Regular expression breaks with \00 in input string

I wrote the following code to illustrate the problem.
Clearly the match should occur in both cases, but it does not occur if \00 is in the input string (within the < > ).
I am not even searching for \00 or \01. This makes the regular expression match rather useless if you have text that contains data in hex form. I am using Labview 8.2. Is this a known bug? Is there a work around?
Tammo
Message Edited by Tammo on 03-06-2008 12:26 PM
Message Edited by Tammo on 03-06-2008 12:26 PM
Attachments:
RegEx1.vi ‏56 KB
BlockDiagram1.jpg ‏18 KB
FrontPanel1.jpg ‏18 KB

There was a brief discussion on this not too long ago.

Similar Messages

Regular Expression to capture user's input string

I am writing a helper class to split user input string into String array according to the following pattern:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class TestDelimiter {
static Pattern p = Pattern.compile("(.*?)[,;]{1}");
static Matcher m;
    public static void main(String[] args) {
        String input = "AAA111111,BBB222222;CCC333333";
        m = p.matcher(input);
        while(m.find()){
            String output = m.group(1);
            System.out.println(output);
}Output:
AAA111111
BBB222222My question is, how can I modify the regular expression string so that the CCC333333 (last element) can also be included?

roamer wrote:
Ok, let's don't argue on this point.Who's arguing?
I think I got the answer. For simplicity, I can just manually add a ";" or "," string after each user input. Just like:You never said anything about that before. You asked how to split "AAA111,BBB222;CCC333" into its components and you were given a correct answer.
Maybe you should rephrase your question including this added requirement. Do you need to have the separator included in the final outputs or why do you want to suddenly add things to user input?

Regular expression help please. (extracting a string subset between two markers)

I haven't used regular expressions before, and I'm having trouble finding a regular expression to extract a string subset between two markers.
The string;
Header stuff I don't want
Header stuff I don't want
Header stuff I don't want
Header stuff I don't want
Header stuff I don't want
Header stuff I don't want
ERRORS 6
Info I want line 1
Info I want line 2
Info I want line 3
Info I want line 4
Info I want line 5
Info I want line 6
END_ERRORS
From the string above (this is read from a text file) I'm trying to extract the string subset between ERRORS 6 and END_ERRORS. The number of errors (6 in this case) can be any number 1 through 32, and the number of lines I want to extract will correspond with this number. I can supply this number from a calling VI if necessary.
My current solution, which works but is not very elegant;
(1) uses Match Regular Expression to the return the string after matching ERRORS 6
(2) uses Match Regular Expression returning all characters before match END_ERRORS of the string returned by (1)
Is there a way this can be accomplished using 1 Match Regular Expression? If so could anyone suggest how, together with an explanation of how the regular expression given works.
Many thanks
Alan
Solved!
Go to Solution.

I used a character class to catch any word or whitespace characters. Putting this inside parentheses makes a submatch that you can get by expanding the Match Regular Expression node. The \d finds digits and the two *s repeat the previous term. So, \d* will find the '6', as well as '123456'.
Jim
You're entirely bonkers. But I'll tell you a secret. All the best people are. ~ Alice

Regular expression substitution with escaped $

I'm looking to perform a substitution (along the lines of String.replaceAll()) that matches with the character $. I have tried $, \$, \\$, and variations thereof in the regex string utilizing String.replaceAll() and with a Pattern.compile(). In both cases I get an "Illegal repetition near index X" error. Anyone have suggestions as to escaping the $ character for regular expression substitutions?
Edited by: dolcraith57 on Dec 6, 2007 11:18 AM

It would help if you'd post details--e.g. original string and desired result, with an explanation of the rules.
str1 = "a$1$2$b$";
str2 = str1.replaceAll("\\$", "X"); // aX1X2XbX
str3 = str1.replaceAll("\\d", "\\$"); // a$$$$$b$

Help with Oracle Connection- "Input string was not in a correct format"

Hello, can some one, anyone please help me. I have a simple VS 2005 C# application that connects to oracle. I've set it up to take the username, password and tnsname as arguments. I am using Client 9i, version 9.2.0.401 of Oracle.DataAccess.dll, and .NET Framework 2.0.
It works fine when I run it from VS 2005, I have also set up a test machine w/o VS 2005 and ran my install package and it runs fine. I sent it to one of my co-workers and when he tries it the OracleConnection obect fails with the error.
"Input string was not in a correct format"
strange in that it is not an ORA error, just the identifed text!?
Here is the code:
## begin code
OracleConnection dbc = new OracleConnection();
string sConnectString = "User Id=" + username.ToString() + ";Password=" + password.ToString() + ";Data Source=" + tnsname.ToString();
dbc.ConnectionString = sConnectString.ToString();
MessageBox.Show("Attempting to Connect to Oracle");
dbc.Open();
MessageBox.Show("Connected to Oracle: " + dbc.ServerVersion);
## end code
Pretty basic, what could be going on?
Since this only happens with an installation, I put in the message boxes to verify exactly where it chokes. I get the first message box, then an error with the identified text. I've seen a number of posts regarding input string format problems, but not a one dealing with OracleConnection.Open(). I added all the ToString() calls just to make sure everything was a string but it did not change the end result.
Anyone? Thanks In advance!
Eric S.

Hello,
well, i got a message "...string not wellformed format...", too.
If you have defined the parameters as string, you don't need the additonal "ToString()". Please ckeck your tnsnames -string. Try to debug this. I believe you have copy the format from the tnsnames.ora and there you have much brackets.
For example, two code snippets:
As datasource i use <Server>:<Port>/<Instance>
try
string FDsn ="User Id="+FDbUser+";Password="+FDbPwd;
FDsn +=";Data Source=wth5:1521/Ora9.wth5";
FConn = new OracleConnection(FDsn);
FConn.Open();
if (FConn.State == ConnectionState.Open )
FConn.Close();
catch (Exception ex)
FStateMsg = "Connection to database failed. Check Configuration in parameter: ConnectionString. " + ex.Message;
Above i connect to a Ora 9.2.0.1
by using
FDsn +=";Data Source=wth5:1522/XE";
i connect to an OraExpress database on the same machine.
My Oracle.dataAccess.dll ist the newest, 10.2.... and it runs under Framework 1.1.x and Framework 2.x
I'll hope it will help you. Best regards!

Regular expression to check the sequence of strings

HI,
I am validating an EDI document like as follows
ISA*XX*XXXXXXXXXXXXXXX*XX*XXXXXXXXXXXXXXX*030130*0912*~IEA*1*000005900~
I need to create a regular expression which checks ISA should always occur before IEA.
Please help me if you have any hints.
Thanks.

Thanks.I had taken that into consideration.
But using regular expression I could say
ISA* only once
IEA* only once
And
ISA before IEA
Any hints how to specify "before"/sequence using Regular expression ?
I agree sometime using basic String operation we can do this.

Regular expression evaluation with logical operator

Hi All,
I am bit confuse with expression evaluation with logical operator. I am trying to understand below expression.
eXa.getTrue() && eXa.getFalse() || eXa.getFalse() && eXa.getTrue() || eXa.getFalse() comes as false and True Count: 1 False Count: 3
As per understanding it should be true with True Count: 1 False Count: 3
it should execute 1st getTrue() and then 1stGetFalse() and then 2nd getfalse() and should skip 2nd getTrue() and execute 3rd fetFalse()
eXa.getTrue() && eXa.getTrue() || eXa.getFalse() && eXa.getTrue() || eXa.getFalse() comes as true and True Count: 2 False Count: 0
As per understanding it should be true with True Count: 3 False Count: 0
it should execute 1st 2 getTrue() and skip 1st getFalse() and then execute 3rd getTrue() and skip last getFalse().
eXa.getTrue() || eXa.getFalse() && eXa.getFalse() || eXa.getTrue() && eXa.getFalse() comes as true and True Count: 1 False Count: 0
As per understanding it should be true with True Count: 2 False Count: 2
it should execute 1st getTrue() and skip 1st getFalse() and then execute 2nd getFalse() and then execute 2nd getTrue() and then 3rd getFalse()
Please help me to understand above expressions.
Here is the methods definition:
private boolean getTrue() {
          trueCount++;
          boolean retrunValue = 4 > 3;
          return retrunValue;
private boolean getFalse() {
          falseCount++;
          boolean retrunValue = 3 > 4;
          return retrunValue;
Thanks for ur help

>
adding parenthesis to make order of ops more obvious. adding "?" to show un-executed calls.
(eXa.getTrue() && eXa.getFalse()) || (eXa.getFalse() && eXa.getTrue()) || eXa.getFalse() comes as false and True Count: 1 False Count: 3
(T && F) = F
(F && ?) = F
(F) = F
F || F || F = F
(eXa.getTrue() && eXa.getTrue()) || (eXa.getFalse() && eXa.getTrue()) || eXa.getFalse() comes as true and True Count: 2 False Count: 0
(T && T) = T
(? && ?) = ?
(?) = ?
T || ? || ? = T
eXa.getTrue() || (eXa.getFalse() && eXa.getFalse()) || (eXa.getTrue() && eXa.getFalse()) comes as true and True Count: 1 False Count: 0
(T) = T
(? && ?) = ?
(? && ?) = ?
T || ? || ? = T

Regular Expression - Select two words after specific string

Hi,
I am trying to select the two words/strings after the first word "door". I am using the search pattern (?<=door).\w+ but in this case I get the complete text after the word "door". I only want to select the two words after the first "door" in the complete text.
Can anybody help me?
Thanks!
Marco Snels

Hi Marco,
I'm relatively handy with RegEx but this seems like a problem where I would employ a little bit of RegEx and CTL, just to make life easier.
You can use the following RegEx (note: I didn't test this in Integrator, only in a RegEx testing tool) to extract the two words after door (but including door, unfortunately):
(?:door)[\s]\w+[\s]\w+
This would give you something like the following in your extracted field:
door is brown
You could then pass through a re-formatter to remove "door" and the whitespace and be on your way. Not the best answer but should perform reasonably well and get you up and going.
Regards,
Patrick Rafferty
http://branchbird.com

Regular Expressions problem with OPA

hi,
I have version OPA 10.2, when I try using regualr expressions it won't let you submit data and move onto the next screen.
I've tried for example using the OPA hel's National Insurance number, I tried for a 6 digit number using:
For a minimum of 6 digits:
^\D*(?:\d\D*){6,}$
and I also tried:
^[0-9]{1,6}$
None of them worked; is there something obvious I'm missing?
Thanks

Hello,
Just tried this out and the two expressions worked as I would expect. I receive the error message as configured in the properties file and cannot progress to the next screen until I enter 6 digits or more. Is that different to the behaviour you are seeing? Also, have you tried using the debugger without screens? The error message would appear at the top of the debug window "Cannot set value of attribute 'test'. Reason: Invalid Value."
Also, are you trying to validate a text attribute or a number attribute? Number attributes can be a little trickier since the regex appears to run against them after they're converted. So if the user enters "99", the regex validates against "99.0".

Dumbfounded by Scanner processing String using regular expression

I was reading Bruce Eckel's book when I came across something interesting: extending Scanner with regular expressions. Unfortunately, I was confronted with an issue that doesn't make much sense to me: if the String that I am scanning contains a hyphen, the Scanner doesn't produce anything. As soon as I take it out, it all works like a charm. Here is my example:
import java.util.Scanner;
import java.util.regex.*;
public class StringScan {
public static void main (String [] args){
     String input = "there's one caveat when scanning with regular expressions";
     Scanner scanner = new Scanner (input);
     String pattern = "[a-z]\\w+";
     while (scanner.hasNext(pattern)){
          scanner.next(pattern);
          MatchResult match = scanner.match();
          String output = match.group();
          System.out.println(output);
}What could be the reason? I imagined it could be because the hyphen for some reason gets given a special meaning but when I tried escaping it, it still didn't work.

Thanks for your prompt reply.
I have figured out what was wrong with my code, by the way. Since a single quote is not a word character, it does not match w+. And as the very first input token does not match, the scanner stops immediately. I rewrote my regex to "[a-z].*" and now it does work.

Help with Understanding Regular Expressions

Hello Folks,
I need some help in understanding the Regular Expressions.
-- This returns the Expected string from the Source String. ", Redwood Shores,"
SELECT
REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                ',[^,]+,', 1, 1) "REGEXPR_SUBSTR"
FROM DUAL;
REGEXPR_SUBSTR
, Redwood Shores,
However, when the query is changed to find the Second Occurrence of the Pattern, it does not match any. IMV, it should return ", CA,"
SELECT
REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                ',[^,]+,', 1, *2*) "REGEXPR_SUBSTR"
FROM DUAL;
REGEXPR_SUBSTR
NULLCan somebody help me in understanding Why Second Query not returning ", CA,"?
I did search this forum and found link to thread "https://forums.oracle.com/forums/thread.jspa?threadID=2400143" for basic tutorials.
Regards,
P.

PurveshK wrote:
Can somebody help me in understanding Why Second Query not returning ", CA,"?With your query...
SELECT
REGEXP_SUBSTR('500 Oracle Parkway, Redwood Shores, CA,aa',
                ',[^,]+,', 1, *2*) "REGEXPR_SUBSTR"
FROM DUAL;You are looking for patterns of "comma followed by 1 or more non-comma chrs followed by a comma."
So, let's manually pattern match that...
'500 Oracle Parkway, Redwood Shores, CA,aa'
                   ^               ^
                   |               |
                           |
                           |
                  Here to here is the
                  first occurence.So the second occurance will start searching for the same pattern AFTER the first occurence.
'500 Oracle Parkway, Redwood Shores, CA,aa'
                                    ^
                                    |
i.e. the search for the second occurence starts heretherefore the first "," from that point is...
'500 Oracle Parkway, Redwood Shores, CA,aa'
                                       ^
                                       |
                                      hereand there is nothing there matching the pattern you are looking for because it only has the
"comma follwed by 1 or more non-comma chrs"... but it doesn't have the "followed by a comma"
...so there is no second occurence,

Extract values with Regular Expression

How to extract values into [ ] using regular expression.
With data As
Select 'AAAAAA[10] AAA: 19C' Txt From Dual Union all
Select 'XX[450]-10A' Txt From Dual Union all
Select '[5]AVC19C' Txt From Dual Union all
Select 'FVD[120]D2AC' Txt From Dual
)I hope this return
10
450
5
120Thanks in advanced

user11118871 wrote:
Thanks for all.
Hi BluShadow, can you explain what this is doing!?
ThanksSure.
Search for ----------------|           /------- Replace the found pattern in the search string with backreference 1
                     /------------\   /\       (the first thing backreferenced in the search string)
regexp_replace(txt, '^.*\[(.*)\].*$','\1')
                     |\/\/\--/\/\/|
                     | \ \ / \ \|
                     | | | | | |\- End of string
                     | | | | | |
                     | | | | | \- Any number of characters
                     | | | | |
                     | | | | \- Right square bracket (escaped)
                     | | | |
                     | | | \- any characters (backreferenced by round brackets)
                     | | |
                     | | \- Left square bracket (escaped)
                     | |
                     | \- Any number of characters
                     |
                     \- Start of stringEdited by: BluShadow on Jun 29, 2009 2:19 PM

Regular expression for 2nd occurance of a substring in a string

Hi,
1)
i want to find the second occurrence of a substring in a string with regular expression so that i can modify that only.
Ex: i have a string like ---> axe,afn,sdk,jdi,afn,mki,mki
in this i want the second occurance of afn and change that one only...
which regular expression i have to use...
Note that ...i have to use regular expression only....no string manipulation methods...(strictly)
2)
How can i apply the multiple regular expressions multiple times on a single string ..i.e in the above instance i have to apply the same 2nd occurrence logic for
substring mki also. for this i have to use a single regular expression string that contains validations for both the sub strings mki and afn.
Thanks in advance,
Venkat

javafreak666 wrote:
Hi,
1)
i want to find the second occurrence of a substring in a string with regular expression so that i can modify that only.
Ex: i have a string like ---> axe,afn,sdk,jdi,afn,mki,mki
in this i want the second occurance of afn and change that one only...
which regular expression i have to use...
Note that ...i have to use regular expression only....no string manipulation methods...(strictly)
2)
How can i apply the multiple regular expressions multiple times on a single string ..i.e in the above instance i have to apply the same 2nd occurrence logic for
substring mki also. for this i have to use a single regular expression string that contains validations for both the sub strings mki and afn.
Thanks in advance,
VenkatWhat do you mean by using a regex to get the index of a second substring? There is not method in Java which uses regex to et the index of a substring.
There are various indexOf(...) methods for this:
String text = "axe,afn,sdk,jdi,afn,mki,mki";
String target = "afn";
int second = text.indexOf(target, text.indexOf(target)+1);
System.out.println("second="+second);Of course you can find the index of a group like this:
Matcher m = Pattern.compile(target+".*?("+target+")").matcher(text);
System.out.println(m.find() ? "index="+m.start(1) : "nothing found");but there is not single method that handles this: you'll have to call the find() and then the start(...) method on the Matcher instance, so the indexOf(...) approach is the favourable one, IMO.

A regular expression to detect a blank string...

Anyone know how to write a regular expression that will detect a blank string?
I.e., in a webservice xsd I'm adding a restriction to stop the user specifying a blank string for an element in the webservice operation call.
But I can't figure out a regular expression that will detect an entirely blank string but that will on the other hand allow the string to contain blank spaces in the text.
So the restriction should not allow "" or " " to be specified but will allow "Joe Bloggs" to be specified.
I tried [^ ]* but this wont allow "Joe Bloggs" to pass.
Any ideas?
Thanks,
Ruairi.

Hi ruairiw,
there is a shortcut for the set of whitespace chars in Java. It is the Expression *\s* which is equal to *[ \t\n\f\r\u000b]*.
With this expression you can test whether a String consists only of whitespace chars.
Expamle:
String regex = "\\s*"; // the slash needs to be escaped
// same as String regex = "[ \t\n\f\r\u000b]";
System.out.println("".matches(regex)); // true
System.out.println(" ".matches(regex)); // true
System.out.println(" \r\n\t ".matches(regex)); // true
System.out.println("\n\nTom".matches(regex)); // false
System.out.println(" Tom Smith".matches(regex)); // falseBesh Wishes
esprimo

Regular Expression string

Hi,
I'm looking for a single regular expression.
E.g. My string is 'get data for comparison'. So when using regular expression as \Wcompar then it returns true.
But I dont want it to return true for string 'get data for comparable' with the same regular epression.
Here, I do not want string comparable and rest all the forms of compar to be selected.

It's not really clear what you need, perhaps something like this?
-- data:
with t as
select 'get data for comparison' txt from dual union all
select 'get date for compar xxxxx' txt from dual union all
select 'compar xx' txt from dual union all
select 'compare xx' txt from dual union all
select 'xx compar' txt from dual union all
select 'xx compare' txt from dual
-- query:
select txt from t
where regexp_like(txt,'(\W|^)compar(\W|$)');
TXT
get date for compar xxxxx
compar xx
xx compar(\W|^) means word-limit or start of the string
(\W|$) means word-limit or end of the string
Edited by: hm on 21.12.2011 05:29

Regular expression breaks with \00 in input string

Similar Messages

Maybe you are looking for