Help: Regular Expression question??

Hello,
How can I extract the following content using Java Regular expression?
<tr bgcolor="#333333">
 <td class="title" colspan="4" height="18"> SUPER_1 - SUPER_2</td>
</tr>
<tr bgcolor="#333333">
 <td class="match-light" width="45" height="18"> </td>
 <td class="match-light" colspan="3" width="286" align="right">March 19 </td>
</tr>
<tr>
 <td colspan="4" height="1"></td>
</tr>
<tr bgcolor="#cfcfcf">
 <td width="45" height="18"> FT</td>
 <td width="118" align="right">SUPER_3</td>
 <td width="50" align="center"><a class="scorelink" target="details" onclick="showDetails();">999 - 888</a></td>
 <td width="118">SUPER_4</td>
</tr>From the above contents, How can I define a regular expression for extract the "*SUPER_1*", "*SUPER_2*", "*March 19*", "*SUPER_3*", "*999*", "*888*" and "*SUPER_4*" ????
Please help.
Best regards,
Eric

Kayaman wrote:
Why not use a better way than regex, like an actual HTML parser (or XML if you have it well-formed)? People seem to love parsing (or rather, asking help how to parse) HTML with regex for some unknown reason.Indeed.
Read this (hilarious):
http://stackoverflow.com/questions/1732348/regex-match-open-tags-except-xhtml-self-contained-tags/1732454#1732454

Similar Messages

Quick regular expression question/help

Can someone help me with two regular expressions I need. I could spend a while trying to figure it out myself, however times short and I really would like to get a fool proof optimal solution (my attempt would be buggy).
Sample sentence
The population, is projected to reach 200,000, or more (by 2020).[7] This is {dummy} text.
The first regular expression
I need all brackets and every thing between them to be removed from a sentence.
Brackets such as: ( ), [ ] and { } .
I.e. Given the above sentence the following would be returned:
The population, is projected to reach 200,000, or more. This is text.
The second regular expression
If a word has a trailing comma character I need to add a whitespace between the word and the comma.
I.e. Given the sentence returned from the first regular expression, this regex would return:
The population *,* is projected to reach 200,000 *,* or more. This is text.
Many thanks to anyonewho can help me with this!
Edited by: Myles on Jan 18, 2008 8:12 AM

http://java.sun.com/docs/books/tutorial/extra/regex/index.html
http://www.regular-expressions.info

Regular Expressions - Questions to SME

Ralph Benzinger presented an online meetup on the topic of Regular Expressions. The presentation (slides only) can be found <a href="http://www.sdn.sap.comhttp://www.sdn.sap.comhttp://www.sdn.sap.com/irj/servlet/prt/portal/prtroot/docs/library/uuid/866072ca-0b01-0010-54b1-9c02a45ba8aa">here</a>
Unfortunately the recording is not going to be available, but Ralph has been generous enough to agree to answer questions posted to this sticky thread.
cheers,
Marilyn

Hello Peter,
You're welcome!
Alas, I was unable to locate the regex documentation on help.sap.com either. In fact, I'm not even sure it has already been updated for 2004s. I recommend that you use the online documentation within the system, e.g., from transactions SE38 or SE80. Do an index search for "regex", and you'll be directed to REGEX, FIND and REGEX, REPLACE, both of which have extensive subsections on regexes.
The class cx_sy_regex is an exception class that is thrown by FIND, REPLACE and cl_abap_regex in case of an invalid regex, such as ".\1" (there is no capture group that back reference \1 can refer to). If the pattern is known statically, the syntax check will report this error, but for statements like "FIND REGEX pat IN text.", the actual pattern is only known at runtime.
The cx_sy_matcher class (and its subclasses) similarly indicate some invalid states, for example trying to call "cl_abap_matcher->replace_found( )" when the matcher has no current match to replace (e.g., replace_found( ) called twice in a row).
Please let me know if I can provide some additional information.
Regards
Ralph

Regular Expression Question

Hi all,
I am suffering in java regular expression, and I hope you guys can help me out. I want to use the String api ".matches" to find out any string pattern like "xxxx.xxxx" where xxx can be only english word(both upper and lower case). Actually I will use this kind of expression to represent the cross join SQL statement in my java class, like "tableA.name = tableB.name", where they should be english letter only. I tried to use MyString.matches("^[A-Z] + \\. + ^[A-Z]") in my java program, but seem it doesn't work. Can you guys figure out the right expression for me ?? Many thanks
Transistor

Thanks for your prompt response, I tried your code, however, it doesn't work out.
I put your code like the following:
if ( searchCriteria.getStringPair().getValue().trim().matches("[A-Za-z]+\\.[A-Za-z]+") {...some action }.
Seems the java program never reach this expression.
Kindly remind that I wan to expression anything like "xxxxx.xxxxx" where xxxx can be a word.
Myriads of thanks
Transistor

Please Help: Regular Expression Gives Stack Overflow

Regular Expressions are the bane of my existence. I hate them and everyone seem to have their own slightly different syntax. Anyhow, I have a problem with the below code, as it gives me a stack overflow:
 public static void main(String[] args) throws IOException {
 System.out.print(parseText(readTextFile("c:\\file.txt"))); // the input file is displayed below
// simple method to test that the regular expression can be matched. it works for simple expression, but not the one below
// this is the code that causes stack overflow
 public static String parseText(String inputString) {
 Pattern p = Pattern.compile("<tr>(\\s)*<td class = 'abc[0-9]*(\\s)*'>(.|\\s)+(\\s)+</td>(\\s)+</tr>");
 Matcher m = p.matcher(inputString);
 m.find();
 return m.group();
// this method just reads in a text file and returns a String with the whole content
// this works, but is posted here for completeness
 public static String readTextFile(String filePath)
 File file = new File(filePath);
 StringBuffer contents = new StringBuffer();
 BufferedReader reader = null;
 try {
 reader = new BufferedReader(new FileReader(file));
 String text = null;
 // repeat until all lines is read
 while ((text = reader.readLine()) != null) {
 contents.append(text).append(
 System.getProperty("line.separator"));
 } catch (FileNotFoundException e) {
 e.printStackTrace();
 } catch (IOException e) {
 e.printStackTrace();
 } finally {
 try {
 if (reader != null) {
 reader.close();
 } catch (IOException e) {
 e.printStackTrace();
 return contents.toString();
 }Below is the content of what I'm feeding the above code as file.txt that results in an error.
 <tr>
 <td class = 'abc1 '>
 <a href = 'http://somedomain.com/bla.html' class="a1">
 something 1
 </a>
 </td>
 <td class = 'abc1 '>
 cars
 </td>
 <td class = 'abc1 ' nowrap align="center">
 ibm
 </td>
 <td class = 'abc1 '>
 <a href = 'http://somedomain.com/bla2.html'>
 Features
 </a>
 </td>
 <td class = 'abc1 '>
 Feb 5, 2009
 </td>
 </tr>
 <tr>
 <td class = 'abc1 '>
 <a href = 'http://somedomain.com/bla2.html' class="b1">
 something 2
 </a>
 </td>
 <td class = 'abc1 '>
 car
 </td>
 <td class = 'abc1 ' nowrap align="center">
 sun
 </td>
 <td class = 'abc1 '>
 <a href = 'http://somedomain.com/bla3.html'>
 Features
 </a>
 </td>
 <td class = 'abc1 '>
 Feb 5, 2009
 </td>
 </tr>
 Here is some lines from the stack overflow output:
Exception in thread "main" java.lang.StackOverflowError
 at java.util.regex.Pattern$GroupTail.match(Unknown Source)
 at java.util.regex.Pattern$BranchConn.match(Unknown Source)
 at java.util.regex.Pattern$CharProperty.match(Unknown Source)
 at java.util.regex.Pattern$Branch.match(Unknown Source)
 at java.util.regex.Pattern$GroupHead.match(Unknown Source)However, if I feed the program this shorter version of the input it works (it matches only once and is just the top portion of the original file.txt):
 <tr>
 <td class = 'abc1 '>
 <a href = 'http://somedomain.com/bla.html' class="a1">
 something 1
 </a>
 </td>
 <td class = 'abc1 '>
 cars
 </td>
 <td class = 'abc1 ' nowrap align="center">
 ibm
 </td>
 <td class = 'abc1 '>
 <a href = 'http://somedomain.com/bla2.html'>
 Features
 </a>
 </td>
 <td class = 'abc1 '>
 Feb 5, 2009
 </td>
 </tr>I did some google searches, but the problem appears to do with how I crafted my regular expression. The solution didn't make sense to me.... Anyone help me please as I spent a big portion of the day already!

@flounder,
Wow, hold your horses...
That comment was meant to denote my level of frustration with this since I had spent a large portion of the day trying to figure it out. It was not intended to designate my issue or time as more important than others.
I certainly appreciate any help considering nobody is being compensated in any way for helping me.
@jschell,
This is intended to be used as a one time deal to automate recapturing data we no longer have in a non-HTML format.
I appreciate the suggestion though.
@JoachimSauer
Thank you for all your help. I appreciate it!

Regular Expression Question, Repetition Operators

These are my success entries for a field;
123456,
123456,123456,
123456,123456,123456,
123456,123456,123456,123456,
"," seperated 6 digits can be repeated unlimited times.
I found on documentation this; "Repetition Operators; {m,} Match at least m times" and for my need i tried this regular expression; "^[[[:digit:]]{6},]{1,}$", but didnt worked :(
Any comments?
Thank you very much :)
Tonguc

repeating exactly 6
{6}
repeating at least 1
+
repeating at least 6
{6,}
ok, your problem is [ instead of (

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("","")

Regular expression question (should be an easy one...)

i'm using java to build a parser. im getting an expression, which i split on a white-space.
how can i build a regular-expression that will enable me to split only on unquoted space? example:
for the expression:
(X=33 AND Y=44) OR (Z="hello world" AND T=2)
I will get the following values split:
(X=33
AND
Y=34)
OR
(Z="hello world"
AND
T=2)
and not:
(Z="
hello
world"
thank you very much!

Instead of splitting on whitespace to get a list of tokens, use Matcher.find() to match the tokens themselves: import java.util.*;
import java.util.regex.*;
public class Test
public static void main(String[] args) throws Exception
 String str = "(X=33 AND Y=44) OR (Z=\"hello world\" AND T=2)";
 List<String> tokens = new ArrayList<String>();
 Matcher m = Pattern.compile("[^\\s\"]+(?:\".*?\")?").matcher(str);
 while (m.find())
 tokens.add(m.group());
 System.out.println(tokens);
}{code} The regex I used is based on the assumptions that there will be at most one run of quoted text per token, that it will always appear in the right hand side of an expression, and that the closing quote will always mark the end of the token. If the rules are more complicated (as sabre150 suggested), a more complicated regex will be needed. You might be better off doing the parsing the old-fashioned way, with out regexes.

Regular Expressions Question

I am using regular expressions to parse through a text file generated by a set of sensors attached to a SeaBird CTD profiler, what matter is that some lines use quotes as a special marker and some dont, but I need to treat both as the same data. Here is the example:
SeaBird Datafile (just some part):
"# name 0 = depS: depth, salt water [m]"
"# name 1 = t068: temperature, IPTS-68 [deg C]"
# name 2 = sal00: salinity, PSS-78 [PSU]
"# name 3 = sigma-t00: density, sigma-t [kg/m^3]"
"# name 4 = flS: fluorometer, sea tech"And I am using this regular expression to read this lines and save those names:
private static final String ColumnName = "(^\"#|^# ) name (\\d)+ = ((.+)\"$|(.+)$)";There is a possiblity that the string either starts with "# and ends with ", or that it only starts with # and no special marker to the end. Can anyone enlighten me on the correct regex because the one I posted up there only works in the case of being surrounded by quotes. Thanks in advance!
Christian A. Sueiras

Try this: private static final String ColumnName = "^(\"?)# name (\\d+) = (.+)\\1$";{code} You match an optional quotation mark at the beginning, and capture it in group #1. At the end, you match whatever was captured in group #1: either a quotation mark, or nothing.

A regular expression question

I have a large body of text which I am breaking into individual words (as part of an experimental indexing project.)
I can break the text into a list by making a paragraph break for every word space (a simple find and replace).
But I want proper names to remain unbroken.
So I am trying to write a regular expression script which will find every occurence of two contiguous words which each begin in a capital letter, and then to replace the space between the two words with an underscore.
So Sigmund Freud becomes Sigmund_Freud.
Does anyone know how I would write this script?
Thanks!!!

You don't need a script, you can do it in the interface:
Find: (\u[-\w]+)\x{20}(?=\u[-\w]+)
Change: $1_
\u[-\w]+ stands for "upper-case letter followed by one or more of hyphen/word character"; here the first name.
\x{20} stands for the space.
followed by another \u[-\w]+, the last name. This one is in a lookahead, so the whole expression paraphrases as "find a word that starts with an upper-case letter followed by a space if it's followed by another word starting with an uc letter".
Peter

Validation on text item as Regular Expression Question(Repetition Operator)

These are my success entries for a field;
123456,
123456,123456,
123456,123456,123456,
123456,123456,123456,123456,
"," seperated 6 digits can be repeated unlimited times.
I found on documentation this; "Repetition Operators; {m,} Match at least m times" and for my need i tried this regular expression; "^[[[:digit:]]{6},]{1,}$", but didnt worked :(
Any comments?
Thank you very much :)
Tonguc

Your expression is a little incorrect because you can't use repetition operators within a bracket expression. How about something like this
"^([[:digit:]]{6},)+$"
Regards,
Peter

Quick regular expression question!

If I have a String such as:
"This is a sentence.[1] This is another."
OR
"This is a sentence.(1) This is another."
I.e. A String has a full stop within it and a non alphanumeric character immediately after it (without whitespace between the full stop and the character)
How can I insert a whitespace character between the full stop and the non alphanumeric character (such as a bracket in the above examples)?
So the above Strings would be transformed into:
"This is a sentence. [1] This is another."
"This is a sentence. (1) This is another."
Thanks

If I understand what you're asking...
str = str.replaceAll("\\.([^\\p{Alnum}\\s])", ". $1");
"This is sentence.[1] This is another.(1) This is a third. [1] This is a fourth.& This is a fifth. This is the last."
"This is sentence. [1] This is another. (1) This is a third. [1] This is a fourth. & This is a fifth. This is the last."For more info:
http://java.sun.com/docs/books/tutorial/extra/regex/index.html
http://www.regular-expressions.info/

Help: Regular Expression - 2

Hi,
Good Day to All..
Database : 11g
Col1
Louie: The Complete First Season [2 Discs] [DVD/Blu-ray]
Snow White and the Seven Dwarfs [Blu-ray/DVD] - Bonus Disc
About Last Night... [WS] [Blu-ray]
High3:[Extended] [3 Discs] [Includes Digital Copy] [BR]
High4: The [Extended] [3 Discs] [BR]
Output should show in the below format:-
Col1
Louie: The Complete First Season [Blu-ray]
Snow White and the Seven Dwarfs [Blu-ray] - Bonus Disc
About Last Night...[Blu-ray]
High3: [BR]
High4: The [BR]
Even if my column contains [Blu-ray/DVD] or [DVD/Blu-ray], i have to replace [Blu-ray] and for other [xxx] braces section has to be removed.
Query which used and worked out for first three column values :-
SELECT col1,
REPLACE(regexp_replace (regexp_replace (col1, '\[.*?Blu.*?ray.*?\]' , '{Blu-ray}' , 1 , 1 ,'i') ,'\[.*?\]') , '{Blu-ray}' , '[Blu-ray]')
new_col1
FROM tablename
WHERE col1 LIKE '%Blu-ray/DVD%' OR col1 LIKE '%[DVD/Blu-ray]%'
OR col1 LIKE '%[Blu-ray]%'
OR col1 LIKE '%[BR]%'Need the solution for [BR] column values.
Thanks ...

One way is to keep nesting your regexps:
SQL> with tab1 as
2 (
3 select 'The Complete Season Three [3 Discs] [Blu-ray] - Disc 1' col1 from dual union all
4 select 'Another movie [5 Discs] [DVD/blu-ray] - Disc 1' col1 from dual union all
5 select '(500) Days of Summer [2 Discs] [Includes Digital Copy] [BLURAY/DVD]' col1 from dual union all
6 select 'High3:[Extended] [3 Discs] [Includes Digital Copy] [BR]' col1 from dual union all
7 select 'High4: The [Extended] [3 Discs] [BR]' col1 from dual
8 )
9 select
10 replace(
11 replace(
12 regexp_replace(
13 regexp_replace(
14 regexp_replace(
15 col1,
16 '\[.*?Blu.*?ray.*?\]', '{Blu-ray}', 1, 1, 'i'
17 ),
18 '\[BR\]', '{BR}', 1, 1, 'i'
19 ),
20 '\[.*?\]'
21 ),
22 '{Blu-ray}',
23 '[Blu-ray]'
24 ),
25 '{BR}',
26 '[BR]'
27 ) test2
28 from tab1
29 ;
TEST2
The Complete Season Three [Blu-ray] - Disc 1
Another movie [Blu-ray] - Disc 1
(500) Days of Summer [Blu-ray]
High3: [BR]
High4: The [BR]The regexp expression in line 16 take [<anything>Blu<anything>ray<anything>] and turn it into '{Blu-ray}. (The original from your last question.)
The regexp in line 18 take [BR] and turn it into {BR} (again the 'i' parameter makes it case in-sensitive so it also catches [Br] [br] and [bR].)
Then line 20 removes all other [] bracket expressions.
Line 22-23 turns {Blu-ray} back into [Blu-ray] and 25-26 turns {BR} into [BR].
You can keep nesting this.
The trick is first to turn [something] into {something}, before you remove all other [], and then turn the {something} into [something].
As with the [<anything>Blu<anything>ray<anything>] expression you can also catch the data even if Blu-ray is not spelt exactly right.
And if you also find cases with for instance [BR/DVD] and wish to turn that into [BR], then you add some suitable .*? in line 18 the same way as line 16 ;-)
Maybe you should consider selecting all bracket expressions in your data in order to get an overview of how many different bracket expression you have to nest?
Edit:
Couldn't resist trying that ;-)
SQL> with tab1 as
2 (
3 select 'The Complete Season Three [3 Discs] [Blu-ray] - Disc 1' col1 from dual union all
4 select 'Another movie [5 Discs] [DVD/blu-ray] - Disc 1' col1 from dual union all
5 select '(500) Days of Summer [2 Discs] [Includes Digital Copy] [BLURAY/DVD]' col1 from dual union all
6 select 'High3:[Extended] [3 Discs] [Includes Digital Copy] [BR]' col1 from dual union all
7 select 'High4: The [Extended] [3 Discs] [BR]' col1 from dual
8 )
9 select
10 bracket_exp,
11 count(*) cnt
12 from (
13 select
14 regexp_substr(tab1.col1,'\[.*?\]',1,r.r) bracket_exp
15 from tab1
16 cross join (
17 select level r
18 from dual
19 connect by level <= 10
20 ) r
21 )
22 where bracket_exp is not null
23 group by bracket_exp
24 order by cnt desc
25 ;
BRACKET_EXP CNT
[3 Discs] 3
[Extended] 2
[BR] 2
[Includes Digital Copy] 2
[Blu-ray] 1
[5 Discs] 1
[2 Discs] 1
[BLURAY/DVD] 1
[DVD/blu-ray] 1
9 rows selected.This is just a piece of one-off code to help you analyze your data and discover which bracket expressions you need to care about.
It is not necessarily optimal, and I have assumed there will be no more than 10 bracket expressions (you could raise that number if you wish.)
Possibly you might consider:
upper(regexp_substr(tab1.col1,'\[.*?\]',1,r.r)) bracket_expto count as identical those bracket expressions that are alike except for lower/upper-case.
Edited by: Kim Berg Hansen on Sep 14, 2011 10:18 AM

Regular expression question - match a template

Hello All!
I've put regex patterns to simple use b4, but this one I'm finding a bit of a challenge.
I need to match lines to the following template:
"9############ ###9,99 XXXXX####9,99 #9"In this template 9s mean digits, Xs mean alphanum characters and the #s mean optional digits.
For example the following strings must mutch the above pattern:
1. "123             18,10 abcde    9,99 1"
2. "1234567890123 1000,99 CCCCC 260,99 2"This is the regex I came up with so far.
"^[ \\d]{13}? [ \\d]{3}?\\d,\\d\\d .{5}?[ \\d]{4}\\d,\\d\\d [ \\d]\\d$"The problem with it of course is that it will match a string like:
1. "9            9 9 ,99 abcde 9 9,99 1"Basically, the question is whether it is possible to ensure that when the optional digits are
not present, they have the required spaces in their positions.
Any help is much appreciated.
P.s. the reason I am trying to use regex instead of splitting the string and validating the tokens
separately is because I have a number of such templates to validate lines against, and each template signals a particular line type. :) Basically I need to decide what type of line is it I just read from a file...
Thnx.

First of all, isn't the first digit mandatory?
That would mean that the first part of your regex should be:"^\\d"followed by 12 characters that can be either ' ' or any digit:"^\\d[ \\d]{12}"followed by space, three spaces or digits and a digit"^\\d[ \\d]{12} [ \\d]{3}\\d"comma, two digits, space, five alphanums"^\\d[ \\d]{12} [ \\d]{3}\\d,\\d{2} [\\w[^_]]{5}"4 space or digit, digit, comma, 2 digits"^\\d[ \\d]{12} [ \\d]{3}\\d,\\d{2} [\\w[^_]]{5}[ \\d]{4}\\d,\\d{2}"space, space or digit, digit, the end"^\\d[ \\d]{12} [ \\d]{3}\\d,\\d{2} [\\w[^_]]{5}[ \\d]{4}\\d,\\d{2} [ \\d]\\d$"And please note that the string you provided"9            9 9 ,99 abcde 9 9,99 1"is not supposed to match but the string"9                9,99 abcde    9,99 9"will.

Regular expression question about punctuation

hello there
i'm trying to check a line in the command line to find the mac address
so i need a little help in the pattern to complie
here is what i've tried
Pattern p4=Pattern.compile("\\d|\\w{0,}\\p{Punct}\\d|\\w{0,}\\p{Punct}\\d|\\w{0,}\\p{Punct}\\d|\\w{0,}\\p{Punct}\\d|\\w{0,}\\p{Punct}\\d|\\w{0,}");but it didn't work so what 's wrong?

thanks uncle_alice it works
but i didn't understand the code well
\\p{XDigit}{2}                    exactly two hexadecimal digits
(?:                               the following as a non-capturing group
[:-]                            followed by a ':' or a '-'
\\p{XDigit}{2}                  followed by exactly two hexadecimal digits
){5}                              the previous (group) exactly five timesHope I got my terminology correct...

Help: Regular Expression question??

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