Java Regex - Find Last Match Using Pattern and Matcher

I'd like to write some regex which would allow me to grab the last occurance of match based on a specified list of items. So for:
Pattern languageRegex = Pattern.compile("(len|end)");
And a string of:
"00| 0lend|"
I want it to extract "end". However, I'd grab "len" using the above regex.
If it was
"00| 0lenend|"
I'd grab "end" which is right.
What regex would allow me to grab "end" rather than "len" from:
"00| 0lend|"
Thanks for your help.

user3940995 wrote:
I have a list of 3 letter codes that I need to check for in a field. The list is finite but about 100 or so items:
len, end, ren, onm, enl, etc.
However, the field I'm checking in has some other data in it which can bleed into the code but the code will always be at the end.
An example would be "000 0rend"
From this I'd want to extract "end". If there is a better way to do this than using regex then I'd be happy to use that, but as I have to process millions of items I'm keen to not loop trying to find a match so I was hoping there would be a regex solution.
Your regex would work for that particular example. I think if I modify it to be
Pattern.compile("len(?!\\D)|end(?!\\D)|enl(?!\\D)"); (which would then be extended for all the list items)
Then I seem to pick up the last occurance as I'd like to.
Thank you for your help!Doesn't sound like you want to use regexp. I would instead build a character graph/tree with my commands in reversed order. I would then search each line backwards and check if it matches something in my tree.

Similar Messages

String.matches vs Pattern and Matcher object

Hi,
I was trying to match some regex using String.matches but for me it is not working (probably I am not using it the way it should be used).
Here is a simple example:
/* This does not work */
String patternStr = "a";
String inputStr = "abc";
if(inputStr.matches( "a" ))
System.out.println("String matched");
/* This works */
Pattern p = Pattern.compile( "a" );
Matcher m = p.matcher( "abc" );
boolean found = false;
while(m.find())
System.out.println("Matched using Pattern and Matcher");
found = true;
if(!found)
System.out.println("Not matching with Pattern and Matcher");
Am I not matches method of String class properly?
Please throw some lights on this.
Thank you.

String.matches looks at the whole string.
bsh % "abc".matches("a");
<false>
bsh % "abc".matches("a.*");
<true>

How to Use Pattern and Matcher class.

HI Guys,
I am just trying to use Pattern and Matcher classes for my requirement.
My requirement is :- It should allow the numbers from 1-7 followed by a comma(,) again followed by the numbers from
1-7. For example:- 1,2,3,4,5 or 3,6,1 or 7,1,3 something like that.
But it should not allow 0,8 and 9. And also it should not allow any Alphabets and special characters except comma(,).
I have written some thing like..
Pattern p = Pattern.compile("([1-7])+([\\,])?([1-7])?");
Is there any problem with this pattern ??
Please help out..
I am new to pattern matching concept..
Thanks and regards
Sudheer

ok guys, this is how my code looks like..
class PatternTest
     public static void main(String[] args)
          System.out.println("Hello World!");
          String input = args[0];
          Pattern p = Pattern.compile("([1-7]{1},?)+");
          Matcher m = p.matcher(input);
          if(m.find()) {
               System.out.println("Pattern Found");
          } else {
               System.out.println("Invalid pattern");
}if I enter 8,1,3 its accepting and saying Pattern Found..
Please correct me if I am wrong.
Actually this is the test code I am presenting here.. I original requirement is..I will be uploading an excel sheets containg 10 columns and n rows.
In one of my column, I need to test whether the data in that column is between 1-7 or not..If I get a value consisting of numbers other than 1-7..Then I should
display him the msg..
Thanks and regards
Sudheer

Lexical search using Pattern and Matcher class

Hi Folks,
Need some help with the following Query. I want to find out how I can implement
the following by using Pattern / Matcher classes.
I have a query that returns the following set of strings,
aa
abc
def
ghi
glk
gmonalaks
golskalskdkdkd
lkaldkdldldkdld
mladlad
n33ieler
What I would like do is to find out any string that starts with g and Lexical occurs after ghi. So this will return
ghi / glk / gmonalaks / golskalskdkdkd
How can I accomplish the above, by using the following two classes.
Pattern datePattern = Pattern.compile();
Matcher dateMatcher = datePattern.matcher();
Thanks a bunch...
_Shoe Maker..

Nothing in your specification requires a regex. Loop though the strings until you find the string "ghi". Then continue looping though the strings outputting those where the first characters is 'g' .

Regular expressions, using pattern and matcher but not include the pattern

Hi all
i have a regular expression but in my matcher it is including the text that is in my regular expression.
ie
String str="-------------------stuff===========";
Pattern mainBody = Pattern.compile("-----(.*?)=====", Pattern.MULTILINE);this matches -------------------stuff=====
now i expect to get some - in there as i match from the start, but i dont want to have the = in the match. how do i do a match that excludes the matching expressions.

nevermind, figured it out
when i do myMatch.group(1) it gives me just my match
sorry for wasting time :)

How to use regular expression using pattern and match concept for this scenario?

Hi Guys,
I have a string "We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle"
I need to replace the numbers based on the below condition.
if more then 5, replace with many
if less then 5, replace with a few
if it is 1, replace with "only one"
below is my code, I am missing the equating part to replace the numbers could any one of you please help me out fixing this.
private static String REGEX="(\\d+)";
private static String INPUT="We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle";
//String pattern= "(.*)(\\d+)(.*)";
private static String REPLACE = "replace with many";
public static void main(String[] args) {
// Create a Pattern object
       Pattern r = Pattern.compile(REGEX);
    // Now create matcher object.
       Matcher m = r.matcher(INPUT);
       //replace the value 7 by the replace string
//How to equate the (\\d+) greater than a number and use it the below code.
              INPUT = m.replaceAll(REPLACE);
       //Print the final Result;
        System.out.println(INPUT);
Thanks and Regards,

Hi,
Try the following which makes use of "appendReplacement" instead with the "start" and "end" methods to locate and check the searched "regExp" string before dynamically setting the "replace" string:
String regExp = "\\d+";
String input = "We have 7 tutorials for Java, 2 tutorials for Javascript and 1 tutorial for Oracle";
String replace;
Pattern p = Pattern.compile(regExp);
// get a matcher object
Matcher m = p.matcher(input);
StringBuffer sb = new StringBuffer();
while (m.find()) {
   Integer x = Integer.valueOf(input.substring(m.start(), m.end()));
   replace = (x >= 5) ? "many" : (x == 1) ? "only one" : "few";
   m.appendReplacement(sb, replace);
m.appendTail(sb);
System.out.println(sb.toString());
HTH.
Regards,
Rajen
P.S: Please mark the post as answered/helpful if it resolves your issue for the benefit of all community members.

Patterns and Matcher.find()

I would be really grateful for some assistance on this. I have been at this for three hours and can't get it correct.
I have a string such as this: "Hello this is a great  little string"
I want to use RegEx, Pattern and Matcher.find() to extract "IMAGINE_SPACIAL_VECTOR" ie. the text between ""
Thanks in advance for your help

import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
     public static void main(String[] args) {
          String data = "Hello this is a great  little string";
          Matcher matcher = Pattern.compile("").matcher(data);
          while (matcher.find()) {
               System.out.println(matcher.group(1));
}Might not be the best way, but it works.
Kaj

Pattern and Matching question

Hey,
I'm trying to use the pattern and matcher to replace all instances of a website
address in some html documents as I process them and post them. I'm
including a sample of some of the HTML below and the code I"m using to
process it. For some reason it doesn't replace the sites in the underlying
images and i can't figure out what I'm doing wrong. Please forgive all the
unused variables, those are relics of another way i may have to do this if i
can't get the pattern thing to work.
Josh
     public static void setParameters(File fileName)
          FileReader theReader = null;
          try
               System.out.println("beginning setparameters guide2)");
               File fileForProcessing=new File(fileName.getAbsolutePath());
               //wrap the file in a filereader and buffered reader for maximum processing
               theReader=new FileReader(fileForProcessing);
               BufferedReader bufferedReader=new BufferedReader(theReader);
               //fill in data into the tempquestion variable to be populated
               //Set the question and answer texts back to default
               questionText="";
               answerText="";
               //Define the question variable as a Stringbuffer so new data can be appended to it
               StringBuffer endQuestion=new StringBuffer();//Stringbuffer to store all the lines
               String tempQuestion="";
               //Define new file with the absolutepath and the filename for use in parsing out question/answer data
               tempQuestion=bufferedReader.readLine();//reads the nextline of the question
               String tempAlteredQuestion="";//for temporary alteration of the nextline
               //while there are more lines append the stringbuffer with the new data to complete the question buffer
               StringTokenizer tokenizer=new StringTokenizer(tempQuestion, " ");//tokenizer for reading individual words
               StringBuffer temporaryLine; //reinstantiate temporary line holder each iterration
               String newToken;   //newToken gets the very next token every iterration? changed to tokenizer moretokens loop
               String newTokenTemp;   //reset newTokenTemp to null each iterration
               String theEndOfIt; //string to hold everything after .com
               char[] characters; //character array to hold the characters that are used to hold the entire link
               char lastCharChecked;
               Pattern thePattern=Pattern.compile("src=\"https:////fakesite.com//ics", Pattern.LITERAL);
               Matcher theMatcher=thePattern.matcher(tempQuestion);
                    while(tempQuestion!=null) //every time the tempquestion reads a newline, make sure you aren't at the end
                         String theReplacedString=theMatcher.replaceAll("https:////fakesite.com//UserGuide/");
                         //          temporaryLine=new StringBuffer();
                         //add the temporary line after processed back into the end question.
                         endQuestion.append(theReplacedString);                              //temporaryLine.toString());
                         //reset the tempquestion to the newline that is going to be read
                         tempQuestion=bufferedReader.readLine();
                         if(tempQuestion!=null)
                              theMatcher.reset(tempQuestion);
                         /*newTokenTemp=null;
                         while(tokenizer.hasMoreTokens())
                              newToken=tokenizer.nextToken(); //get the next token from the line for processing
                              System.out.println("uhhhhhh");
                              if(newToken.length()>36) //if the token is long enough chop it off to compare
                                   newTokenTemp=newToken.substring(0, 36);
                              if(newTokenTemp.equals("src=\"https://fakesite.com"));//compare against the known image source
                                   theEndOfIt=new String(); //intialize theEndOfIt
                                   characters=new char[newToken.length()]; //set the arraylength to the length of the initial token
                                   characters=newToken.toCharArray(); //point the character array to the actual characters for newToken
                                   lastCharChecked='a'; // the last character that was compared
                                   int x=0; //setup the iterration variable and go from the length of the whole token back till you find the first /
                                   for(x=newToken.length()-1;x>0&&lastCharChecked!='/';x--)
                                        System.out.println(newToken);
                                        //set last char checked to the lsat iterration run
                                        lastCharChecked=characters[x];
                                        //set the end of it to the last char checked and the rest of the chars checked combined
                                        theEndOfIt=Character.toString(lastCharChecked)+theEndOfIt;
                                   //reset the initial newToken value to the cut temporary newToken root + userguide addin, + the end
                                   newToken=newTokenTemp+"//Userguide"+theEndOfIt;
                              //add in the space aftr the token to the temporary line and the new token, this is where it should be parsed back together
                              temporaryLine.append(newToken+" ");
                         //add the temporary line after processed back into the end question.
                         endQuestion.append(temporaryLine.toString());
                         //reset the tempquestion to the newline that is going to be read
                         tempQuestion=bufferedReader.readLine();
                         //reset tokenizer to the new temporary question
                         if(tempQuestion!=null)
                         tokenizer=new StringTokenizer(tempQuestion);
               //Set the answer to the stringbuffer after converting to string
               answerText=endQuestion.toString();
               //code to take the filename and replace _ with a space and put that in the question text
               char theSpace=' ';
               char theUnderline='_';
               questionText=(fileName.getName()).replace(theUnderline, theSpace);
          catch(FileNotFoundException exception)
               if(logger.isLoggable(Level.WARNING))
               logger.log(Level.WARNING,"The File was Not Found\n"+exception.getMessage()+"\n"+exception.getStackTrace(),exception);
          catch(IOException exception)
               if(logger.isLoggable(Level.SEVERE))
               logger.log(Level.SEVERE,exception.getMessage()+"\n"+exception.getStackTrace(),exception);
          finally
               try
                    if(theReader!=null)
                         theReader.close();
               catch(Exception e)
<SCRIPT language=JavaScript1.2 type=text/javascript></SCRIPT>
<H1><IMG class=img_whs1 height=63 src="https://fakesite.com/ics/header4.jpg" width=816 border=0 x-maintain-ratio="TRUE"></H1>
<H1>Associate Existing Customers</H1>
<P>blahalbalhblabhlab blabhalha blabahbablablabhlablhalhab.<SPAN style="FONT-WEIGHT: bold"><B><IMG class=img_whs2 height=18 alt="Submit a

If you use just / it misinterprets it and it ruins
your " " tags for a string. I don't think so. '/' is not a special character for Java regex, nor for Java String.
The reason i used
literal is to try to force it to directly match,
originally i thought that was the reason it wasn't
working.That will be no problem because it enforces '.' to be treated as a dot, not as a regex 'any character'.
Message was edited by:
hiwa

Pattern and Matcher of Regular Expressions

Hello All,
MTMISRVLGLIRDQAISTTFGANAVTDAFWVAFRIPNFLRRLFAEGSFATAFVPVFTEVK
ETRPHADLRELMARVSGTLGGMLLLITALGLIFTPQLAAVFSDGAATNPEKYGLLVDLLR
LTFPFLLFVSLTALAGGALNSFQRFAIPALTPVILNLCMIAGALWLAPRLEVPILALGWA
VLVAGALQLLFQLPALKGIDLLTLPRWGWNHPDVRKVLTLMIPTLFGSSIAQINLMLDTV
IAARLADGSQSWLSLADRFLELPLGVFGVALGTVILPALARHHVKTDRSAFSGALDWGFR
TTLLIAMPAMLGLLLLAEPLVATLFQYRQFTAFDTRMTAMSVYGLSFGLPAYAMLKVLLP
I need some help with the regular expressions in java.
I have encountered a problem on how to retrieve two strings with Pattern and Matcher.
I have written this code to match one substring"MTMISRVLGLIRDQ", but I want to match multiple substrings in a string.
Pattern findstring = Pattern.compile("MTMISRVLGLIRDQ");
Matcher m = findstring.matcher(S);
while (m.find())
outputStream.println("Selected Sequence \"" + m.group() +
"\" starting at index " + m.start() +
" and ending at index " m.end() ".");
Any help would be appreciated.

Double post: http://forum.java.sun.com/thread.jspa?threadID=726158&tstart=0

"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

Searching and Matching - Difference between 'Match Pattern' and 'Match Geometric Pattern'?

I was wondering if someone can explain to me the difference between 'Match Pattern' and 'Match Geometric Pattern' VIs? I'm really not sure which best to use for my application. I'm trying to search/match small spherical particles in a grey video in order to track their speed (I'm doing this after subtracting two subsequent frames to get rid of background motion artifacts).
Which should I use?
Thank you!
Solved!
Go to Solution.

Hi TKassis,
1.You may find from this link for the difference between these two,
Pattern Match : http://zone.ni.com/reference/en-XX/help/370281P-01/imaqvision/imaq_match_pattern_3/
Geometric Match : http://zone.ni.com/reference/en-XX/help/370281P-01/imaqvision/imaq_match_geometric_pattern/.
2. I always prefer match pattern because of its execution speed, and incase of geometric pattern match it took lot of time to match your result. You may find in the attached figure for same image with these two algorithm execution time.
Sasi.
Certified LabVIEW Associate Developer
If you can DREAM it, You can DO it - Walt Disney

How to find last accessed/updated tables and the query text?

I am using :
Oracle8i Enterprise Edition Release 8.1.7.4.0 - Production
With the Partitioning option
JServer Release 8.1.7.4.0 - Production
How to find last accessed/updated tables and the query text?
Regards
LEE1212

Check DBA_TBALES view there you find one date column that indicate last update
One option is as follows:
(1) Turn the auditing on: AUDIT_TRAIL = true in init.ora
(2) Restart the instance if its running.
(3) Audit the table:
     AUDIT INSERT,SELECT,DELETE,UPDATE on TableName
     by ACCESS WHENEVER SUCCESSFUL
(4) Get the desired information using :
     SELECT OBJ_NAME,ACTION_NAME ,to_char(timestamp,'dd/mm/yyyy , HH:MM:SS')
     from sys.dba_audit_object.
Cheer,
Virag Sharma
http://virag.sharma.googlepages.com/
http://viragsharma.blogspot.com/
Message was edited by:
virag_sh

Regular Expression Character Sets with Pattern and Matcher

Hi,
I am a little bit confused about a regular expressions I am writing, it works in other languages but not in Java.
The regular expressions is to match LaTeX commands from a file, and is as follows:
\\begin{command}([.|\n\r\s]*)\\end{command}
This does not work in Java but does in PHP, C, etc...
The part that is strange is the . character. If placed as .* it works but if placed as [.]* it doesnt. Does this mean that . cannot be placed in a character range in Java?
Any help very much appreciated.
Kind Regards
Paul Bain

In PHP it seems that the "." still works as a all character operator inside character classes.
The regular expression posted did not work, but it does if I do:
\\begin{command}((.|[\n\r\s])*)?\\end{command}
Basically what I'm trying to match is a block of LaTeX, so the \\begin{command} and \\end{command} in LaTeX, not regex, although the \\ is a single one in LaTeX. I basically want to match any block which starts with one of those and ends in the end command. so really the regular expression that counts is the bit in the middle, ((.|[\n\r\s])*)?
Am I right it saying that the "?" will prevent the engine matching the first and last \\bein and \\end in the following example:
\\begin{command}
some stuff
\\end{command}
\\begin{command}
some stuff
\\end{command}

Pattern and Matcher problem. Help Please!

I am trying to make the user enter a correct US$ in HTML(jsp format) ie 12.34 but not 12.3456. As far as i know, the regular expression below is correct for $$.
but the there is no error even if the user types 12.34567
can anyone help me please?
this is the code i wrote...
public static final int DATA_ENTRY = 1;
public static final int INVALID_CURRENCY = 2;
public static final int PROCESS_INPUT = 3;
int state;
String a; //data input from user
Pattern p = Pattern.compile("\\d{1,3}(?:(?:,\\d\\d\\d)*|\\d*)(?:\\.\\d\\d)?");
Matcher m = p.matcher(a);
state = DATA_ENTRY; // this is default
if (m.find() {
     state = PROCESS_INPUT;
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Here's two pattern strings that both require a two-place fraction for each entry but do not permit more than two places.
import java.util.regex.*;
public class snggun {
public static void main(String[] args) {
    String input = (args.length > 0 ? args[0] : "10.00");
//    String mask = "\\d{1,3}(?:(?:,\\d\\d\\d)*|\\d*)(?:\\.\\d\\d)?";
    // requires two-place fraction
//    String mask = "^(\\d{1,3})(,(\\d\\d\\d))*(\\.\\d\\d)\\z";
    // requires two-place fraction
    String mask = "\\d{1,3}(?:(?:,\\d{3})*|\\d*)(\\.\\d{2})\\z";
    Pattern pattern = Pattern.compile(mask);
    Matcher match = pattern.matcher(input);
    System.out.println("input = " + input + " qualifies " + match.find());
    if(match.find())
      for(int j = 0; j <= match.groupCount(); j++)
        System.out.println("group " + j + " " + match.group(j));
}

How to find out the used space and free space in the DB for attachments

Hi,
In CRM 5.2 web UI, we can save a transaction by saving attachments like work documents or text files.
Could someone help me find out the used space, free space, maximum capacity on the CRM DB for these attachments.
Thanks & Best Regards,
Ramesh.

Hi,
check with below table
TNAPR --- Processing programs for output
and NACE Transaction code
NACE -- out types
Regards,
Madhu

Java Regex - Find Last Match Using Pattern and Matcher

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