PACKAGE BASICS.UTIL..REGEX

I have problem with my perogram!
is package basic util.regex need to be download or it's already built in java!for information i use j2sdk 1.4.1!
thanks a lot

you're probably thinking of java.util.regex, which is part of jdk 1.4 and above.

Similar Messages

Regular Expressions (java.util.regex)

I am developing using a product that must
use java 1.2.2_05a but I want to use regular
expressions, does anybody know where I can
get of the package java.util.regex without
having to download the whole java 1.4 release.
Or does someone know of an alternative that
I can use ?

There is another regex pack for java available from Apache Foundation Project. You can try it.
Take a look at http://jakarta.apache.org/

WHERE to find [b]java.util.regex.*[\b] package?

Does anyone know where to obtain a copy of the java.util.regex.* package (separate package)? This is a new package included in version 1.4.0.

Does anyone know where to obtain a copy of the
java.util.regex.* package (separate package)?
This is a new package included in version 1.4.0.Simple:
Go to your java sdk directory
$ jar xvf src.jar
$ cd java/util/regex
$ javac ASCII.java
$ javac Matcher.java
$ javac Pattern.java
$ javac PatternSyntaxException.java
$ cd ../../..
$ jar cvf regex.jar java/util/regex/*.class
et voila; how difficult can that be ? There is no native, JVM depending stuff in there, although I did not check for dependencies on other new stuff inside the jdk.
I'm also unaware if this isn't illegal under the agreement with Sun.

Parsing xhtml using java.util.regex

I am parsing an XHTML file using the java.util.regex package and I am perplexed at why the following doesn�t work.
The lines I wish to match are either like this:
Some String.</td>
or
Some String.</td>
The code I use to try to achieve this is:
Pattern somePattern = Pattern.compile(".*()?(.*)[.]()?</td>.*");
String s = null;
while((s = br.readLine()) != null) {
if(somePattern.matcher(s).matches()) {
System.out.println("0:" + eventMatcher.group(0));
System.out.println("1:" + eventMatcher.group(1));
System.out.println("2:" + eventMatcher.group(2));
System.out.println("3:" + eventMatcher.group(3));
I expect to get as output
0:Some String.</td> 1:
2:Some String
3:
or
0:Some String.</td>
1:null
2:Some String
3:null
depending on which lines provide the match as mentioned above. Instead I get:
0:Some String.</td>
1:null
2:(empty string)
3:
or
0:Some String.</td>
1:null
2:(empty string)
3:null
Any ideas? Thanks in advance.

Consider the terms of ".*()?(.*)[.]()?</td>.*"
.* - greedily collect characters
()? - optionallly collect information taht will always be matched by the previous .* pattern so will be empty.
(.*) - greedily collect characters that will also have been swallowed by the first .* so will be empty
[.] - a single .
()? - optionally collection
</td> - must be there
.* - collect the rest of the charcters in the line.
Therefore in general groups 1 and 2 will be empty because the first .* will have collected the information you wanted to capture!
You could just make the first .* non-greedy by using .*? but this may fail for other reasons.
So, in general terms, what are you trying to extract?

Please help on java.util.regex.*

Hi all,
My RTF file looks like this:
Project Num\tab N/A
\par Project Name\tab Hook-up Installation and Service
\par
My intension is to read the file until the \tab and store Project Num as a string into a
variable. Similarly read until \par and store the value of Project Num into another variable.
So that i can use those variables further in my program.
I used java.util.regex.* package for this purpose. I could successfully split the sentence whenever it sees \tab and \par but don't know how to get the text before and after the delimeters and store them into variables.
The code which i wrote is:
import java.util.regex.*;
import java.io.*;
import java.nio.*;
import java.nio.charset.*;
import java.nio.channels.*;
public class RegexDemo{
public static void main(String[] args){
// Create a pattern to match breaks
Pattern p = Pattern.compile("\\\\.[a-z][a-z]",Pattern.DOTALL);
try{
File file = new File("sample.rtf");
FileInputStream fis = new FileInputStream(file);
FileChannel fc = fis.getChannel();
// Get a CharBuffer from the source file
ByteBuffer bb = fc.map(FileChannel.MapMode.READ_ONLY, 0, (int)fc.size());
Charset cs = Charset.forName("8859_1");
CharsetDecoder cd = cs.newDecoder();
CharBuffer cb = cd.decode(bb);
// Run some matches
Matcher m = p.matcher(cb);
while (m.find())
System.out.println("Found comment: "+m.group());
}catch(Exception e){
e.printStackTrace();
Please somebody help me in this regard. I have spent lot of time searching the forums but couldn't find any solution.
Thanks in advance
rnallu

Just put target inside parenthesis with delimiters at boundaries.
Example: "(\\w+)\\t(\\d)\\s" will match occurrences of a word followed with a tab char then a digit followed with a whitespace. If target string matches pattern then m.group(1) contains the word and m.group(2) contains the digit.

Java.util.regex error

Hello,
I checked JavaDoc multiple times but do not see what is wrong with
myString.replaceAll("D:\\web\\mars","")which results in
java.util.regex.PatternSyntaxException: Illegal/unsupported escape squence near index 7
D:\web\mars
 ^
 at java.util.regex.Pattern.error(Unknown Source)
 at java.util.regex.Pattern.escape(Unknown Source)
 at java.util.regex.Pattern.atom(Unknown Source)
 at java.util.regex.Pattern.sequence(Unknown Source)
 at java.util.regex.Pattern.expr(Unknown Source)
 at java.util.regex.Pattern.compile(Unknown Source)
 at java.util.regex.Pattern.<init>(Unknown Source)
 at java.util.regex.Pattern.compile(Unknown Source)
 at java.lang.String.replaceAll(Unknown Source)
 at ArticleImageImportProcessor.main(ArticleImageImportProcessor.java:40)
Exception in thread "main" please, every suggestion/hint is most appeciated

You have to "encode" backslash twice, first for String purpose and second time because of special meaning of '\' in regular expressions.
It should looks like
myString.replaceAll("D:\\\\web\\\\mars","")

Who use sql-mapping with java.util.regex?

Hi everyone:
I use the IBatis SQL-Mapping and I think it is very good.Now I want to add the search function to my BBS forum.I also want to display the content high light like jive.I mean that if I want to find the string "ibatis",then the search result "ibatis" will be high light displayed.
So I must use the java.util.regex in jsdk1.4.But the problem is that what I get is a List if I use sql-mapping.For example:
          String resource="conf/XML/sql/lyo-sql-map.xml";
          Reader reader=Resources.getResourceAsReader(resource);
          sqlmap=XmlSqlMapBuilder.buildSqlMap(reader);
          List articlelist=sqlmap.executeQueryForList("selectSiteArticle","%"+icontent+"%");
The result I get is a List and I have no time to use regex.
I don't know whether I could do this:
Iterate the List,use the regex and later place all the object back to the List.
It's right?
How to use regex with sql-mapping?Thks

Any idea? :(

Import the classes in the package java.util

I haven't taken a class in two years so I have ideas but am unsure of myself, and I most likely am way off base, but is this how you import the classes in the package java.util
import java.util.*;
I would appreciate anyone's help. I have a lot of questions.
Thanks,
rp

My assignment is below so that you will know exactly what I am asking. I am being asked to create an instance of LinkedList. I know that this is a part of the java.util. So when you create an instance of LinkedList would that be similar to instance variables? or am I misunderstanding something. I have several ideas, so I will start with the first one.
LinkList();
I really appreciate your taking the time to answer my questions to help me to rebuild my confidence in doing this.
Thanks,
// 2. import the classes in the package java.util
public class MyProgram
     public static void main(String [] args)
          // 3. Create an instance of LinkedList
          // 4. add 5 entries to the list so that it contains
          //     Ann, Bart, Carl, Dirk, Zak
          // 5. display the list on one line using a loop and ADT list methods
          System.out.println("\n\n3. ");
          // 6. display the list using one println
          System.out.println("\n\n4. ");
          // 7. create a ListIterator object for the list
          // 8. display the list on one line using the iterator
          System.out.println("\n\n6. ");
          // 9. restore the iterator to the first item in the list
          // 10. retrieve and display the first item in the iteration,
          //      then advance the iterator to the next item
          System.out.println("\n\n8. ");
          // 11. advance the iterator to the end of the list without displaying anything;
          //      do not use the length of the list
          // 12. display the last item in the list
          System.out.println("\n\n10. ");
          // 13. move back to the beginning of the list without displaying anything;
          //      do not use the length of the list
          // 14. display the first item in the list
          System.out.println("\n\n12. ");
          // 15. advance the iterator to the end of the list without displaying anything;
          // 16. advance the iterator - what happens?
          System.out.println("\n\n14. ");
          // 17. advance the iterator within a try block; catch the exception, display a message,,
          //      and set the iterator back to the beginning of the list.
          System.out.println("\n\n15. ");
          // 18. what does nextIndex and previousIndex return?
          System.out.println("\n\n16. ");
          // 19. move the iterator to the third item in the list; if you were to
          //      execute next(), the third item would be returned and the iterator
          //      would advance to the 4th item.
          System.out.println("\n\n17. ");
          // 20. add the string "New" to the list; where is it inserted relative to
          //      the current item in the iteration?
          System.out.println("\n\n18. ");
          // 21. remove the current item; what happens?
          System.out.println("\n\n19. ");
          // 22. display the current item in the list without advancing the iteration
          //      in 2 ways, as follows:
          // 22A - using only iterator methods
          System.out.println("\n\n20A. ");
          // 22B using an iterator method and list methods
          System.out.println("\n\n20B. ");
          // 23. advance the iterator and remove the current item; which one is removed?
          System.out.println("\n\n21. ");
          // 24. move the iterator back and remove the current item; which one is removed?
          System.out.println("\n\n22. ");
          // 25. move the iterator to the first item in the list and change it to "First"
          System.out.println("\n\n23. ");
     } // end main
} // end MyProgram

Doubt in Regular Expressions : java.util.regex

I want to identify words starting with capital letters in a sentence and I want to replace the matched word with "#" added in front of it.... For example, if my input sentence is
"Christopher Williams asked Mike Muller a question"
my output should be,
"#Christopher #Williams asked #Mike #Muller a question"
How do I do that using java.util.regex ?
In perl we can can use *"back references"* in *"replacement string"* . Perl replacement accepts back references whereas java replacement method accepts only strings....
Please help me.....

Your replacement is swallowing the space before the uppercase character, and won't match at the beginning of the line.
Also, it's unnecesarily verbose. String has a replaceAll method (that calls the same methods of Pattern and Matcher under the covers)sentence = s.replaceAll("(^| )([A-Z])", "$1#$2");Disclaimer: I'm no prome, sabre or u/a :-) That can probably be simplified.
db

Regular expressions with java.util.regex

Hello Guys,
I wrote last time this
* Uses split to break up a string of input separated by
* commas and/or whitespace.
* See: http://developer.java.sun.com/developer/technicalArticles/releases/1.4regex/
* Change: I have slightly modified the source
import java.util.regex.*;
public class Splitter {
public static void main(String[] args) throws Exception {
// Create a pattern to match breaks
Pattern p = Pattern.compile("[<>\\s]+");
// Split input with the pattern
String[] result =
p.split("<element attributname1 = \"attributwert1\" attributname2 = \"attributwert2\">");
for (int i=0; i<result.length; i++)
if (result.equals(""))
System.out.println("EMPTY");
else
System.out.println(result[i]);
int res = result.length - 1;
System.out.println("\nStringlaenge: " + res);
I wonder, why I got an empty element in reult[0]. Have anyone an idea?
We'll come together next time
... �nhan Inay ([email protected])

What is wrong with this Pattern?
Pattern p = Pattern.compile("^<[a-zA-Z0-9_\\"=]+[\\s]*$>");
This time i used following Split:
p.split("<element attributname1=\"attributwert1\" attributname2=\"attributwert2\">");
I've got a compilation error:
U:\qms_neu\htdocs\inay\Source\myWork\Regex-Samples>javac Splitter.java
Splitter.java:14: illegal start of expression
Pattern p = Pattern.compile("^<[a-zA-Z0-9_\\"=]+[\\s]*$>");
^
Splitter.java:14: illegal character: \92
Pattern p = Pattern.compile("^<[a-zA-Z0-9_\\"=]+[\\s]*$>");
^
Splitter.java:14: illegal character: \92
Pattern p = Pattern.compile("^<[a-zA-Z0-9_\\"=]+[\\s]*$>");
^
Splitter.java:14: unclosed string literal
Pattern p = Pattern.compile("^<[a-zA-Z0-9_\\"=]+[\\s]*$>");
^
Splitter.java:17: ')' expected
p.split("<element attributname1=\"attributwert1\" attributname2
=\"attributwert2\">");
^
Splitter.java:14: unexpected type
required: variable
found : value
Pattern p = Pattern.compile("^<[a-zA-Z0-9_\\"=]+[\\s]*$>");
^
Splitter.java:18: cannot resolve symbol
symbol : variable result
location: class Splitter
for (int i=0; i<result.length; i++)
^
Splitter.java:19: cannot resolve symbol
symbol : variable result
location: class Splitter
if (result.equals("")){
^
Splitter.java:21: cannot resolve symbol
symbol : variable result
location: class Splitter
System.out.println(result[0]);
^
Splitter.java:24: cannot resolve symbol
symbol : variable result
location: class Splitter
System.out.println(result[i]);
^
Splitter.java:25: cannot resolve symbol
symbol : variable result
location: class Splitter
int res = result.length - 1;
^
11 errors

Java.util.regex vs jarkata-oro on regular expression

hi, all
Both java utility (java.util.regex) and jarkata-oro (org.apache.oro.text.perl) provide supports of Perl5 compatible regular expressions. What are your opinions of which one do you choose?
I've searched in this forum and others online. I don't see this topic. Maybe I miss something...
Anyway, your opinions/experences are more than welcome.
thx
Tim

One big difference:
java.util.regex is JAVA 1.4+ only.
jarkata-oro is JAVA 1.2+

Where are the standard java packages io,util etc found in JRE

hello,
can u guys tell me where the standard java packages io,util found in the JRE.
is it the rt.jar file??
as i need to set the path in my applicaion to it coz, my application after running generates a .java file and then compiles it and creates a .class file on the local machine every time it is run.

You want the <JRE Home>\bin directory as your path if you use the JRE to run the class. Eyeball the SDK bin directory, find the same directory in the JRE structure, and set your PATH to it.

Util.regex.Pattern documentation

The 1.5 documentation for util.regex.Pattern defines quantifiers that are greedy, reluctant, or possessive. The definitions of these quantifiers seem to be the same. For example, X?, X??, and X?+ are each defined as "X, once or not at all." Is this a mistake? If not, what's that difference among greedy, reluctant, and possessive?

It's not a mistake, it's just incomplete. A normal (greedy) quantifier matches as many times as it can, but will back off if necessary to achieve an overall match. A reluctant quantifier matches the minimum number of times that it has to, and only tries to match more if that's the only way to achieve an overall match. A greedy quantifier matches as many times as it can and never backs off, even if that makes an overall match impossible. Here's a demonstration:import java.util.regex.*;
public class Test
public static void main(String[] args)
    String input = "XXXXX";
    Pattern p1 = Pattern.compile("(X+)(X+)");
    Pattern p2 = Pattern.compile("(X+?)(X+)");
    Pattern p3 = Pattern.compile("(X++)(X+)");
    Matcher m = p1.matcher(input);
    if (m.matches())
       System.out.println("p1:\t" + m.group(1) + "\t" + m.group(2));
    m = p2.matcher(input);
    if (m.matches())
       System.out.println("p2:\t" + m.group(1) + "\t" + m.group(2));
    m = p3.matcher(input);
    if (m.matches())
       System.out.println("p3:\t" + m.group(1) + "\t" + m.group(2));
p1:     XXXX    X
p2:     X       XXXXIn p1, the X+ in the first group initially matches all five X's, then hands off to the second group. The X+ there has to match at least one X, but there are none left. So the first group gives up one of its X's, the second group matches it, and Bob's your uncle.
In p2, the X+? has to match at least one X, so it does, then hands off to the second group, which happily gobbles up the rest of the input.
In p3, the X++ matches all the X's, but refuses to back off and give the X+ in the second group the one X it needs, so the match fails.

Ava.util.regex.pattern and * - + /

hi...
i'm korean... so I can't speak english.. sorry..^^
but i hava a problem..
import java.util.regex.*;
public class Operator
/     public static void main(String args[])
          String operator="/";
////////////////////////////////////////////////////////////// error point..
          Pattern pattern=Pattern.compile(operator);
          Matcher m=pattern.matcher("- ----* / */* /+");
          int count=0;
          while(m.find()) {
     count++;
          System.out.println(count);
Exception in thread "main" java.util.regex.PatternSyntaxException: Dangling meta character '+' near index 0
+
operator : / - : ok...
operator : * + : error...
i had to use + *..
what's problem??

Are you using matches()? Then keep in mind that it requires that the entire String is matched by the RE.
pattern.matcher("about:foobar").matches(); //will return false, as "foobar" is not matched by your pattern
pattern.matcher("about:").matches(); //will return true
pattern.matcher("about:foobar").find(); //will return true
pattern.matcher("notabout:foobar").find(); // will return false

[bug]Jdev 11g:NullPointerException at java.util.regex.Matcher.getTextLength

Hi,
Jdev 11.1.1.0.31.51.56
If somebody of you get the following trace stack when running a jspx using ViewCriteriaRow.setOperator :
There is bug 7534359 and metalink note 747353.1 available.
java.lang.NullPointerException
at java.util.regex.Matcher.getTextLength(Matcher.java:1140)
at java.util.regex.Matcher.reset(Matcher.java:291)
at java.util.regex.Matcher.<init>(Matcher.java:211)
at java.util.regex.Pattern.matcher(Pattern.java:888)
at oracle.adfinternal.view.faces.model.binding.FacesCtrlSearchBinding._loadFilter
CriteriaValues(FacesCtrlSearchBinding.java:3695)
Truncated. see log file for complete stacktrace
Workaround:
If you use
 vcr.setAttribute("Job",job);
or
 vcr.setAttribute("Job","="+job);
than add following line of code:
 vcr.setOperator("Job","="); regards
Peter

Hi,
useful to mention that this happens when setting the equal operator or LIKE operator
vcr.setAttribute("Job","= '"+job+"'");
or
vcr.setOperator("Job","=");
Frank

PACKAGE BASICS.UTIL..REGEX

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