Regular expressions and back references

Just wanted to know if anyone else noticed that.
In the javadoc of java.util.regex.Pattern in the "Back references" section it says that you need to use \n to match capturing group but it does not work. To match a capturing group one need to use a "$" sign which is not standard for this type of operation.
For example, the following code should work according to the API and most other regular expression engines:
Pattern.compile("([A-Z])").matcher("ThisIsATestString").replaceAll(" \1");
But to make this work you need to use:
Pattern.compile("([A-Z])").matcher("ThisIsATestString").replaceAll(" $1");
So, is this just a doc bug or am I missing something?
Someone have any idea why Sun choose to use the "$" sign instead of the regular "\" sign??
TIA,
Shaul

The doc you're referring to is talking about using back-refereneces within the regex, not in the replacement string. For instance, if you wanted to find all instances of things like "foo-foo" or "bar-bar", you would use a Pattern like Pattern p = Pattern.compile("([a-z]+)-\\1");For the most part, they've made the syntax the same a Perl's regexes, and that's why they use $n instead of \n in the replacement string. The replacement string is described in the Matcher javadoc.

Similar Messages

[DW CC] Regular Expression and Back Reference in "Replace with": Escape

I have a problem with escaping a character in "Replace with".
Let's assume the following situation:
Content:
foobar
Search for:
(foo)(bar)
Replace by:
$1zot$2
Result:
foozotbar
Everything is fine.
But I want to insert the number "2" in between foo and bar.
When I use
Replace by: $12$2
The result is:
$12$2
It seems that the regex engine interpretes the "$12" as a whole. And because there's no back reference with the count "12" DW cannot work correctly.
The question:
How can I "escape" the "2"?
Thanks.

Dreamweaver uses JavaScript for its Find and Replace with regex.
I've just checked the Regular Expressions Cookbook by Jan Goyvaerts and Steven Levithan (O'Reilly). It deals with exactly this sort of situation where capturing groups can be ambiguous. This is what it says:
"Java and JavaScript try to be clever with $10 [and above]. If a capturing group with the two-digit number exists in your regular expresssion, both digits are used for the capturing group. If fewer capturing groups exist, only the first digit is used to reference the group, leaving the second as a literal. Thus $23 is the 23rd capturing group, if it exists. Otherwise, it is the second capturing group followed by a literal 3."
In other words, there is no way to escape the 2 in the Replace field. It would appear there's a bug in Dreamweaver's use of capturing groups.

Regular expressions and backreference

Hello!
I am trying to use backreferences in REGEXP in the PERL-style, where I want to match my regular expression and later refer to the grouped values. I can read that those are referecenced with \1 .. \9, but I simply cant get it to work. Here is an example in PL/SQL:
SELECT REGEXP_SUBSTR(l_users.adresse,'([A-Z]+)\s+(\d+)')
INTO l_dummy_varchar2
FROM dual;
OR I could do things like:
l_dummy_varchar2 := REGEXP_SUBSTR(l_users.adresse,'([A-Z]+)\s+(\d+)');
It seems to work, but I cant figure out how to get the backreferenced value.
I would love to do things like:
dbms_output.put_line('my value ='||\1)
but this doesnt work.
Help is very much appreciated.
Best regards
Dannie

Likewise you can extract things using the
REGEXP_SUBSTR, but you don't need back
referencing...backreferencing is better than additional function (ltrim) use, and BTW be careful with this "ltrims":
SQL> set serveroutput on
SQL>
SQL> DECLARE
2       v_txt VARCHAR2(100);
3     BEGIN
4       v_txt := ltrim(regexp_substr('HERE IS AN ASCII CHARACTER', 'IS AN [[:alnum:]]*'),'IS AN ');
5       DBMS_OUTPUT.PUT_LINE('Word after IS AN: '||v_txt);
6 END;
7 /
Word after IS AN: CII
PL/SQL procedure successfully completed
SQL>
SQL> DECLARE
2       v_txt VARCHAR2(100);
3     BEGIN
4       v_txt := regexp_replace('HERE IS AN ASCII CHARACTER', 'IS AN ([[:alnum:]]*)|.','\1');
5       DBMS_OUTPUT.PUT_LINE('Word after IS AN: '||v_txt);
6 END;
7 /
Word after IS AN: ASCII
PL/SQL procedure successfully completed
SQL> -----------
VB
http://volder-notes.blogspot.com/

Can somebody help me in getting some good material for Regular Expressions and IP Community list

can somebody help me in getting some good material for Regular Expressions and IP Community list

I'm not sure what you mean by "IP Community list", but here are 3 reference sites for Regular Expressions:
Regular Expression Tutorial - Learn How to Use Regular Expressions
http://www.regular-expressions.info/tutorial.html
Regular Expressions Cheat Sheet by DaveChild
http://www.cheatography.com/davechild/cheat-sheets/regular-expressions/
Regular Expressions Quick Reference
http://www.autohotkey.com/docs/misc/RegEx-QuickRef.htm

Regular expressions and sql

I have working regular expressions and a working sql connection, but I don�t know how to stop the info from getting into the database when input doesent match the regular expression.
For instans, you put in an e-mail without an "@" and my program writes and error message. But the info still gets in to the database.
Any help would be much apreciated as I dont know where to start. If you have links or code examples that would be great to.
Thanx.

Well, the obvious answer is "only write the data to the database if the input doesn't match the regular expression."
Presumably you're really asking how to do that - but it depends upon how your application is structured in the first place, and you haven't told us anything at all about that.

Regular Expression and PL/SQL help

I am using Oracle 9i, does 9i support regular expression? What functions are there?
My problem is the birth_date column in my database comes from teleform ( a scan program that reads what people wrote on paper), so the format is all jacked up.... 50% of them are 01/01/1981, 10% are 5/14/1995, 10% are 12/5/1993, 10% are 1/1/1983, 10% are 24-JUL-98. I have never really used regular expression and pl/sql, can anybody help me convert all of them to 01/01/1998?
Does Oralce 9i support regular expression? What can I do if oralce 9i does not support regular expression? Thank you very much in advance.

9i doesn't support regular expressions (at least not in the 10g regular expressions sense. There is an OWA_PATTERN_MATCH package that has some facilities for regular expressions). But it doesn't look like this is a regular expressions problem.
Instead, this is probably a case where you need to
- enumerate the format masks you want to try
- determine the order you want to try them
- write a small function that tries each format mask in succession until one matches.
Of course, there is no guarantee that you'll ever be able to convert the data to the date that the user intended because some values will be ambiguous. For example, 01/02/03 could mean Feb 1, 2003 or Jan 2, 2003 or Feb 3, 2001 depending on the person who entered the data.
Assuming you can define the order, your function would just try each format mask in turn until one generated a valid date, i.e.
BEGIN
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_1 );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_2 );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_3 );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
BEGIN
    l_date := TO_DATE( p_string_value, format_mask_N );
    RETURN l_date;
EXCEPTION
    WHEN OTHERS THEN
      NULL;
END;
RETURN NULL;
END;Justin

Juniper MX Regular expressions and user permissions ACS 5.4

Hi everyone!
Im having some trouble with regular expressions and permissions on our Juniper MX routers through ACS 5.4, and i would like some insight/help/poitners!!
We have a team of engineers that should only have read only permissions (important: show configuration) and also be able to just change the description on interfaces.
Thus far with the following regular expressions set for the shell profile they are going through i have managed the above, however the problem is when an engineer inputs "Show configuration", only the interfaces descriptions configuration is shown! The rest of the configuration will not be printed.
deny-commands1=.*.
allow-commands1=configure
deny-configuration1=.*.
allow-commands2=interfaces .*. description .*$
allow-configuration1=interfaces .*. description .*$
allow-commands2=show configuration.*
allow-commands3=show configuration
(some of these regex i know that are not needed, i was just playing around to check everything before posting)
Any pointers as to why or how to resolve this?
example output with the above:
show configuration
## Last commit: 2014-01-09 09:34:44 EET by someone
interfaces {
    xe-0/0/0 {
    xe-0/0/1 {
        description xxxx;
    xe-0/1/0 {
        description xxxx;
    xe-0/1/1 {
        description xxxx;
    xe-0/2/0 {
        disable;
    xe-0/2/1 {
        description xxxx;
    xe-0/3/0 {
        description xxxx;
    xe-0/3/1 {
        description xxxx;
    ae0 {
        description "xxxx";
    ae1 {
        description xxxx;
    demux0 {
    lo0 {
{master}
Thanks in advance!
Spyros

You are absolutely right!! I was doing research online after posting the above. The correct RADIUS attribute to use is actually CVPN3000/ASA/PIX7.x-Group-Based-Address-Pools. Then create the pool in ASA, and call that pool's name in ACS under that RADIUS attribute. Someone explained this perfectly in this community before. Much appreciate your answer!
Here's from another post last year:
ACS 5 does not have the feature of IP pools. Logically its always good to setup pools locally on vpn server and if you want user to pick ip from specific local pool you can configure acs to push that attribute.
On ACS Go to > Policy Elements -> Network Access ->   Authorization Profiles -> Create ->
Name of the Policy ->Dictionary Type: Radius-Cisco VPN 3000/ASA/PIX7.x
Attribute Type : CVPN3000/ASA/PIX7.x-Group-Based-Address-Pools
Attribute Type: String
Attribute Value : Static MYPOOL (Name of the Pool which is defined on the ASA)
Access Policies ->Default Network Access -> Authorization -> Create -> Under result section call the Authorization p

Regular expression and XML

Hello,
I have an XML file containing regular expressions and i parse the file, extract the pattern from it and search for it using java regex package. The problem is it works fine when patterns are words but when the pattern is something like
write \\d+ (write followed by a space followed by one or mre digits) it doesn't work.
I wrote the same code but with the pattern embedded in it,ie. without using XML and it worked. But when extracting with XML it fails.
Also if the pattern is write[0-9] it only extracts write[0-9 and gives an error of no closing bracket.
Could anyone please tell me what i am missing out
Thank you

thank you for your replies. Well i have still no got over the problem so i am posting my code here and hoping it can get solved
import org.xml.sax.*;
import org.xml.sax.helpers.*;
import java.io.*;
import java.util.regex.*;
class textextractor extends DefaultHandler{
     boolean regex=false;
public void startElement(String namespaceURI,String localName,String qn,Attributes attr)
          if(localName.equals("REGEX"))
           regex=true;
public void characters(char [] text,int start,int length)throws SAXException {
          String t=new String(text,start,length);
          boolean flag=false;
          if(regex==true)
            Pattern pattern;
            String w=new String(t);
          pattern = Pattern.compile(w);
          Matcher matcher;
          matcher=pattern.matcher("there is a bat   read write 13    error at line ");
          while(matcher.find())
           flag=true;
           System.out.println("I found the text \"" + matcher.group() +"\" starting at index "
           + matcher.start() +"and ending at index " + matcher.end() + ".");
         if(!flag)
           System.out.println("not found");
         regex=false;
public class saxt2 {
     public static void main(String args[]) {
          try {
                XMLReader parser= XMLReaderFactory.createXMLReader();
                ContentHandler handler=new textextractor();
                parser.setContentHandler(handler);
                                parser.parse("d:\\regex.xml");
              }catch (Exception e) {
               System.err.println(e);
}The xml file is
                  <RegularExpression>
                  <REGEX>write</REGEX>
                  <REGEX>write \\d+</REGEX>
                  <REGEX>read[0-9]</REGEX>
                  </RegularExpression>by running the code you can see that write is found,write \\d+ doesn't match write 13 in the string and read[0-9] gives and error.
Any help will be greatly appreciated

Find text using regular expression and add highlight annotation

Hi Friends
Is it possible to find text using regular expression and add highlight annotation using plugin

A plugin can use the PDWordFinder to get a list of the words on a page, and their location. That's all that the API offers for searching. Of course, you can use a regular expression library to work with that word list.

Regular expression and output format

hi all,
i have following scenario-
regular expression: [0-9]{3}-[0-9]{3}-[0-9]{4}
generated value by the above regular expression: 123-234-6789
output format to display the generated above value: xxx-xxx-$1
now i need to display the generated value (123-234-6789) in the specified output format (xxx-xxx-$1) and the final output will be xxx-xxx-6789
how is it possible?
Note: here regular expression and output format can vary
br,
bashar

Hi, Bashar
You can solve this problem by using the Data Masking Technique.
Masking data means replacing certain fields with a Mask character (such as an X). This effectively disguises the data content while preserving the same formatting on front end screens and reports. For example, a column of credit card numbers might look like:
4346 6454 0020 5379
4493 9238 7315 5787
4297 8296 7496 8724
and after the masking operation the information would appear as:
4346 XXXX XXXX 5379
4493 XXXX XXXX 5787
4297 XXXX XXXX 8724
The masking characters effectively remove much of the sensitive content from the record while still preserving the look and feel. Take care to ensure that enough of the data is masked to preserve security.
It would not be hard to regenerate the original credit card number from a masking operation such as: 4297 8296 7496 87XX since the numbers are generated with a specific and well known checksum algorithm.
Best Regards,
Mahfuz Khan

"Match Regular Expression" and "Match Pattern" vi's behave differently

Hi,
I have a simple string matching need and by experimenting found that the "Match Regular Expression" and "Match Pattern" vi's behave somewhat differently. I'd assume that the regular expression inputs on both would behave the same. A difference I've discovered is that the "|" character (the "vertical bar" character, commonly used as an "or" operator) is recognized as such in the Match Regular Expression vi, but not in the Match Pattern vi (where it is taken literally). Furthermore, I cannot find any documentation in Help (on-line or in LabVIEW) about the "|" character usage in regular expressions. Is this documented anywhere?
For example, suppose I want to match any of the following 4 words: "The" or "quick" or "brown" or "fox". The regular expression "The|quick|brown|fox" (without the quotes) works for the Match Regular Expression vi but not the Match Pattern vi. Below is a picture of the block diagram and the front panel results:
The Help says that the Match Regular Expression vi performs somewhat slower than the Match Pattern vi, so I started with the latter. But since it doesn't work for me, I'll use the former. But does anyone have any idea of the speed difference? I'd assume it is negligible in such a simple example.
Thanks!
Solved!
Go to Solution.

Yep-
You hit a point that's frustrated me a time or two as well (and incidentally, caused some hair-pulling that I can ill afford)
The hint is in the help file:
for Match regular expression "The Match Regular Expression function gives you more options for matching
strings but performs more slowly than the Match Pattern function....Use regular
expressions in this function to refine searches....
Characters to Find
Regular Expression
VOLTS
VOLTS
A plus sign or a minus sign
[+-]
A sequence of one or more digits
[0-9]+
Zero or more spaces
\s* or * (that is, a space followed by an asterisk)
One or more spaces, tabs, new lines, or carriage returns
[\t \r \n \s]+
One or more characters other than digits
[^0-9]+
The word Level only if it
appears at the beginning of the string
^Level
The word Volts only if it
appears at the end of the string
Volts$
The longest string within parentheses
The first string within parentheses but not containing any
parentheses within it
$[^()]*$
A left bracket
A right bracket
cat, cag, cot, cog, dat, dag, dot, and dag
[cd][ao][tg]
cat or dog
cat|dog
dog, cat
dog, cat cat dog,cat
cat cat dog, and so on
((cat )*dog)
One or more of the letter a
followed by a space and the same number of the letter a, that is, a a, aa aa, aaa aaa, and so
on
(a+) \1
For Match Pattern "This function is similar to the Search and Replace
Pattern VI. The Match Pattern function gives you fewer options for matching
strings but performs more quickly than the Match Regular Expression
function. For example, the Match Pattern function does not support the
parenthesis or vertical bar (|) characters.
Characters to Find
Regular Expression
VOLTS
VOLTS
All uppercase and lowercase versions of volts, that is, VOLTS, Volts, volts, and so on
[Vv][Oo][Ll][Tt][Ss]
A space, a plus sign, or a minus sign
[+-]
A sequence of one or more digits
[0-9]+
Zero or more spaces
\s* or * (that is, a space followed by an asterisk)
One or more spaces, tabs, new lines, or carriage returns
[\t \r \n \s]+
One or more characters other than digits
[~0-9]+
The word Level only if it begins
at the offset position in the string
^Level
The word Volts only if it
appears at the end of the string
Volts$
The longest string within parentheses
The longest string within parentheses but not containing any
parentheses within it
([~()]*)
A left bracket
A right bracket
cat, dog, cot, dot, cog, and so on.
[cd][ao][tg]
Frustrating- but still managable.
Jeff

Assistance with Regular Expression and Tcl

Assistance with Regular Expression and Tcl
Hello Everyone,
I recently began learning Tcl to develop scripts for automating network switch deployments.
In my script, I want to name the device with a location and the last three octets of the base mac address.
I can get the Base MAC address by :
show version | include Base
Base ethernet MAC Address       : 00:00:00:DB:CE:00
And I can get the last three octets of the MAC address using the following regular expression.
([0-9a-f]{2}[:-]){2}([0-9a-f]{2}$)
But I have not been able to figure out how to call the regular expression in the tcl script.
I have checked several resources but have not been able to figure it out. Suggestions?
Ultimately, I want to set the last three octets to a variable (something like below) and then call the variable when I name the switch.
set mac [exec "sh version | i Base"] (include the regular expression)
ios_config "hostname location$mac"
Thanks for any assistance in advance.
Chris

This worked for me.
Switch_1(tcl)#set result [exec show ver | inc Base]
Base ethernet MAC Address       : 00:1B:D4:F8:B1:80
Switch_1(tcl)#regexp {([0-9A-F:]{8}\r)} $result -> mac
1
Switch_1(tcl)#puts $mac
F8:B1:80
Switch_1(tcl)#ios_config "hostname location$mac"
%Warning! Hostname should contain at least one alphabet or '-' or '_' character
locationF8:B1:80(tcl)#

The Regular Expressions and GUID.

Hello gurus, I hope you can help me!
I am needing to select GUIDS from a table and for this, I need the Regular Expression. My
Perl is not good and not good Regular Expression. My database is Oralce 11.2.0.2.0 and
the Operating System of the Machine is Linux (Oracle Version 6). If you require the further information,
I will watch closely. Thank you. Jehangir.

>
Hi Jehangir and welcome to the forums.
I am needing to select GUIDS from a table and for this,Well, the first thing you should do is read the forum FAQ and also the post by BluShadow
at the top of the posts on the forum home page. You should have provided code (DDL
and DML) showing your particular problem, but since this is your first time, I'll be gentle ;)
We have done this - sometimes clients have GUIDs as PKs and we need to send data to
their systems, but it isn't as simple as it may first appear.
GUIDs can come in three formats.
The Oracle one - SELECT Sys_GUID() from DUAL which is just a string of 32 Hexadecimal characters.
Then the string with hyphens and then the string with hypens and {} at beginning and end (see
sample data).
with datax as
SELECT '79A864CCD8E44CD8B0A2765DF9EF337B' as guid FROM DUAL UNION ALL
SELECT '79A864CFD8E44CD7B0A2765DF9EF337B' FROM DUAL UNION ALL
SELECT '8gdfsgsgfdg' FROM DUAL UNION ALL -- dummy for testing
SELECT '21EC2020-3AEA-1069-A2DD-08002B30309D' FROM DUAL UNION ALL
SELECT '21EC5550-3AEA-1069-A2FF-08002B30309D' FROM DUAL UNION All
SELECT '{21CC2020-3AFA-1A69-A2DD-08002B30309D}' FROM DUAL
-- first one is the Oracle format
select * from datax where regexp_like(guid, '[0-9a-fA-F]{32}'); -- Oracle select sys_guid();
-- second one is with hyphens
select * from datax where regexp_like(guid, '[0-9a-fA-F]{8}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{12}');
-- third one is with hyphens and curly brackets.
select * from datax where regexp_like(guid, '^\{[0-9a-fA-F]{8}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{4}\-[0-9a-fA-F]{12}\}$');
-- This converts both of the last two formats back into Oracle format, which is what
-- we use. Notice, that I haven't used regualar expressions to do this. Regexes are
-- computationally expensive, and you should use Oracle's string furnctions if possible
SELECT REPLACE(REPLACE(REPLACE(GUID, '{', ''), '}', ''), '-', '') FROM Datax;HTH,
Paul...
Jehangir.

Java Regular Expressions and Pattern

I have a file that i first want to get all the lines that match a given pattern. Then from these lines that match i want to extract two values.
Example line for the pattern to match
INFO | jvm 1 | 2006/11/07 15:14:09 | INFO | Tue Nov 07 15:14:09 CET 2006 | XLDB PPS Data Dumper: MESSAGE:- 406 Processing .. '[ /opt/nexus/horizon/raw_data/network/pp_CE01S4H_sta_20050703T015717_SYDP3001_546.bdf ]'
So all the lines that are like these i want to extract two variables
2006/11/07 15:14:09
and
/opt/nexus/horizon/raw_data/network/pp_CE01S4H_sta_20050703T015717_SYDP3001_546.bdf
so i can store these variables in a database.
Can someone help me with writing the pattern to match and the regular express to extract? Also if anyone else has a better way of doing this i am all ears and i have a lot of log files to go through.

import java.util.regex.*;
class Main
public static void main(String[] args)
    String txt="INFO | jvm 1 | 2006/11/07 15:14:09 | INFO | Tue Nov 07 15:14:09 CET 2006 | XLDB PPS Data Dumper: MESSAGE:- 406 Processing .. '[ /opt/nexus/horizon/raw_data/network/pp_CE01S4H_sta_20050703T015717_SYDP3001_546.bdf ]'";
    String re1=".*?";     // Non-greedy match on filler
    String re2="((?:2|1)\\d{3}(?:-|\\/)(?:(?:0[1-9])|(?:1[0-2]))(?:-|\\/)(?:(?:0[1-9])|(?:[1-2][0-9])|(?:3[0-1]))(?:T|\\s)(?:(?:[0-1][0-9])|(?:2[0-3])):(?:[0-5][0-9]):(?:[0-5][0-9]))";     // Time Stamp 1
    String re3=".*?";     // Non-greedy match on filler
    String re4="((?:\\/[\\w\\.]+)+)";     // Unix Path 1
    Pattern p = Pattern.compile(re1+re2+re3+re4,Pattern.CASE_INSENSITIVE | Pattern.DOTALL);
    Matcher m = p.matcher(txt);
    if (m.find())
        String timestamp1=m.group(1);
        String unixpath1=m.group(2);
        System.out.print("("+timestamp1.toString()+")"+"("+unixpath1.toString()+")"+"\n");
}

Regular expressions and limiting matched input

Hi everyone :) I am trying to put together a regular expression that matches strings that contain elements of the form;
{<some text>}
However, each piece of text may contain multiple embedded instances of this pattern. I want to ensure that I am always getting the first (or outermost) instance.
So, if I had;
{OneStart}{TwoStart}{TwoEnd}{OneEnd}
I want to make sure that I get 'One' first and 'Two' second. So I have to place a stipulation in my regular expression to match this pattern only if it has not located the patten previously.
At the moment, I have this -
([^\\{]*?)(\\{TagStart\\})(.*?)(\\{TagEnd\\})(.*)
What I think that I have to do is modify the first capture group '([^\\{]*?)' which at the moment only does not match if a preceding '{' is found to match only if a preceding '{<text>}' sequence is not found.
Anyone got any idea how to do this?
Thanks in advance.
Ben

Doesn't it work anyway, since you're using greedy operators? If not, won't it work if you remove the .* at the end and use find() rather than matches()? And finally, what's the (.*?) supposed to match? Looks to me like that should be .*

Regular expressions and back references

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