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
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Pattern matching using Regular expression
Hi,
I am working on pattern matching using regular expression. I the table, I have 2 columns A and B
A has value 'A499BPAU4A32A386KBCZ4C13C41D20E'
B has value like '*CZ4*M11*7NQ+RDR+RSM-R9A-R9B'
the requirement is that I have to match the columns of B in A. If there is a value with * sign, this must be present in A like 'CZ4' should exit in string A.
The issue I am facing is that there are 2 values with * sign. The code works fine for first match (CZ4) but it does not look further as M11 does not exist in A.
I used the condition
AND instr(A,substr(REGEXP_SUBSTR(B, '*[^*]{3}'),2) ,1)=0
First of all, is this possible to match multiple patterns in one condition?
If yes, please suggest.
Thanksuser2544469 wrote:
Thanks a lot Frank. This query worked wonderful for the test data I have provided however I have some concerns:
- query doesnot include the column BOOK which is a mandatory check.Sorry, that was my mistake. It was a very easy mistake to make, since you posted sample data where it didn't matter. Instead of doing a cross-join between vn and got_must_have_cnt, do an inner join, using book. That means book will have to be in got_must_have_cnt, and all the sub-queries from which it descends. Look for comments that say "March 22".
If you want to treat '+' in test_cat.codes as '*', then the simplest thing is probably just to use REPLACE, so that when the table has '+', you use '*' instead.
WITH got_token_cnt AS
SELECT cat
, book -- Added March 22
, REPLACE (codes, '+', '*') AS codes -- If desired. Changed March 22
, LENGTH (codes) - LENGTH ( TRANSLATE ( codes
, 'x*+-'
, 'x'
) AS token_cnt
FROM test_cat
, cntr AS
SELECT LEVEL AS n
FROM ( SELECT MAX (token_cnt) AS max_token_cnt
FROM got_token_cnt
CONNECT BY LEVEL <= max_token_cnt
, got_tokens AS
SELECT t.cat
, t.book -- Added March 22
, REGEXP_SUBSTR ( t.codes
, '[*+-]'
, 1
, c.n
) AS token_type
, SUBSTR ( REGEXP_SUBSTR ( t.codes
, '[*+-][^*+-]*'
, 1
, c.n
, 2
) AS token
FROM got_token_cnt t
JOIN cntr c ON c.n <= t.token_cnt
, got_must_have_cnt AS
SELECT cat, book -- Changed March 22
, COUNT (CASE WHEN token_type = '*' THEN 1 END) AS must_have_cnt
FROM got_tokens
GROUP BY cat, book -- Changed March 22
SELECT mh.cat
, vn.vn_no
FROM got_must_have_cnt mh
JOIN vn ON mh.book = vn.book -- Changed March 22
LEFT OUTER JOIN got_tokens gt ON mh.cat = gt.cat
AND INSTR (vn.codes, gt.token) > 1
GROUP BY mh.cat
, mh.must_have_cnt
, vn.vn_no
HAVING COUNT (CASE WHEN gt.token_type = '*' THEN 1 END) = mh.must_have_cnt
AND COUNT (CASE WHEN gt.token_type = '-' THEN 1 END) = 0
ORDER BY mh.cat
- query is very slow with 60000 records in vn table. Cost is somewhere around 36000.See these threads:
When your query takes too long ...
HOW TO: Post a SQL statement tuning request - template posting
Relational databases were designed to have (at most) one piece of information in each column. If you decide to have multiple items in the same column (as you have a variable number of tokens in the codes column), don't be surprised if that makes things slower and more complicated. Most of the query I posted, and perhaps most of the time needed, is jsut to normalize the data. If you stored the data in a narmalized form, perhaps something like got_tokens, then you wouldn't need the first 3 sub-queries that I posted.
Edited by: Frank Kulash on Mar 22, 2011 12:04 PM -
XML node name matching with regular expressions
Hello,
If i have an xml file that has the following:
<parameter>
<name>M2-WIDTH</name>
<value column="09" date="2004-10-31T19:56:30" row="03" waferID="PUK444150-20">10.4518</value>
</parameter>
<parameter>
<name>M2-GAP</name>
<value column="29" date="2004-10-31T19:56:30" row="06" waferID="PUK444150-03">2.864</value>
</parameter>
<parameter>
<name>RES-LENGTH</name>
<value column="29" date="2004-10-31T19:56:30" row="06" waferID="PUK444150-03">2.864</value>
</parameter>
Is there anyway i can get a list of nodes that match a certain pattern say where name=M2* ?
I cant seem to find any information where i can match a regular expression. I see how you can do:
String expression=/parameter[@name='M2-LENG']/value/text()";
NodeList nodes = (NodeList) xPath.evaluate(expression, inputSource, XPathConstants.NODESET);
But i want to be able to say:
String expression=/parameter[@name='M2-*']/value/text()";
Is this possible? if so how can i do this?
Thanks!As implemented in Java, XPath does not support regular expressions, but in most cases there are workarounds thanks to XPath functions. Correct me if I'm wrong, but setting your expression against the XML document (i.e. because there are no "name" attributes in the whole document) I think you mean to get the value of the <value> elements that have a <parameter> parent element and a <name> sibling element whose value starts with "M2-". If that is the case, you can use the following query expression:String expression = "//parameter/value[substring(../name,1,3)='M2-']";Sorry if I misunderstood the meaning of your expression, but I hope this will help you get the hang of using XPath functions as a substitute for regular expressions.
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Logical AND in Java Regular Expressions
I'm trying to implement logical AND using Java Regular Expressions.
I couldn't figure out how to do it after reading Java docs and textbooks. I can do something like "abc.*def", which means that I'm looking for strings which have "abc", then anything, then "def", but it is not "pure" logical AND - I will not find "def.*abc" this way.
Any ideas, how to do it ?
BakenFirst off, looks like you're really talking about an "OR", not an "AND" - you want it to match abc.*def OR def.*abc right? If you tried to match abc.*def AND def.*abc nothing would ever match that, as no string can begin with both "abc" and "def", just like no numeric value can be both 2 and 5.
Anyway, maybe regex isn't the right tool for this job. Can you not simply programmatically match it yourself using String methods? You want it to match if the string "starts with" abc and "ends with" def, or vice-versa. Just write some simple code. -
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,
beroetzI 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 -
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><!-- if( typeof( kadovInitEffects ) != 'function' ) kadovInitEffects = new Function();if( typeof( kadovInitTrigger ) != 'function' ) kadovInitTrigger = new Function();if( typeof( kadovFilePopupInit ) != 'function' ) kadovFilePopupInit = new Function();if( typeof( kadovTextPopupInit ) != 'function' ) kadovTextPopupInit = new Function(); //--></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 aIf 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 -
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;
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found = true;
if(!found)
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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> -
Patterns and Matcher.find()
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I have a string such as this: "Hello this is a great <!-- @@[IMAGINE_SPACIAL_VECTOR]@@ --> little string"
I want to use RegEx, Pattern and Matcher.find() to extract "IMAGINE_SPACIAL_VECTOR" ie. the text between "<!-- @@[" and "]@@ -->"
Thanks in advance for your helpimport java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Example {
public static void main(String[] args) {
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Matcher matcher = Pattern.compile("<!-- @@\\[(.*?)]@@ -->").matcher(data);
while (matcher.find()) {
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How to Use Pattern and Matcher class.
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Please help out..
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Please correct me if I am wrong.
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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).
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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/.
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Sasi.
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ie
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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 $$.
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public static final int DATA_ENTRY = 1;
public static final int INVALID_CURRENCY = 2;
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state = INVALID_CURRENCY;Here's two pattern strings that both require a two-place fraction for each entry but do not permit more than two places.
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public class snggun {
public static void main(String[] args) {
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} -
String.matches() question - regular expression help
How come the following code's if condition returns false?
String someFile="Dr. Phil.pdf";
if (someFile.matches("[.][Pp][Dd][Ff]$")) {
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matches(".*[.][Pp][Dd][Ff]$")Thanks.The documentation is your friend.
[String.matches(regex)|http://java.sun.com/javase/6/docs/api/java/lang/String.html#matches(java.lang.String)] says:
An invocation of this method of the form str.matches(regex) yields exactly the same result as the expression
Pattern.matches(regex, str)And [Pattern.matches(regex, str)|http://java.sun.com/javase/6/docs/api/java/util/regex/Pattern.html#matches(java.lang.String, java.lang.CharSequence)] says
behaves in exactly the same way as the expression
Pattern.compile(regex).matcher(input).matches()And [Matcher.matches()|http://java.sun.com/javase/6/docs/api/java/util/regex/Matcher.html#matches()] says
Attempts to match the entire region against the pattern. -
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abc
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Database access denied for MCT6 in Multisim 10 Student Edition
I am getting this error for a component that I was able to use over the last few days. The component is the MCT6 optocoupler. Luckily I had saved an older version of my circuit, or I would be screwed. I am using Multisim Student Edition Version 10.0
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I dont really think that the math header has syntax errors.
Hi, This is a very serious accusation from make. Mini:Evaluator develop$ make bison -d grammar.y grammar.y: conflicts: 24 shift/reduce flex rules.l cc -0 -o Evaluator grammar.tab.c lex.yy.c -ly -ll -lm In file included from grammar.y:3: /usr/include/