Best Regular expression ?

What is the best regular expression to represent the following string in java ?
<?xml version = '1.0' encoding = 'UTF-8'?>
Thanks in advance !

What about that exact same string? depending on your regular expression implementation you'll probably need to escape some characters.. So something like: <\?xml version = '1.0' encoding = 'UTF-8'\?>

Similar Messages

Help in regular expression matching

I have three expressions like
1) [(y2009)(y2011)]
2) [(y2008M5)(y2011M3)] or [(y2009M5)(y2010M12)]
3) [(y2009M1d20)(y2011M12d31)]
i want regular expression pattern for the above three expressions
I am using :
REGEXP_LIKE(timedomainexpression, '???[:digit:]{4}*[:digit:]{1,2}???[:digit:]{4}*[:digit:]{1,2}??', 'i');
but its giving results for all above expressions while i want different expression for each.
i hav used * after [:digit:]{4}, when i am using ? or . then its giving no results. Please help in this situation ASAP.
Thanks

I dont get your question Can you post your desired output? and also give some sample data.
Please consider the following when you post a question.
1. New features keep coming in every oracle version so please provide Your Oracle DB Version to get the best possible answer.
You can use the following query and do a copy past of the output.
select * from v$version 2. This forum has a very good Search Feature. Please use that before posting your question. Because for most of the questions
that are asked the answer is already there.
3. We dont know your DB structure or How your Data is. So you need to let us know. The best way would be to give some sample data like this.
I have the following table called sales
with sales
as
 select 1 sales_id, 1 prod_id, 1001 inv_num, 120 qty from dual
 union all
 select 2 sales_id, 1 prod_id, 1002 inv_num, 25 qty from dual
select *
from sales 4. Rather than telling what you want in words its more easier when you give your expected output.
For example in the above sales table, I want to know the total quantity and number of invoice for each product.
The output should look like this
Prod_id sum_qty count_inv
1 145 2 5. When ever you get an error message post the entire error message. With the Error Number, The message and the Line number.
6. Next thing is a very important thing to remember. Please post only well formatted code. Unformatted code is very hard to read.
Your code format gets lost when you post it in the Oracle Forum. So in order to preserve it you need to
use the {noformat}{noformat} tags.
The usage of the tag is like this.
<place your code here>\
7. If you are posting a *Performance Related Question*. Please read
 {thread:id=501834} and {thread:id=863295}.
 Following those guide will be very helpful.
8. Please keep in mind that this is a public forum. Here No question is URGENT.
 So use of words like *URGENT* or *ASAP* (As Soon As Possible) are considered to be rude.

URL paths and regular expressions in ASDM

Some background info - I've recently switched to an ASA 5510 on 8.4(3) coming from a Checkpoint NGX platform (let's say fairly quickly and without much warning ). I have a couple questions and they're kind of similar so I'll post them up. I've read docs about regex and creating them both via command line and ASDM, but the examples always seem to include info I don't need or honestly something I don't understand yet (mainly related to defining class\inspect maps). If someone could provide a simple example of how to do these in ASDM that would help a lot in understanding how regular expressions are properly configured. So here we go.
I know this is basic but I need to make sure I understand this properly - I have a single web server (so this won't be a global policy) where I need to allow access to a specific URL path\file and that's it. So we'll call it \test\testfile.doc. Any other access to any other path should be dropped. What's the best way to do this in ASDM (6.4)? I think if I saw a basic example for this I could figure out next few questions but I'll post them as well just in case.
I have another single public web server (again this won't be a global policy) where I'd like to specify blocking file types, like .php, .exe., etc... again a basic example would be great.
Lastly, and this is kind of related, but we have a single office/domain and sometimes we get spam from forged addresses appearing to be from our domain. On Checkpoint I used to use its built-in SMTP security server and could define if it received mail from *@mydomain.com to drop it because we would never receive mail externally from our own domain name. I saw something similar with ESMTP in ASDM and it looks kind of like how you set up the URL access mentioned above. Can I configure this in ASDM as well, and if so how?
TIA for your help,
Jordan

/bump

Regular Expression question I think

My application is receiving HTML as a string and I'm trying to simplify it before displaying. The string could contain one or more substrings similar to this:
<span style="cursor:pointer" onmouseout="hideTooltip()" onmouseover="createTooltip( this,'The quartile that your firm\'s value falls into. Each quartile contains 25% of the values in the Peer Group. The 1st Quartile is always the best. The 4th Quartile is always the worst.', ( findPosX( this ) - 150))">QuartileThe text will not be consistent and it may be in the string as many as 4 times. What I'd like to do is replaceAll so that I end up with
QuartileIs there a way to do this with regular expression? I've tried replaceAll("","") But that takes out everything to the end of the String. I want it to stop at >.
Any way?

replaceAll("","")

Use regular expressions to extract .llb filename from path

I am trying to be clever (always a dangerous thing) and use a regular expression to extract the name of a library from a filepath converted to a string. Whilst I appreciate there are other ways to do this, regex would appear to be a very powerful neat way, should I be able to get it to work.
i.e if I have a string of the type, C:\applications\versions\library.llb\toplevel.vi, I want to be able to extract library.llb from the string, given that it will be of variable length, may include numbers & spaces and may be within a file hierarchy of variable depth. In other words, I want to extract the portion of the string between the last \ that ends with .llb
The best I have managed so far, is \\+.*llb which returned everything minus the drive letter and the toplevel.vi
Can anyone help me achieve this, or am i better using an alternative method (e.g filepath to array string, search for .llb)
Thanks
Matt
Solved!
Go to Solution.

Hi Matt,
attached you'll find two other options.
Mike
Message Edited by MikeS81 on 04-13-2010 01:30 PM
Attachments:
Path.PNG ‏7 KB

Regular Expressions in CS5.5 - something is wrong

Hello Everybody,
Please correct me, but I think, I found a serious problem with regular Expressions in Indesign CS5.5 (and possibly in other apps from CS5.5).
Let's start with simple example:
var range = "a-a,a,a-a,a";
var regEx = /(a+-a+|a+)(,(a+-a+|a+))*/;
alert( "Match:" +regEx.test(range)+"\nLeftContext: "+RegExp.leftContext+"\nRightContext: "+RegExp.rightContext );
What I expected was true match and the left and the right context should be empty. In Indesign CS3 that is correct BUT NOT in CS5.5.
In CS 5.5 it seems that the only first "a-a" is matched and the rest is return as the rightContext - looks like big change (if not parsing error in RegExp engine).
Please correct me if I am wrong.
The second example - how to freeze ID CS5.5:
var range = "a-a,a,a-a,a";
var regEx = /(a+-a+|a+)(,(a+-a+|a+)){8,}/;
alert( "Match:" +regEx.test(range)+"\nLeftContext: "+RegExp.leftContext+"\nRightContext: "+RegExp.rightContext );
As you can see it differs only with the {8,} part instead of *
Run it in CS5.5 and you will see that the ID hangs (in CS3 of course it runs flawlessly}.
The third example - how to freeze ID 5.5 in one line (I posted it earlier in Photoshop forum because similiar problem was called earlier):
alert((/(n|s)? /gmi).test('s') );
As you can guess - it freezes the CS5.5 (CS3 passes the test).
Please correct me if I am doing something wrong or it's the problem of Adobe.
Best regards,
Daniel Brylak

Hi Daniel,
Thanks for sharing. Really annoying indeed.
Just to complete your diagnosis, what you describe about CS.5 is the same in CS5, while CS4 behaves as CS3.
var range = "aaaaa";
var regEx = /(a+-a+|a+)(,(a+-a+|a+))*/;
alert([
    "Match:" +regEx.test(range),
    "LeftContext: "+RegExp.leftContext.toSource(),        // => CS3/4: EMPTY -- CS5+: EMPTY
    "RightContext: "+RegExp.rightContext.toSource()        // => CS3/4: EMPTY -- CS5+: ",a,a-a,a"
    ].join('\r'));
So there is a serious implementation problem of the RegExp object from ExtendScript CS5.
I don't think it's related to the greedy modes. By default, JS RegExp quantifiers are greedy, and /a*/ still entirely captures "aaaaaa" in CS5+.
By the way, you can make any quantifier non-greedy by adding ? after the quantifier, e.g.: /a*?/, /a+?/, etc.
I guess that Adobe ExtendScript has a generic issue in updating the RegExp.lastIndex property in certain contexts—see http://forums.adobe.com/message/3719879#3719879 —which could explain several bugs such as the Negative Class bug —see http://forums.adobe.com/message/3510078#3510078 — or the problems you are mentioning today.
@+
Marc

Regular expression and pattern matching/replacing

I have a list of key words. It has around 1000 key word now but can grow to 5000 keywords.
My web application displays lot of texts which are stored in the database. My requirement is to scan each text for the occurance of any of the above keywords. If any keyword is present I have to replace that with some custom values, before showing it to the user.
I was thinking of using using regular expression for replacing the keyword in the text using matcher.replaceAll method as follows:
Pattern pattern = Pattern.compile(patternStr);
Matcher matcher = pattern.matcher(inputStr);
String output = matcher.replaceAll(replacementStr);
But My pattern string will have around 5000 keywords with the 'OR' Logical Operator like- keyword1| keyword2 I keyword3 | ..........
Will such a big pattern string adversly affect the performance? What can I do to speed up the performance? (Since my keyword list is not static i would prefer to do the replacement just before showing the text to the user)
Any suggestions are most welcome.

I don't think a pure regex approach would be that slow, but it would be a maintenance nightmare. I think a combined regex/table-lookup approach would be best: use a regex to identify potential keywords, then look them up in the table to confirm. For instance, to find all Java keywords you could use the regex "\\b[a-z]{2,12}+\\b" to filter out anything that can't possibility be a keyword.
What are you going to replace the keywords with? Will it vary depending on which keyword is found? If so, you'll have to use a table--and you won't be able to use the replaceAll method, because it can't handle dynamically generated replacement values. You would have to use the lower-level appendReplacement and appendTail method instead.

Using regular expressions to find and replace code.

Hi! Semi-newbie coder here.
I'm trying to strip out code from multiple pages, I've tried regular expressions but I'm struggling to understand them. I also need to do it across a LOT of pages, so I need an automated way of doing it.
The best way I can explain is with an analogy:
I want to delete any string of characters that start with c, ends with t and includes anything inbetween, so it would pick up "cat, cut, chat, coconut, can do it" whatever appears in the middle of those.
Except, instead of c and t, I want it to find strings of code starting with <div class="advert" and ending with Vote while picking up everything in between, (including spaces, code, comments, etc.). Then, deletes that whole string including the starting and ending.Is there a regular expression I could use in dreamweaver that could do this? Is there a way to do this at all?

Let me begin by saying I'm a complete idiot with DW's Reg Ex. I use Search Specific Tag whenever possible. See screenshot below.
Try this on your Current Document to see if it works. Then make a back-up copy of site before attempting it on Entire Local Site as you cannot "Undo" this process.
Good luck,
Nancy O.

Matches from regular expression into collection

Hello,
I need to do the following:
I have a long string with some similar repeated data. I would like, using a regular expression, to extracts all matches in a collection. Is there a way of performing this task?
I have look through the owa_pattern package, but as far as I found out, I can extract only a simple match. Here is an exact quote:
"If multiple overlapping strings can match the regular expression, this function takes the longest match. " - http://download.oracle.com/docs/cd/B28359_01/appdev.111/b28419/w_patt.htm
So what can I do if I want to get all the matches?
Thank you in anticipation. Any help would be appreciated.
Best regards,
beroetz

I think your need a tokenizer-function.
If the string +:in_str+ is delimited by +:in_delimiter+ you could try this:
SELECT REGEXP_REPLACE(REGEXP_SUBSTR( :in_str || :in_delimiter, '(.*?)' || :in_delimiter, 1, LEVEL ), :in_delimiter, '') TOKEN
BULK COLLECT INTO :my_nested_table
FROM DUAL
CONNECT BY REGEXP_INSTR( :in_str || :in_delimiter, '(.*?)' || :in_delimiter, 1, LEVEL ) > 0
ORDER BY LEVEL ASC;
I wrote a string-to-textarray-tokenizer (and it's pendant) some times ago, being able to cut from certain positions within the string using regular expressions and return the elements into an nested table of varchar2. It looks like:
TYPE pos_arraytype IS TABLE OF POSITIVE ;
TYPE text_arraytype IS TABLE OF VARCHAR2(2000);
FUNCTION stringToTextarray(in_str IN VARCHAR2, in_pos_arr IN pos_arraytype, in_regexp_arr IN text_arraytype DEFAULT NULL, in_trim_strings IN BOOLEAN DEFAULT TRUE)
RETURN text_arraytype ;
in_str is the string to be tokenized
in_pos_arr is a table of positive values of positions in the string to be cut
in_regexp_arr is a table of regular expressions to use at each position declared by in_pos_arr
in_trim_strings is a flag, if the cutted element should be trimmed
using above for example:
in_str = 'Markus van Muster 347651234XY Musterdaam ABCDE'
in_pos_arr = (1, 13, 35, 35, 42)
in_regexp_arr = ('(.?){12}', '([^[:digit:]]?){22}', '[[:digit:]]{4}', '[[:alpha:]]{2}', '(.?){14}')
in_trim_strings = TRUE
RETURN collection ('Markus','van Muster','1234','XY','Musterdaam')
If you need the code, then tell me! I'm looking for....
Cheers,
Martin
Edited by: Nuerni on 17.10.2008 08:49

Query regarding Regular expressions

Help me in regular expression for “one or two digits must followed by : and one or two digits”

user8701050 wrote:
thanqI assume you meant "thank you." Please use real words.
In any case, you're welcome. So now that you know, your best course of action would be to study that tutorial, and/or this one: http://www.regular-expressions.info/tutorial.html, take your best shot, then post again if you get stuck, showing what you tried and explaining clearly the problems you encountered.
Good luck!

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/

Urgent help regarding Java regular expressions.

hello everyone,
I am trying to parse a html file which contains
dyn.Img("http://www.boston.com/news/nation/articles/2007/06/21/bill_clinton_takes_bigger_campaign_role&h=306&w=410&sz=13&hl=en&start=43","","QFo9lqKeMR7uzM:","http://cache.boston.com/resize/bonzai-fba/AP_Photo/2007/06/21/1182410228_1931/410w.jpg","125","93","\x3cb\x3eBill Clinton\x3c/b\x3e takes bigger campaign \x3cb\x3e...\x3c/b\x3e","","","410 x 306 - 13k","jpg","www.boston.com","","","http://tbn0.google.com/images","1")
the given above function many times. I have to fetch the whole functions into an array. So i have to write a regular expression which recognises the whole above string.
Can anyone please help me.
Thank you,
chaitanya

well if this is all you want
http://cache.boston.com/resize/bonzai-fba/AP_Photo/2007/06/21/1182410228_1931/410w.jpg
You can always substring it like chuck said
***BUT all the images would have to be .jpg for this to work***
back = we.indexOf(".jpg");
int x = 0;
while (back < web.lastIndexOf(".jpg"))
 back = web.indexOf("http",back+1);
 picture[x] = web.substring(front, back);
 x++;
 front = back;
 } Might not be the best code but it worked with a website i had to parse
Message was edited by:
mark07

Help with Java and Regular Expression

Hello,
I have one line of Java and regular expression that I got from the web :
String[] opt;
opt = line.split("\\s+(?=([^\"]*\"[^\"]*\")*[^\"]*$)");
If "line" contains :
1 "Hello World"
this code would give me :
opt[0] : 1
opt[1] : "Hello World"
which is almost what I would like to do except for the double quotes. I wonder if someone could please help me change the regular expression so that it would return :
opt[0] : 1
opt[1] : Hello World
Thank you and Best Regards,
Chris

It looks to me like this is a line from a space delimited file so you should use one of the free CSV parsers such as http://opencsv.sourceforge.net/ . These will handle the quotes properly even if a field actually contains a quote.

Regular Expressions find and replace

Hi ,
I have a question on using Regular Expressions in Java(java.util.regex).
Problem Description:
I have a string (say for example strHTML) which contains the whole HTML code of a webpage. I want to be able to search for all the image source tags and check whether they are absolute urls to the image source(for eg. <img src="www.google.com/images/logo.gif" >) or relative(for eg. <img src="../images/logo.gif" >).
If they are realtive urls to the image path, then I wish to replace them with their absolute urls throughout the webpage(in this case inside string strHTML).
I have to do it inside a servlet and hence have to use java.
I tried . This is the code. It doesn't match and replace and goes inside an infinite loop i.e probably the pattern matches everything.
//Change all images to actual http addresses FOR example change src="../images/logo.gif" to src="http://www.google.com/../images/logo.gif"
 String ddurl="http://www.google.com/";
String strHTML=" < img src=\"../images/logo.gif\" alt=\"Google logo\">";
Pattern p = Pattern.compile ("(?i)src[\\s]*=[\\s]*[\"\']([./]*.*)[\"\']");
Matcher m = p.matcher (strHTML);
while(m.find())
m.replaceAll(ddurl+m.group(1));
what is wrong in this?
Thanks,
Rajiv

Right, here's the full monte (whatever that means):import java.util.regex.*;
public class Test1
public static void main(String[] args)
 String domain = "http://www.google.com/";
 String strHTML =
 " < img src=\"images/logo.gif\" alt=\"Google logo\">\n" +
 " <img alt=\"Google logo\" src=images/logo.gif >\n" +
 " <IMG SRC=\"/images/logo.gif\" alt=\"Google logo\">\n" +
 " <img alt=\"Google logo\" src=../images/logo.gif>\n" +
 " <img src=http://www.yahoo.com/images/logo.gif alt=\"Yahoo logo\">";
 String regex =
 "(<\\s*img.+?src\\s*=\\s*) # Capture preliminaries in $1. \n" +
 "(?: # First look for URL in quotes. \n" +
 " ([\"\']) # Capture open quote in $2. \n" +
 " (?!http:) # If it isn't absolute... \n" +
 " /?(.+?) # ...capture URL in $3 \n" +
 " \\2 # Match the closing quote \n" +
 " | # Look for non-quoted URL. \n" +
 " (?!http:) # If it isn't absolute... \n" +
 " /?([^\\s>]+) # ...capture URL in $4 \n" +
 Pattern p = Pattern.compile(regex, Pattern.CASE_INSENSITIVE | Pattern.COMMENTS);
 Matcher m = p.matcher(strHTML);
 StringBuffer sbuf = new StringBuffer();
 while (m.find())
 String relURL = m.group(3) != null ? m.group(3) : m.group(4);
 m.appendReplacement(sbuf, "$1\"" + domain + relURL + "\"");
 m.appendTail(sbuf);
 System.out.println(sbuf.toString());
}First off, observe that I'm using free-spacing (or "COMMENTS") mode to make the regex easier to read--all the whitespace and comments will be ignored by the Pattern compiler. I also used the CASE_INSENSITIVE flag instead of an embedded (?i), just to remove some clutter. By the way, your second (?i) was redundant; the first one would remain in effect until "turned off" with a (?-i). Another way to localize a flag's effect by using it within a non-capturing group, e.g., (?i:img).
As jaylogan said, the best way to filter out absolute URL's is by using a negative lookahead, and that's what I've done here. The problem of optional quotes I addressed by trying to match first with quotes, then without. The all-in-one approach might work with URL's, since they can't (AFAIK) contain whitespace anyway, but the alternation method can be used to match any attribute/value pair. It's also, I feel, easier to understand and maintain. Unfortunately, it also means that you can't use replaceAll(), since you have to determine which alternative matched before doing the replacement, but the long version is still pretty simple (especially when you can just copy it from the javadoc for the appendReplacement() method, as I did).

Migrating Regular Expression from apache to Java1.4

Did any one out there migrate from apache regular expressions to the new Java regular expressions? Are the two compatible? How much work is it to do this? What are the pitfalls? Are there any resources on this topic?
Thanks,

I did migrate some Jakarta regexp to java 1.4's. If ur old codes just used match() method to see if a string matches the pattern, then search/replace using any decent text edit could quickly migrate all of them. However, if you used lots of the Apache regexp's APIs, the fastest way is to write a wrap class for each class you used in the regexp pacakage, and use the new regexp package to implement all the Apach regexp APIs, then, the only thing you need to do is to replace the import statement to import those wrapper classes. I personally think this is the best and common way to migrate from an old API package to a new one.

Best Regular expression ?

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