Beginner question about Regular expression

Hi all !
I'd like to use a regular expression to parse a string like this:
*<ID>4</ID><GROUP>5</GROUP>....*
So for example to retrieve the ID I have built the following regular expression:
Pattern p = Pattern.compile("<ID>(.*?)</ID>");
Matcher m = p.matcher(handle);
if (m.find()) {
      System.out.println("->"+m.group());
} else {
System.out.println("No match!");
}The function m.group returns "<ID>4</ID>" but I want just the value (4) between the tag. Is there
a way to get it ?
thanks a lot
mark

fmarchioniscreen wrote:
thank you very much, that's exactly what I needed.
But it looks like you're parsing some XML like data: probably better to use a proper parser on it. Well it's a very short string containing XML tags. it's used in a marginal area of the application so I prefer just using a regular expression to fetch the values
thanks again
MarkYou could use XPath to get the value.

Similar Messages

Basic question about regular expressions

Hello,
I am a beginner to regular expressions. I want to rewrite the following expression:
public static final String REGULAR_EXP_SOFTWARE_PART_NUMBER = "([0-9]{7}[a-z]{1})(\\-{1})([a-z]{1})";I want THIS match
(\\-{1})to occur EITHER if a hyphen is encountered OR if a space is encountered (instead of just the hyphen).
How do I rewrite this?
Thanks in advance,
Julien.

Hello and thanks for your feedback,
I have created a small class as follows:
package regExpr;
import java.util.regex.Matcher;
import java.util.regex.Pattern;
* @author Martin
public class RegExprTest {
     private static String stringToBeParsed = "3800157w-e26";
     public static void main(String[] args) {
          Pattern pattern = Pattern.compile("" +
                    "([0-9]{7})" +
                    "([a-z]{1})" +
                    "(( |-){1})" +
                    "([a-z]{1})" +
                    "([0-9]{2})" +
          Matcher matcher = pattern.matcher(stringToBeParsed);
          while(matcher.find()){
               System.out.println(matcher.group(1));
               System.out.println(matcher.group(2));
               System.out.println(matcher.group(3));
               System.out.println(matcher.group(4));
               System.out.println(matcher.group(5));
               System.out.println(matcher.group(6));
}the class is trying tobreak down the following string "3800157w-e26" as follows:
3800157(seven digits)
w(one letter)
-(hyphen)
e(one letter)
26(two digits)
Oddly enough the output of the class is as follows:
3800157
w
e
26
I have to call the group method six times and I get two hyphens!
Can anyone help?
Thanks in advance,
Julien

Question about Regular Expressions, please help!

I have created an app which reads files and extracts certain data using regular expressions in JDK1.4 using Pattern and Matcher classes.
However it needs to run on JDK1.2.2 (dont ask). The regular expression classes are not available in 1.2.2 (the Pattern and Matcher class) so i am looking for something similiar which i can use?
I need something that loops through all the matches found in the file like how Matcher works i.e.
while (matcher.find())
// do this
Help!

http://jakarta.apache.org/regexp/

Simple question about regular expressions

Hi,
Using Java's regular expression syntax, what is the correct pattern string to detect strings like the following :-
AnnnnnA
where A = a single (fixed) alphabetic character and
n = at least one but possibly many digits [0-9].
Example strings to be searched :-
A45A (this should match)
A3A (this should match)
A3446655577A (this should match)
A hello world A (this should NOT match as no digits are present between the A's).
Thanks.

A least one digit "A.*\\d.*A"
Only digits "A\\d+A"

One question about Regular Expression!!!

I need to creat such a regular expression to match the format "[ ][ ][ ]".
For example, there is a context,
(1), " The project manager defines [1][0.400][+goals] for iterations."
Suppose that there are some spaces or "\n" characters in this way,
(2), " The project manager defines [    1 ] [ 0.400   ]
[   +goals] for iterations."
If the pattern match the format succefully, (2) strings should be replaced by (1)strings, in order words, the format of (1) is what I need finally,
I had ever tried creating a regular expression likes \\[([^\n\s]]+)\\]\\[([^\n\s]]+)\\]\\[([^\n\s]]+)\\] , but it does not work well!
DO YOU HOW TO IMPLEMENT IT IN JAVA?
Thanks for your any reply!

What I really need is that, via the regular
expression, all the spaces and \n characters in
square brackets [ and ], ] and [, will be thrown
away.
For example,
Original:
1) "The project manager defines [   1 ] [
0.400 ]
[   +goals] for iterations with the support"
After matching:
2) "The project manager defines [1][0.400][ [+goals]
for iterations with the support"
String 2) is what I need finally!
Thanks for your any reply!Well I gave you the answer to that one already :-)
If you need to preserve the spaces in between words use this one. I'm sure there's a better way to do it, I'm no RegEx master.
    public static void main(String[] args)
        String s = "[ 1 ] [ 0.400 ]\n[ +go als]";
        System.out.println( "Before: " + s );
        System.out.println( "\n\n" );
        s = s.replaceAll( "\\[\\s+", "[" );
        s = s.replaceAll( "\\s+\\]", "]" );
        s = s.replaceAll( "\\]\\s+\\[", "][" );
        System.out.println( "After: " + s );
    }

An additional question about regular expressions with String.matches

does the String.matches() method match expressions when some substring of the String matches, or does it have to match the entire String? So, if i have the String "123ABC", and i ask to match "1 or more letters" will it fail because there are non-letters in the String, but then pass if i add "1 or more letters AND 1 or more digits"? so, in the latter every character in the String is accounted for in the search, as opposed to the first. Is that correct, or are there ways to JUST match some substring in the String instead of the whole thing? i WILL make some examples too... but does that make sense?

It has to match the whole String. Use Matcher.find() to match on just a sub-string()

Simple question about regular expression

Hi
I have a little problem with
select regexp_substr('123 Mapla Avenue','[a-z]') my_test from dual;
answer: M
I excecute this query in SQLPlus and SQL Developer result is this same.
select regexp_substr('123 Mapla Avenue','[M]') my_test from dual;
answer: M
select regexp_substr('123 Mapla Avenue','[a]') my_test from dual;
answer: a
I used oracle 10g
Thanks for your help

hm wrote:
In the oracle documentation of regexp_substr you can find:Do not confuse pattern and sort. Pattern [a-z] means any lowercase letter. REGEXP_SUBSTR parameter match_param value i tells REGEXP to treat uppercase letters same as lowercase letters and vice versa. And setting NLS_SORT can do the same. As you can see it is not that straight-forward. To make it transparent use exact pattern you need. In this particular case use:
select regexp_substr('123 Mapla Avenue','[[:alpha:]]') my_test from dual;where class [:alpha:] is POSIX predefined class of all letters (regardless of case). This way you are not dependent of client side settings like NLS_SORT and the above will always return first letter within a string. If you want first uppercase letter use:
select regexp_substr('123 Mapla Avenue','[[:upper:]]') my_test from dual;Or, for first lowercase letter:
SQL> alter session set nls_sort=binary;
Session altered.
SQL> select regexp_substr('123 Mapla Avenue','[a-z]') my_test from dual;
M
a
SQL> select regexp_substr('123 Mapla Avenue','[[:lower:]]') my_test from dual;
M
a
SQL> alter session set nls_sort=binary_ci;
Session altered.
SQL> select regexp_substr('123 Mapla Avenue','[a-z]') my_test from dual;
M
M
SQL> select regexp_substr('123 Mapla Avenue','[[:lower:]]') my_test from dual;
M
a
SQL> SY.

Question about Regular Expressions

Hi averyone!
Could any one help me to create RegEx for string: <object>
Thanks!
Kind Regards, Dmitry.

"<object>"

Off Topic: Books about Regular Expression

Hi
Somebody can to indicate books about Regular Expression in Oracle ?
Thanks

Regex tag of Blog of Volder.
http://volder-notes.blogspot.com/search/label/Regular%20Expressions
This entry mentions my regex solution :-)
http://volder-notes.blogspot.com/2007/10/removing-duplicate-elements-from-string.html
By the way
My regex homepage mentions regex problems of perl like regex (regex of EmEditor).
http://www.geocities.jp/oraclesqlpuzzle/regex/
example questions (written by Japanese language)
http://www.geocities.jp/oraclesqlpuzzle/regex/regex-2-1.html
http://www.geocities.jp/oraclesqlpuzzle/regex/regex-3-5.html
http://www.geocities.jp/oraclesqlpuzzle/regex/regex-4-4.html

Two Questions about Airport Express

All, I am a recent convert to using a Mac so bear with me.
I have two questions about the Airport Express.
Question if I buy one in the USA, I will visiting this week on holiday,can I use it the UK with a UK adaptor/apple plug?
Can the Express be used with my current wireless network, using a Netgear DG834G V4, I have read lots in the forums about putting a $ sign in front of the WPA password?
Any help and advice gratefully received.

Tesserax wrote:
Hello andixbox. Welcome to the Apple Discussions!
Question if I buy one in the USA, I will visiting this week on holiday,can I use it the UK with a UK adaptor/apple plug?
Yes, but at least two things to consider:
o The US version of the 802.11n AirPort Express Base Station (AXn) only supports radio channels 1-11 in the 802.11g radio mode. Channels 12-14, used in other countries outside the US will not be available.
o The AXn supports both the 2.4 and 5 GHz radio bands for 802.11n. The 5 GHz band is restricted in certain countries (like the UK) so operating in this mode may get you unwanted attention.
That should be okay then because the Netgear broadcasts on channels 1-11. I hadn't appreciated the 5 Ghz band, but having checked both US and UK Apple sites the techspecs both mention 2.4 Ghz and 5 Ghz.
Can the Express be used with my current wireless network, using a Netgear DG834G V4, I have read lots in the forums about putting a $ sign in front of the WPA password?
If your intent is to have the AXn join the Netgear as a wireless client, then yes it should work. The "$" requirement was for WEP. This shouldn't be necessary for WPA or WPA2.
I intended to use the AXn as an range extender, but you mention wireless client, is that something different?

Question in regular expressions

Hi,
I have this string (abdcerpabdcerpabdcerpaabdcerpabdcerp)
and i want to this string abdcerp and may be followed by one or more a's
So i want to get these results:
abdcerp
abdcerp
abdcerpa
abdcerp
abdcerp
I know regular expressions very well but i failed to generate one that can do so. I tried using this regex (abdcerpa*?) but it failed. it's not working and i dont know why, it's not getting abdcerpa. It only gets abdcerp
can anyone help me with that telling me the reson why did this regex is not good or tell me a regex for doing so. But I need it to be tested cause I already know about the concepts and tried different ways and regexs but failed
thanks
bye

That forum was retired and is now read-only. According to the announcement;
Any future posts on this topic should be put in the
.NET Framework Class Libraries forum.
Regards, Dave Patrick ....
Microsoft Certified Professional
Microsoft MVP [Windows]
Disclaimer: This posting is provided "AS IS" with no warranties or guarantees , and confers no rights.

Strange result about regular expressions

Hello everybody,
I write these codes to try regular expressions in Java, but there are some strang results. I read the reference like Sun Java Tutorials. however, I cann't find the problem.
Environnement:
WindowsXP Home + NetBeans IDE 5.0 + JDK 1.5
Input String:
"I write these codes to try regular expressions in Java, but it doesn't work. I read some reference like Sun Java Tutorials. Then, always cann't find the problem. Could you help me? Thanks."
My codes:
public static void main(String[] args) throws Exception, IOException {
P.rintln("Let's go!");
Date start = new Date();
if(args.length != 1) {
P.rintln("Input Error! Input format: java javaclass [directory path]");
System.exit(0);
StringBuffer sb = new StringBuffer();
String input = TextFile.read(args[0]);
sb = addSectionEelement(input, "re");
P.rintln(sb.toString());
P.rintln("Ok, it's over");
Date end = new Date();
System.out.println("It spends " + (end.getTime() - start.getTime()) + " ms.");
public static StringBuffer addSectionEelement(String input, String regex) {
Matcher m = Pattern.compile(regex).matcher(input);
StringBuffer sb = new StringBuffer();
int count = 0;
while(m.find()) {
count++;
P.rintln(m.group());
P.rintln("Found " + count + " fois.");
return sb;
Output:
run:
Let's go!
Found 0 fois.
Ok, it's over
It spends 16 ms.
BUILD SUCCESSFUL (total time: 0 seconds)
However if I change the Bold line by
sb = addSectionEelement(input, "r");
The resultats become:
run:
Let's go!
r
r
r
r
r
r
r
r
r
r
r
Found 11 fois.
Ok, it's over
It spends 15 ms.
BUILD SUCCESSFUL (total time: 0 seconds)
I have no idea about it. And you?
Thanks

Hi guys,
I re-examine the codes. In fact, it's the problem of encodings of the input file.
See u

About regular expressions

This question was posted in response to the following article: http://help.adobe.com/en_US/ColdFusion/10.0/Developing/WSc3ff6d0ea77859461172e0811cbec0a38 f-7fff.html

"ColdFusion supplies four functions that work with regular expressions" should be "ColdFusion supplies six functions that work with regular expressions,"

Help About Regular Expression.

Hello,
I am trying to parse string buffer by using Regular Expression.
Suppose my string buffer is:
Hi , How are you?
Hello: abc
hurrey : [ this is test msg
Pls reply to this mail
Hello: xyz
Test1
I want to search string: "Hello: anystring till end of line" which is
not included in [].
So In above example my Regular expression should only find
first "Hello: abc".
Is it possible by using Regular expression?

Can we have Regular Expression which will get both "Hello: string"
suppose my string buffer is:
Hi , How are you?
Hello: abc
hurrey : [ this is test msg
Pls reply to this mail
Hello: xyz
Test1
happy: [ test2
my test
Hello: abc
then result should be :
Hello: abc
[ this is test msg
Pls reply to this mail
Hello: xyz
Test1
[ test2
my test
Hello: abc
]

Question about FC Express

Hi all.
I have no FC experience.....just some iMovie putzing with the family video's. I do use Photoshop CS5 and have years of experience with Pro Tools and work in film/TV production so I am not a total stranger to filmmaking and audio production, using difficult software etc etc. I spend hours every day in my music studio or editing photo's and minor video editing. Nothing deep like FCP though. I hear the FCP learning curve is brutal.
Anyway, I am dabbling with writing music aimed at soundtrack stuff for film, TV and commercials and am wondering about FC Express.
Would it be a good program to learn about dropping music into a video timeline?
I was thinking about trying to put together some stuff I shoot, Mini DV and Flip HD stuff and then write music for it to practice with. I'd also like to try and come up with markers and hit points and all so that as I get better I don't look like a total bozo when I try to get a short or low budge Indie or student film to score.
What I would really like to do is get some footage with dialogue and audio effect already mixed and then write to that.
I have friends who work in post who have agreed to look around for some stuff for me to work with. If they were to give me sections of FCP sessions, would I be able to open those in Final Cut Express?
Also, I read that FCE 4 does not have Soundtrack that 3.5 HD has. Is this a detriment and can one still buy 3.5 or has it been replaced with v4?
Thanks.
Message was edited by: Fumblyfingers

FCE should do much of what you require.
However, be warned. It is exactly the same app as FCP with a few items removed.
So for "normal" editing the learning curve is identical.
You will not be able to open FCP projects in it.
If you need Soundtrack you should be able to buy a cheap FCE 3.5 Upgrade on eBay etc. and just install Soundtrack.

Beginner question about Regular expression

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