Matching Quotes With Regular Expressions

Hi, I have been attempting to develop an app which extract texts from pdfs then applies regular expression to the text. In one instance I attempt to match a curly open quote symbol which matches the . construct as well as the \W. However, it does not match \p{Punct} or \p{P} - does anyone know why this might be?
I have read that a pattern \p{Pi} exists for matching opening brackets but am told when I run Java that no such characters class exists - is there anywhere where I can get info on all of the character classes available (not the Pattern javadoc)?
Thanks very much,
Ross

. construct match anything and \W matches any non alphanumaric
to match a curly bracket use \{
Basicaly ( { [ has special meanings in regex so when you match them you have to use \{, \(, \[ in java strings it means \\{, \\(, \\[

Similar Messages

Match pattern: change regular expression search via front panel?

Hello,
I have an application that I am developing which is making use of Serial VISA.
I am scanning the output of a serial port which is constantly spitting out a long string of data.
Data is being pulled from the string with several combinations of SCAN FROM STRING functions and MATCH PATTERN Functions.
Question:
How can I use a button or TEXT box on the FRONT PANEL that can change the MATCH PATTERN Functions Regular expression?
for example the string may spit out the following:
Weight\s\s\s\s\s\s\s\s\s\s\s\s-0.00\slb\s\s\s\s\s\s-16\sbits\s+74.40\s\B0F\sCorrected\s\s\s\s\s\s-0.00\slb\s+0.999987\s%\s\r
in this case the serial device is spitting out data in LB.
It could be possible for the device to spit out data in KG ... thus I need to change the REGULAR EXPRESSION.
Thank you for your time

Hi,
just a quick example.
there are other way of doing this but I think the ComboBox is quite an easy one.
hope this helps
When my feet touch the ground each morning the devil thinks "bloody hell... He's up again!"
Attachments:
ComboBox.vi ‏12 KB

How to search with regular expression

I make pdx files so that I can search text quickly. But Acrobat doesn't provide a way to search with regular expression. I'm wondering if there is a way that I don't know to search for regular expression in Acrobat Pro 9?

First, Acrobat must "mount" the PDX.
As "Find" does not use the cataloged index, use Shift+Ctrl+F to open the advanced search dialog.
It may be helpful to first enter Acrobat Preferences and for the Search category tick "Always use advanced search options".
Back to the Search dialog - use the drop down menu for "Look In" to pick "Select Index" then, if no PDXs show, click the Add button.
In the Open Index File dialog, browse to the location of the desired PDX and select it.
OK out and use "Return results containing" to pick a "Match ..." requirement or Boolean.
To become familiar with query syntax, for Acrobat, it is good to review Acrobat Help.
http://help.adobe.com/en_US/Acrobat/9.0/Professional/WS58a04a822e3e50102bd615109794195ff-7 c4b.w.html
Be well...

Problem with Regular Expression

Hi There!!
I have a problem with regular expression. I want to validate that one word and second word are same. For that I have written a regex
Pattern p=Pattern.compile("([a-z][a-zA-Z]*)\\s\1");
Matcher m=p.matcher("nikhil nikhil");
boolean t=m.matches();
if (t)
          System.out.println("There is a match");
     else
          System.out.println("There is no match");
The result I am getting is always "There is no match
Your timely help will be much appreciated.
Regards

Ram wrote:
ErasP wrote:
You are missing a backward slash in the regex
Pattern p = Pattern.compile("([a-z][a-zA-Z]*)\\s\\1");
But this will fail in this case.
Matcher m = p.matcher("Nikhil Nikhil");It is the reason for that *[a-z]*.The OP had [a-z][a-zA-Z]* in his code, so presumably he know what that means and wants that String not to match.

Grouping & Back-references with regular expressions on Replace Text window

I really appreciate the inclusion of the Regular Expressions in the search & replace feature. One thing I am missing is back-references in the replacement expression. For instance, in the unix tools vi or sed, I might do something like this:
s/$firstPart$ $secondPart$ $oldThirdPart$/\2 \1 newThirdPart/g
which would allow me to switch the places of firstPart and secondPart, and totally replace thirdPart. If grouping and back-references are already present in the Replace Text window, how does one correctly invoke them?

duplicate of Grouping & Back-references with regular expressions on Replace Text window

Assistance with Regular Expression and Tcl

Assistance with Regular Expression and Tcl
Hello Everyone,
I recently began learning Tcl to develop scripts for automating network switch deployments.
In my script, I want to name the device with a location and the last three octets of the base mac address.
I can get the Base MAC address by :
show version | include Base
Base ethernet MAC Address       : 00:00:00:DB:CE:00
And I can get the last three octets of the MAC address using the following regular expression.
([0-9a-f]{2}[:-]){2}([0-9a-f]{2}$)
But I have not been able to figure out how to call the regular expression in the tcl script.
I have checked several resources but have not been able to figure it out. Suggestions?
Ultimately, I want to set the last three octets to a variable (something like below) and then call the variable when I name the switch.
set mac [exec "sh version | i Base"] (include the regular expression)
ios_config "hostname location$mac"
Thanks for any assistance in advance.
Chris

This worked for me.
Switch_1(tcl)#set result [exec show ver | inc Base]
Base ethernet MAC Address       : 00:1B:D4:F8:B1:80
Switch_1(tcl)#regexp {([0-9A-F:]{8}\r)} $result -> mac
1
Switch_1(tcl)#puts $mac
F8:B1:80
Switch_1(tcl)#ios_config "hostname location$mac"
%Warning! Hostname should contain at least one alphabet or '-' or '_' character
locationF8:B1:80(tcl)#

Get the string between li tags, with regular expression

I have a unordered list, and I want to store all the strings between the li tags (<li>.?</li>)in an array:
<ul>
<li>This is String One</li>
<li>This is String Two</li>
<li>This is String Three</li>
</ul>
This is what have so far:
<li>(.*?)</li>
but it is not correct, I only want the string without the li tags.
Thanks.

No one?
Anoyone here experienced with Regular Expression?

Match beginning of line with Regular Expression

I'm confused about dreamweaver's treatment of the characters
^ and $ (beginning of line, end of line) in regex searches. It
seems that these characters match the beginning of the file, not
the beginning of the various lines in the file. I would expect it
to work the other way around. A search like:
(^.)
should match every line in the file, so that a find/replace
could be performed at the beginning of each line, like this:
HELLO$1
which would add 'HELLO' at the start of each line in the
file.
Instead, this action only matches the first character of the
file, sticks 'HELLO' in front of it, and then quits (or moves on to
the next file). The endline character $ behaves in a similar
fashion, matching only the end of the file, not the end of each
line.
I've searched, and all the literature about regular
expressions in dreamweaver seems to indicate that I'm expecting the
correct behavior:
www.adobe.com/devnet/dreamweaver/articles/regular_expressions_03.html
quote:
^ Beginning of input or line ^T matches "T" in "This good
earth" but not in "Uncle Tom's Cabin"
$ End of input or line h$ matches "h" in "teach" but not in
"teacher"
Thanks for any insight, folks.

Hi Winston,
I am still digesting the material from the regular expression book and will take sometime to become proficient with it.
It seems that using groupCount() to eliminate the unwanted text does not work in this case, since all the lines returned the same value. Ie 3 posted earlier. This may be because the patterns are complex and only a few were grouped together. Otherwise, could you provide an example using the string posted as opposed to a hyperthetic one. In the meantime, at least one solution have been found by defining an additional special pattern “\\A[^%].*\\Z”, before combining / intersecting both existing and the new special pattern to get the best of both world. Another approach that should also work is to evaluate the size of String.split() and only accept those lines with a minimum number of tokens.
Anyhow, I have come a crossed another minor stumbling block in the mean time with the following line, where some hidden characters is preventing the existing pattern from reading it:
o;?Mervan Bay 40 Boyde St 7 br t $250,000 X West Park AE
Below is the existing regular expression that works for other lines with the same pattern but not for special hidden characters such as “o;?”:
\\A([A-Z][a-z]*){1,2} [0-9]{0,4}/?[0-9]{0,4}-?[0-9]{0,4} ([A-Z][a-z]*){1,2} St|Rd|Av|Sq|Cl|Pl|Cr|Gr|Dr|Hwy|Pde|Wy|La [0-9] br [h|u|t] \\$\\d+,\\d+|\\$\\d*\\,\\d+,\\d+ ([A-Z][a-z]*){1,}\\ZIs it possible to come up with a regular expression to ignore them so that this line could be picked up? Would also like to know whether I could combine both the special pattern “\\A[^%].*\\Z” with existing one as opposed to using 2 separate patterns altogether?
Many thanks,
Jack

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.

Problems with regular expressions.

Hi all.
I want to substitute within an xml file a string with the following structure:
Referencer="//@Model/@OwnedElement.0/@OwnedElement.1/@OwnedElement.0/@OwnedElement.1/@OwnedElement.33 ..."
where the dots mean that the string within the quotes may be repeated (separated by spaces) many times.
I've tried to match such expression with java regex, with no luck. I thought running
line.contains("Referencer=\"[^\"\\r\\n]*\"");
would return true, but it didn't.
Could anybody point me the right way?.
Thank you all in advance.

Method String.contains() does not take a regular expression - you need to use method String.matches() or class Pattern.
This looks to be a good problem to solve using http://elliotth.blogspot.com/2004/07/java-implementation-of-rubys-gsub.html .

Help with Regular Expression for field validation

I'm fairly new to using regular expressions and using Acrobat. This is probably a simple question, but I've been unable to figure it out.
I have a text field on a PDF that I would like to be 9 characters in length. The first 2 characters can only be alphanumeric, the last 7 characters can only be numeric.
At first I was using the following, which allows all the characters to be alphanumeric:
var re = /^[A-Za-z0-9 :\\_]$/;
if (event.change.length >0) {
if (event.willCommit == false) {
    if (!re.test(event.change)) {
        event.rc = false
That works fine, but it's not quite what I needed. With some assistance I changed it (see below) to fit what I was looking for. However, this didn't work; it prevents anything from being entered in the field:
var re = /^[A-Za-z0-9]{2}\d{7}$/;
if (event.change.length >0) {
if (event.willCommit == false) {
    if (!re.test(event.change)) {
        event.rc = false
Any help would be greatly appreciated.
Thanks...

Here's a function you can call form the field's custom Format script. It should be placed in a document-level JavaScript:
function custom_ks1() {
    // Define non-commited regular expression
    var re = /^[A-Za-z0-9]{0,2}([0-9]{0,7})?$/;
    // Get all of the characters the user has entered
    var value = AFMergeChange(event);
    // Allow field to be cleared
    if(!value) return;
    if (event.willCommit) {
        // Define commited regular expression
        var re = /^[A-Za-z0-9]{2}[0-9]{7}$/;
        if (!re.test(value)) { // If final value doesn't match, alert user
            app.alert("Your error message goes here.");
            // event.rc = false
    } else { // not commited
        // Only allow characters that match the regular expression
        event.rc = re.test(value);
Call it like this:
// Custom Keystroke script
custom1_ks();

Help with regular expression needed

Hi,
Perhaps someone here can help me with my regular expression I'm trying to build in my Java code.
The regular expression that I'm looking to build consists of any non-whitespace character up until it finds one or two <>= symbols and then any character thereafter. So both these Strings would match the expression:
City 1==London
Age>=18
The regular expression that I'm using is as follows:
(\\S+)([><=]){1,2}(.+)However, group 1 always retrieves the first <>= symbol as in "City 1=". How can I make the <>= part greedy so that it retrieves both operator symbols?
Thanks.

Make the first group, the non-spaces, reluctant:
"(\\S+?)([<>=]{1,2})(.+)"

Searching in reverse with regular expressions

I have recently constructed a "Find" dialog for my editor and would like to support the use of regular expressions (for people who know more about them than I do.) I have radio buttons that allow the user to search down (from the current caret position to the end of the document) or up (from the current caret position to the beginning of the document.)
I have it working fine with literal text via String.indexOf() and String.lastIndexOf(), but I'm not sure how to implement an "upwards" search using regular expressions. It seems that java.util.regex only provides for searching from the current caret position downwards. Any suggestions?

two not-very-good-for-big-document suggestions
1) search up by searching down (from the beginning) to the last match before the position searched from
2) when the user asks for a search find all matches, number them and keep the index in the doc where they were found, then look up the largest index less than the carets index
you might try constructing succesively larger strings that end at the current caret position, and matching them one by one. If you're near the bottom, and theres no matches, this is also likely to be sloow..
asjf

Looping through files with Regular expressions ?

Hi,
My Question is:
if i have the following Regular Expression,
String regrex = "tree\\s\\w{1,4}.+\\s=\\s(.*;)";
The file in which i am looking for the string has multiple entries, is it
possible to do another regular expression on the captured group (.*;)
which is in the original Regular expression ?
The text that is captured by the RE is of the type "(1,(2,((3,5),4)));" for
each entry, and different entries in the file have slightly different syntax
is it possible to loop through the file and first of all check for the presence
of the original RE in each entry of the file
and then secondly, check for the presence of another RE on the captured group?
[ e.g. to check for something like, if the captured group has a 1 followed by a 3
followed by a 5 followed by a and so on ].
Thanks Very much for any help, i've been struggling with this for a while!!
Much appreciated
The code that i have so far is as follows:
import java.util.*;
import java.util.regex.*;
import java.io.*;
import java.lang.*;
import javax.swing.*;
public class ExpressReg {
public String Edit;
public ExpressReg(String editorEx){
Edit = editorEx; // Edit = JTextArea
String regrex = "tree\\s\\w{1,4}.+\\s=\\s(.*;)";
//String regrex1 = "(.*;)";
Pattern p = Pattern.compile(regrex);
Matcher m = p.matcher(editorEx); // matcher can have more than one argument!
boolean result = m.find();
if(result){
JOptionPane.showMessageDialog(null, "String Present in Editor");
else if(!result){
JOptionPane.showMessageDialog(null, "String Not Present In Editor");

if i have the following Regular Expression,
String regrex = "tree\\s\\w{1,4}.+\\s=\\s(.*;)";
The file in which i am looking for the string has multiple entries, is it
possible to do another regular expression on the captured group (.*;)
which is in the original Regular expression ?Yes, the capturing group is $1 (the only one) referenced in source code as m.group(1).
m.group() will return entire matching.
simply use this way:
String result = m.group(1);
// your stuff: could be another validation
The text that is captured by the RE is of the type "(1,(2,((3,5),4)));" for
each entry, and different entries in the file have slightly different syntax
is it possible to loop through the file and first of all check for the presence
of the original RE in each entry of the file
and then secondly, check for the presence of another RE on the captured group?Again "Yes", no limits!
Don't need to create another Matcher, just use m.reset(anotherSourceString)..loop using the same pattern.
Note: Take care with ".*" because regex nature is "greedy", be more specific, eg.: "\\d" just matches digits (0-9).
Can you give us some sample of "slight difference" ?

How to find sunstring with regular expression?

How can I find a substring in a string with a regular expression?
Example:
I have a original string "<tr><th>RecordId: </th><td valign=middle>A4711</td></tr>"
Now i want to extract the value "A4711" from this string with a regular expression. Everything except "A4711" is fixed, the id "A4711" itself is dynamic. How is it possible to get the substring "A4711" of the original string with a regular expression?

i wrote a little method with the infos above to get such results:
     * Get all substrings of a string that matches a regular expression.
     * @param original String to inspect.
     * @param regExp Regular expression as search criteria.
     * @return All matches of <i>regExp</i> or null if one input parameter is null.
    public static String[] getSubstrings(String original, String regExp) {
        String[] result = null;
        if (original != null && regExp != null) {
            Pattern pattern = Pattern.compile(regExp);
            Matcher matcher = pattern.matcher(original);
            boolean matchFound = matcher.find();
            Vector matches = new Vector();
            while (matchFound) {
                String match = matcher.group();
                matches.addElement(match);
                matchFound = matcher.find();
            }//next match
            int count = matches.size();
            result = new String[count];
            for (int i = 0; i < count; i++) {
                result[i] = (String) matches.elementAt(i);
            }//next match
        }//else: input unavailable
        return result;
    }//getSubstrings()

Matching Quotes With Regular Expressions

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