouldll work here, but obviously
it will get ugly factorially fast as you add more
terms.You got that right: AND means the regex operands can appear in any order.
That's why I'm looking for some regex library that does all this ugly work for me. Again, from a theoretical point of view, it IS possible to express the described semantics of AND with regular expressions, although they will get rather obfuscated.
Unless somebody has done something similar in java (e.g., for C++, there's Ragel: http://www.cs.queensu.ca/~thurston/ragel/) , I will probably use some finite-state-machine libraries and compile the complex regex's into automata (which can be minimized using well-defined operations on FSMs).
>
You'd probably just be better off doing multiple
calls to matches() or whatever. Yes, that's another possibility, do the boolean operators in Java itself.
Of course, if you
really are just looking for literals, then you can
just use str.contains(a) && !str.contains(b) &&
(str.contains(c) || str.contains(d)). You don't
seem to need regex--at least not from your example.OK, bad example, I do have "real" regexp's in there :)

Trying to use regular expressions to convert names to Title Case

I'm trying to change names to their proper case for most common names in North America (esp. the U.S.).
Some examples are in the comments of the included code below.
My problem is that *retName = retName.replaceAll("( [^ ])([^ ]+)", "$1".toUpperCase() + "$2");* does not work as I expect. It seems that the toUpperCase method call does not actually do anything to the identified group.
Everything else works as I expect.
I'm hoping that I do not have to iterate through each character of the string, upshifting the characters that follow spaces.
Any help from you RegEx experts will be appreciated.
{code}
* Converts names in some random case into proper Name Case. This method does not have the
* extra processing that would be necessary to convert street addresses.
* This method does not add or remove punctuation.
* Examples:
* DAN MARINO --> Dan Marino
* old macdonald --> Old Macdonald <-- Can't capitalize the 'D" because of Ernst Mach
* ROY BLOUNT, JR. --> Roy Blount, Jr.
* CAROL mosely-BrAuN --> Carol Mosely-Braun
* Tom Jones --> Tom Jones
* ST.LOUIS --> St. Louis
* ST.LOUIS, MO --> St. Louis, Mo <-- Avoid City Names plus State Codes
* This is a work in progress that will need to be updated as new exceptions are found.
public static String toNameCase(String name) {
* Basic plan:
* 1. Strategically create double spaces in front of characters to be capitalized
* 2. Capitalize characters with preceding spaces
* 3. Remove double spaces.
// Make the string all lower case
String retName = name.trim().toLowerCase();
// Collapse strings of spaces to single spaces
retName = retName.replaceAll("[ ]+", " ");
// "mc" names
retName = retName.replaceAll("( mc)", " $1");
// Ensure there is one space after periods and commas
retName = retName.replaceAll("(\\.|,)([^ ])", "$1 $2");
// Add 2 spaces after periods, commas, hyphens and apostrophes
retName = retName.replaceAll("(\\.|,|-|')", "$1 ");
// Add a double space to the front of the string
retName = " " + retName;
// Upshift each character that is preceded by a space
// For some reason this doesn't work
retName = retName.replaceAll("( [^ ])([^ ]+)", "$1".toUpperCase() + "$2");
// Remove double spaces
retName = retName.replaceAll(" ", "");
return retName;
Edited by: FuzzyBunnyFeet on Jan 17, 2011 10:56 AM
Edited by: FuzzyBunnyFeet on Jan 17, 2011 10:57 AM

Hopefully someone will still be able to provide a RegEx solution, but until that time here is a working method.
Also, if people have suggestions of other rules for letter capitalization in names, I am interested in those too.
* Converts names in some random case into proper Name Case. This method does not have the
* extra processing that would be necessary to convert street addresses.
* This method does not add or remove punctuation.
* Examples:
* CAROL mosely-BrAuN --> Carol Mosely-Braun
* carol o'connor --> Carol O'Connor
* DAN MARINO --> Dan Marino
* eD mCmAHON --> Ed McMahon
* joe amcode --> Joe Amcode <-- Embedded "mc"
* mr.t --> Mr. T <-- Inserted space
* OLD MACDONALD --> Old Macdonald <-- Can't capitalize the 'D" because of Ernst Mach
* old mac donald --> Old Mac Donald
* ROY BLOUNT,JR. --> Roy Blount, Jr.
* ST.LOUIS --> St. Louis
* ST.LOUIS,MO --> St. Louis, Mo <-- Avoid City Names plus State Codes
* Tom Jones --> Tom Jones
* This is a work in progress that will need to be updated as new exceptions are found.
public static String toNameCase(String name) {
 * Basic plan:
 * 1. Strategically create double spaces in front of characters to be capitalized
 * 2. Capitalize characters with preceding spaces
 * 3. Remove double spaces.
 // Make the string all lower case
 String workStr = name.trim().toLowerCase();
 // Collapse strings of spaces to single spaces
 workStr = workStr.replaceAll("[ ]+", " ");
 // "mc" names
 workStr = workStr.replaceAll("( mc)", " $1 ");
 // Ensure there is one space after periods and commas
 workStr = workStr.replaceAll("(\\.|,)([^ ])", "$1 $2");
 // Add 2 spaces after periods, commas, hyphens and apostrophes
 workStr = workStr.replaceAll("(\\.|,|-|')", "$1 ");
 // Add a double space to the front of the string
 workStr = " " + workStr;
 // Upshift each character that is preceded by a space and remove double spaces
 // Can't upshift using regular expressions and String methods
 // workStr = workStr.replaceAll("( [^ ])([^ ]+)", "$1"toUpperCase() + "$2");
 StringBuilder titleCase = new StringBuilder();
 for (int i = 0; i < workStr.length(); i++) {
 if (workStr.charAt(i) == ' ') {
 if (workStr.charAt(i+1) == ' ') {
 i += 2;
 while (i < workStr.length() && workStr.charAt(i) == ' ') {
 titleCase.append(workStr.charAt(i++));
 if (i < workStr.length()) {
 titleCase.append(workStr.substring(i, i+1).toUpperCase());
 } else {
 titleCase.append(workStr.charAt(i));
 return titleCase.toString();
{code}

Using regular expressions in java

Does anyone of you know a good source or a tutorial for using regular expressions in java.
I just want to look at some examples....
Thanks

thanks a lot... i have one more query
Boundary matchers
^      The beginning of a line
$      The end of a line
\b      A word boundary
\B      A non-word boundary
\A      The beginning of the input
\G      The end of the previous match
\Z      The end of the input but for the final terminator, if any
\z      The end of the input
if i want to use the $ for comparing with string(text) then how can i use it.
Eg if it is $120 i got a hit
but if its other than that if should not hit.

Changeparticular characters in a string by using regular expressions ...

Hello Everyone,
I am trying to write a function by using oracles regular expression function REGEXP_REPLACE but I could not succed till now.
My problem as follows, I have a text in a column for example let say 'sdfsdf Sdfdfs Sdfd' I want replace all s and S characters with X and make the text look like 'XdfXdf XdfdfX Xdfd'.
Is it possible by using regular expressions in oracle ?
Can you give me some clues ?
Thank you

SSU wrote:
Hello Everyone,
I am trying to write a function by using oracles regular expression function REGEXP_REPLACE but I could not succed till now.
My problem as follows, I have a text in a column for example let say 'sdfsdf Sdfdfs Sdfd' I want replace all s and S characters with X and make the text look like 'XdfXdf XdfdfX Xdfd'.
Is it possible by using regular expressions in oracle ?
Can you give me some clues ?
Thank you
SQL> SELECT
2 regexp_replace('sdfsdf Sdfdfs Sdfd','s|S','X') from dual;
REGEXP_REPLACE('SD
XdfXdf XdfdfX XdfdRegards,
Achyut

Using regular expressions for validating time fields

Similar to my problem with converting a big chunk of validation into smaller chunks of functions I am trying to use Regular Expressions to handle the validation of many, many time fields in a flexible working time sheet.
I have a set of FormCalc scripts to calculate the various values for days, hours and the gain/loss of hours over a four week period. For these scripts to work the time format must be in HH:MM.
Accessibility guidelines nix any use of message box pop ups so I wanted to get around this by having a hidden/visible field with warning text but can't get it to work.
So far I have:
var r = new RegExp(); // Create a new Regular Expression Object
r.compile ("^[00-99]:\\] + [00-59]");
var result = r.test(this.rawValue);
if (result == true){
true;
form1.flow.page.parent.part2.part2body.errorMessage.presence = "visible";
else (result == false){
false;
form1.flow.page.parent.part2.part2body.errorMessage.presence = "hidden";
Any help would be appreciated!

Date and time fields are tricky because you have to consider the formattedValue versus the rawValue. If I am going to use regular expressions to do validation I find it easier to make them text fields and ignore the time patterns (formattedValue). Something like this works (as far as my very brief testing goes) for 24 hour time where time format is HH:MM.
// form1.page1.subform1.time_::exit - (JavaScript, client)
var error = false;
form1.page1.subform1.errorMsg.rawValue = "";
if (!(this.isNull)) {
var time_ = this.rawValue;
if (time_.length != 5) {
    error = true;
else {
    var regExp = /^([01]?[0-9]|2[0-3]):[0-5][0-9]$/;
    if (!(regExp.test(time_))) {
      error = true;
if (error == true) {
form1.page1.subform1.errorMsg.rawValue = "The time must be in the format HH:MM where HH is 00-23 and MM is 00-59.";
form1.page1.subform1.errorMsg.presence = "visible";
Steve

Using Regular Expressions To Remove Characters JDK 1.4

I want to write a regular expression to remove all commas in a string of text.
string is:
1,000
or 1,000,000
I want it to return 1000 and 1000000.
I have tried some but I am just starting with Regular Expressions.
Please Help!

Try this tutorial: Linux : Education : Tutorials
Using regular expressions
David Mertz
President, Gnosis Software, Inc.
September 2000
http://www-105.ibm.com/developerworks/education.nsf/linux-onlinecourse-bytitle/6C2B4863702F592B8625696200589C5B?OpenDocument

Using Regular Expressions to replace Quotes in Strings

I am writing a program that generates Java files and there are Strings that are used that contain Quotes. I want to use regular expressions to replace " with \" when it is written to the file. The code I was trying to use was:
String temp = "\"Hello\" i am a \"variable\"";
temp = temp.replaceAll("\"","\\\\\"");
however, this does not work and when i print out the code to the file the resulting code appears as:
String someVar = ""Hello" i am a "variable"";
and not as:
String someVar = "\"Hello\" i am a \"variable\"";
I am assumming my regular expression is wrong. If it is, could someone explain to me how to fix it so that it will work?
Thanks in advance.

Thanks, appearently I'm just doing something weird that I just need to look at a little bit harder.

Using regular expressions to get a customized output

Hi,
I have a string/varchar variable with the data ',a,b,c,' in it.
I want the display as follows:
a
b
c
I would like to get the similar output using regular expressions.
How do I get this output using REGEXP_REPLACE or REGEXP_SUBSTR?
Please do the needful.
Thanks & Regards,
Rakshit

I remember that, however if we look closer, that one has a little flaw: The 2nd row should be null, because ",," indicates an empy field. The MODEL clause solution works just fine in this case:
with t as (select 'aaaa,,bbbb,cccc,dddd,eeee,ffff' col1 from dual)
-- end of sample data
SELECT col_new
FROM t
MODEL
 PARTITION BY (ROWNUM rn)
 DIMENSION BY (0 dim)
 MEASURES(col1, col1 col_new)
 RULES ITERATE(99) UNTIL (ITERATION_NUMBER = LENGTH(REGEXP_REPLACE(col1[0], '[^,]')))
 (col_new[ITERATION_NUMBER] = REPLACE(REGEXP_SUBSTR(col1[0], '(^|,)[^,]*', 1, ITERATION_NUMBER+1), ','))
COL_NEW
aaaa
bbbb
cccc
dddd
eeee
ffff
7 Zeilen ausgewählt.Update: I had this nagging feeling that I missed something, and there it was. If you want to see what the problem with my solution is, change the example to
with t as (select ',aaaa,,bbbb,cccc,dddd,eeee,ffff' col1 from dual)So I went back and tried to fix BlueShadows approach. Here it is:
with t as (select 'aaaa,,bbbb,cccc,dddd,eeee,ffff' txt from dual)
-- end of sample data
SELECT REPLACE(REGEXP_SUBSTR(',' || txt, ',[^,]*', 1, level), ',') col_new
FROM t
CONNECT BY level <= length(regexp_replace(txt,'[^,]*'))+1
;C.

How to fetch substring using regular expression

Hi,
I am new to using regular expression and would like to know some basic details of how to use them in Java.
I have a String example= "http://www.google.com/foobar.html#*q*=database&aq=f&aqi=g10&fp=c9fe100d9e542c1e" and would like to get the value of "q" parameter (in bold) using regular expression in java.
For the same example, when we tried using javascript:
match = example.match("/^http:\/\/(?:(?!mail\.)[^\.]+?\.)?google\.[^\?#]+(?:.*[\?#&](?:as_q|q)=([^&]+))?/i}");
document.write('
' + match);
We are getting the output as: http://www.google.com/foobar.html#q=database,*database* where the bold text is the value of "q" parameter.
In Java we are trying to get the value of the q parameter separately or atleast resembles the output given by JavaScript. Please help me resolving this issue.
Regards
Praveen

BalusC wrote:
Regex is a cumbersome solution for fixed patterns like URL's. String#substring() in combination with String#indexOf would most likely already suffice.I usually agree, although, in this case, finding the exact parameter might be difficult without a small regex, perhaps:
"\\wq=\\s*"in conjunction with Pattern/Matcher, used similarly to an indexOf() to find the start of the parameter value.
Winston

Matching substrings between square brackets using regular expressions

Hello,
I am new at Java and have a problem with regular expressions. Let me describe the issue in 3 steps:
1.- I have an english sentence. Some words of the sentence stand between square brackets, for example "I [eat] and [sleep]"
2- I would like to match strings that are in square brackets using regular expressions (java.util.regex.*;) and here is the code I have written for the task
+Pattern findStringinSquareBrackets = Pattern.compile("\\[.*\\]");+
+ Matcher matcherOfWordInSquareBrackets = findStringinSquareBrackets.matcher("I [eat] and [sleep]");+
+//Iteration in the string+
+ while ( matcherOfWordInSquareBrackets.find() )+
+{+
+ System.out.println("Patter found! :"+ outputField.getText().substring(matcherOfWordInSquareBrackets.start(), matcherOfWordInSquareBrackets.end())+""); +
+ }+
3- the result I have after running the code described in 2 is the following: *Patter found!: [eat] and [sleep]*
That is to say that not only words between square brackets are found but also the substring "and". And this is not what I want.
What I would like to have as a result is:
*Patter found!: [eat]*
*Patter found!: [sleep]*
That is to say I want to match only the words between the square brackets and nothing else.
Does somebody know how to do this? Any help would be great.
Best regards,
Abou

You can find the words by looping through the sentence and then return the substring within the indexes.
int start=0;
int end=0;
for(int i=0; i<string.length(); i++)
 if(string.substring(i,i+1).equals("[");
start=i;
if(start!=0)
if(string.substring(i,i+1).equlas("]");
end=i;
return string.substring(start,end+1);
}something like that. This code will only find the firt word however. I do not know much about regex so I cannot help anymore.
Edited by: elasolova on Jun 16, 2009 6:45 AM
Edited by: elasolova on Jun 16, 2009 6:46 AM

Checking valid e-mail's using Regular expressions

Hey buddies ,
I desperately need some help here. I need to develop a generic method that wiull use regular expressions and patterns to validate an e-mail.
Does anyonw have any idea on how to do this. And if possible please share some code with me.
Thanks a lot

You can do regular expresions in java using java.util.regex.*:
import java.util.regex.*;
   Pattern p = Pattern.compile("\\S++\\s++");
   Matcher m = p.matcher(sInputLine);
   if(m.find()){
      sUser = m.group().trim();
   }For more info on regular expressions in java, check out the javadocs:
http://java.sun.com/j2se/1.4.2/docs/api/java/util/regex/Pattern.html

Requirement to update a column by using Regular Expression

Hi All,
    I have a requirement to update a column which is having values like below code.
based on the conditinos I need to update from ‘E’ to ‘Z’. Few positions I need to update and remaining positions I need leave as it is.
How I can achive this requirement by using regular expression regexp_replace.
Actual value --> 'AEAAAEAA    EE    AA    EE    AA    EE    EEEAA    AA    AA       '
After update --> 'AZAAAZAA    EE    AA    EE    AA    EE    EEEAA    AA    AA       '

below is my requirement. I am adding the conditions dynamically as per the conditions. I dont know the position of the E. If 'E' is in any position I need to update with 'Z' if that 'E' satisfy the condition.
I dont want to update all the E's to Z's.
I want to update specific E's which satisfy the condition.
    IF l_kwhhilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,1,1) = ''E'' OR ';
        l_exp := l_exp||'Z';
    ELSE
        l_exp := l_exp||'\1';
    END IF;
    IF l_kwhilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,2,1) = ''E'' OR ';
         l_exp := l_exp||'Z';
    ELSE
        l_exp := l_exp||'\2';
    END IF;
    IF l_kvahilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,3,1) = ''E'' OR ';
         l_exp := l_exp||'Z';
    ELSE
        l_exp := l_exp||'\3';
    END IF;
    IF l_kvarhilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,4,1) = ''E'' OR ';
         l_exp := l_exp||'Z';
    ELSE
        l_exp := l_exp||'\4';
    END IF;
    IF l_todkwh1hilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,5,1) = ''E'' OR ';
         l_exp := l_exp||'Z';
    ELSE
        l_exp := l_exp||'\5';
    END IF;
    IF l_todkwh2hilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,6,1) = ''E'' OR ';
         l_exp := l_exp||'Z';
    ELSE
        l_exp := l_exp||'\6';
    END IF;
    l_exp := ''''||l_exp||'''';
    l_exp1 := '\1\2\3\4'||l_exp1;
    IF l_todkw1hilow >= 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,11,1) = ''E'' OR ';
         l_exp1 := l_exp1||'Z';
    ELSE
        l_exp1 := l_exp1||'\5';
    END IF;
    IF l_todkw2hilow = 1 THEN
        l_where := l_where||' SUBSTR(OVERRIDEFIELD,12,1) = ''E'' OR ';
         l_exp1 := l_exp1||'Z';
    ELSE
        l_exp1 := l_exp1||'\6';
    END IF;
    l_exp1 := ''''||l_exp1||'''';
IF i give 10 in the regexp_replace it is not working.
SET OVERRIDEFIELD =     REGEXP_REPLACE(SUBSTR(overridefield,1,6), ''(.)(.)(.)(.)(.)(.)'','||l_exp||')'||
                                          ' ||REGEXP_REPLACE(SUBSTR(overridefield,7,6), ''(.)(.)(.)(.)(.)(.)'','||l_exp1||')'||
                                          ' ||SUBSTR(overridefield,13)'||

Using Regular Expression

I do have a table with one varchar2 column with the following values:
COL_NAME
1.1.3.TECH.SG.1.Action.2
1.1.3.TECH.SG.1.Action.3
1.1.3.TECH.SG.1.Action.4
1.1.3.TECH.SG.1.Action.5
1.1.3.TECH.SG.1.Action.6
I need the last digit to be reduced by 1 like below:
COL_NAME NEW_VALUE
1.1.3.TECH.SG.1.Action.2 1.1.3.TECH.SG.1.Action.1
1.1.3.TECH.SG.1.Action.3 1.1.3.TECH.SG.1.Action.2
1.1.3.TECH.SG.1.Action.4 1.1.3.TECH.SG.1.Action.3
1.1.3.TECH.SG.1.Action.5 1.1.3.TECH.SG.1.Action.4
1.1.3.TECH.SG.1.Action.6 1.1.3.TECH.SG.1.Action.5
How can this be achieved by a SQL statement using Regular Expression ?
Pls suggest.
Regards
MS

You don't really need regexps, you can just use normal instr() and substr():
1 with v as (
2     select '1.1.3.TECH.SG.1.Action.2' as val from dual union all
3     select '1.1.3.TECH.SG.1.Action.3' as val from dual union all
4     select '1.1.3.TECH.SG.1.Action.4' as val from dual union all
5     select '1.1.3.TECH.SG.1.Action.5' as val from dual union all
6     select '1.1.3.TECH.SG.1.Action.6' as val from dual
7 )
8 select val
9 , substr(val, 1, instr(val, '.', -1))
10 || to_char(to_number(substr(val, instr(val, '.', -1) + 1)) + 1) new_val
11* from v
SQL> /
VAL                      NEW_VAL
1.1.3.TECH.SG.1.Action.2 1.1.3.TECH.SG.1.Action.3
1.1.3.TECH.SG.1.Action.3 1.1.3.TECH.SG.1.Action.4
1.1.3.TECH.SG.1.Action.4 1.1.3.TECH.SG.1.Action.5
1.1.3.TECH.SG.1.Action.5 1.1.3.TECH.SG.1.Action.6
1.1.3.TECH.SG.1.Action.6 1.1.3.TECH.SG.1.Action.7
5 rows selected.cheers,
Anthony

WSA S370 CLI - Use regular expression with "tail"

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