Regular Expression Required for Checksum variable

I am wanting to create a regular expression that extracts a variable checksum value (cs=) which is unique to a given server response string.
The issue I am having is that a simple regex being name="cs" value="(.+?)" just does not work because there are 18 different checksum values being returned per session id/server response.
So a simple regex just picks up any "cs" value and I guess gets confused since there are 18 different ones contained within the server response (all user related)
What i need is a regex to use say this value from the server response (JONHONEYMAN%40YAHOO.COM) or any other unique value which is contained within the string response from the server then extract the cs (Checksum) value from that string and make it variable :)
Here is the server response string:-
"fp=220:1200:2977638312763704:: NO::P1200_EMAIL_ADDRESS,P1200_ORGANISATION_NAME,P1200_UCRN,P1200_ORG_ID,P1200_CALLING_PAGE:JONHONEYMAN%40YAHOO.COM%2C john%20Honeyman%2C260%2C220%2C1190&cs=336788D01EC6E80B1877B3EE982E8B2D8" >Select</td></tr
can anyone help?
Thanks

You need an XML Parser.

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Regular Expression Search for Case Statement in VBA

Hi,
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resuse the same code for many different servers.
I have the mainServer variable set as a Variant and in my current example it is set as "INTPXY001" (internal proxy server 001). I have tried different regex statements for the potential to have INTPXY001 - INTPXY999, EXTPXY001 - EXTPXY999, and
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'========================================
Set mainServer As Variant
Set AppVisio = CreateObject("visio.application")
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Select Case mainServer
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You cannot declare variables As Variant in VBScript. All variables in VBScript are implicitly variants.
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I have the following requirement:
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28 Oct 2008 12:27:48 | INFO | SearchController | Exception while fetching search results in controller:Unclosed character class near index
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WebTest - VS2013 - Extract Regular Expression - Required=False

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When I run my WebTest, it tells me that it passes the extraction rule but it never created the Context Parameter so the condition is skipped.
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How to add regular expression in viewobject bind variables

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also
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Hari

Perfect... Thanks Arun.
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Hari

How about the statement of regular expression like for this

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How to use regular expression replace for this special characters?

hi,
I need to replace the below string, but i couldnt able to do if we use the special charaters '+', '$' . can anyone suggest a way to do this?
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you should use escape character \.
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Hi,
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DATA:
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With rgds,
Anil Kumar Sharma .P

Regular Expression Fails for Large Inputs

I've got a problem on mu RegEx
This is happenning for only when I input large files to it.
My code is shown brlow...
public static boolean isComment(String line) {
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public static boolean isNewLine(String line) {
Pattern pattern = Pattern.compile("^[\\s\n]+$", Pattern.CASE_INSENSITIVE);
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^else^
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^}^
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^}^
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^Matcher matcher = pattern.matcher(line);^
^String s = line;^
^while (matcher.find()) {^
^s = matcher.replaceAll(replaceStr(matcher.group()));^
^}^
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^Matcher matcher = pattern.matcher(line);^
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^Matcher matcher = pattern.matcher(line);^
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^}^
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s += temp;+
+if (isHaveEnding(matcher.group(1))) {+
+notEnded = true;+
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+String str = readFile("test.txt");+
+System.out.println("Read Compleate");+
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{code}
This code is currently parse SQL statements and add '/' characters finding the missed places.
When i give this 'test.txt' as a long file (with 10000 LOC) gives belov error.
*Exception in thread "main" java.lang.StackOverflowError*
* at java.lang.Character.codePointAt(Character.java:2335)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2873)*
* at java.util.regex.Pattern$Branch.match(Pattern.java:4530)*
* at java.util.regex.Pattern$GroupHead.match(Pattern.java:4570)*
* at java.util.regex.Pattern$Loop.match(Pattern.java:4697)*
* at java.util.regex.Pattern$GroupTail.match(Pattern.java:4629)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2876)*
* at java.util.regex.Pattern$Branch.match(Pattern.java:4530)*
* at java.util.regex.Pattern$GroupHead.match(Pattern.java:4570)*
* at java.util.regex.Pattern$Loop.match(Pattern.java:4697)*
* at java.util.regex.Pattern$GroupTail.match(Pattern.java:4629)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2876)*
* at java.util.regex.Pattern$Branch.match(Pattern.java:4530)*
* at java.util.regex.Pattern$GroupHead.match(Pattern.java:4570)*
* at java.util.regex.Pattern$Loop.match(Pattern.java:4697)*
* at java.util.regex.Pattern$GroupTail.match(Pattern.java:4629)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2876)*
* at java.util.regex.Pattern$Branch.match(Pattern.java:4530)*
* at java.util.regex.Pattern$GroupHead.match(Pattern.java:4570)*
* at java.util.regex.Pattern$Loop.match(Pattern.java:4697)*
* at java.util.regex.Pattern$GroupTail.match(Pattern.java:4629)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2876)*
* at java.util.regex.Pattern$Branch.match(Pattern.java:4530)*
* at java.util.regex.Pattern$GroupHead.match(Pattern.java:4570)*
* at java.util.regex.Pattern$Loop.match(Pattern.java:4697)*
* at java.util.regex.Pattern$GroupTail.match(Pattern.java:4629)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2876)*
* at java.util.regex.Pattern$Branch.match(Pattern.java:4530)*
* at java.util.regex.Pattern$GroupHead.match(Pattern.java:4570)*
* at java.util.regex.Pattern$Loop.match(Pattern.java:4697)*
* at java.util.regex.Pattern$GroupTail.match(Pattern.java:4629)*
* at java.util.regex.Pattern$BitClass.match(Pattern.java:2876)*
This is not hapenning for small files.
Can anyone help!

Let's see if I can get this to display as intended: {noformat}(?:/\*(?:[^*]|(?:\*+[^/*]))*\*+/)|(?:--.*)|((?:[^/-]++|/(?!\*)|-(?!-))*+){noformat} Okay, when I try that regex (in EditPad Pro, against the source code of this web page), I see this result: The second alternative, {noformat}(?:--.*){noformat} matches anything from a double hyphen to the end of that line, and the third alternative: {noformat}((?:[^/-]++|/(?!\*)|-(?!-))*+){noformat} matches everything else (up to the next double hyphen, that is). These pages happen to contain a lot of SGML comments, so there are plenty of double hyphens to match. In a large document with only a few double hyphens, it would be easy to blow the stack trying to find the next stopping point.
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Regular Expression needed for a password validator

Business Rules:
Password must be at least 8 characters
Contain at least 1 non alpha character and no spaces.
Here is what I got so far:
String regex = "(?=^.{8,}$)((?!.*\\s)(?=.*[^a-zA-Z])(?=.*[a-zA-Z0-9]))^.*$";
 String [] password = new String [7];
 password [0] = "H@ffman1";
 password [1] = "hoffman1";
 password [2] = "Hoffman1";
 password [3] = "Hoffman 1";
 password [4] = "hoffman 1";
 password [5] = "hoffmans";
 password [6] = "123456789";
 for(int i=0; i<password.length; i++){
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System.out.println(password[i] + " == " + matcher.matches());
}Output:
H@ffman1 == true
hoffman1 == true
Hoffman1 == true
Hoffman 1 == false
hoffman 1 == false
hoffmans == true // (This is a problem)
123456789 == true // (This is a problem)

YoungWinston wrote:
prometheuzz wrote:
which is pretty much what our OldWinston suggested...Actually, I was thinking more along the lines of
System.out.println( password.matches("^[^\\s]{8,}$")
&& password[i].matches("[^a-zA-Z]") );
(I may have got the number of backslahes wrong; 'always forget that stuff).Ah, I see. But you probably meant:
password.matches(".*[^a-zA-Z].*")and they're cryptic enough as it is :-).
Cryptic? Nah...

Regular Expression match for BOM char .

i am trying to match BOM char in file with regex but \\ufeff seem to be not working .
please suggest solution on this

This line String srcStr = new String(b);will only produce the String you expect if the BOM bytes can be converted to characters under your default character encoding. They probably can't.
I use this class
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 0x2B, 0x2F, 0x76, 0x38
 0x2B, 0x2F, 0x76, 0x39
 0x2B, 0x2F, 0x76, 0x2B
 0x2B, 0x2F, 0x76, 0x2F
 0xDD, 0x73, 0x66, 0x73
 0xEF, 0xBB, 0xBF
 0x0E, 0xFE, 0xFF
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 0xFF, 0xFE
 static private int testForBOM(int[] bom, int[] bytes)
 for (int index = 0; index < bom.length; index++)
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 final int[] bytes =
 read(), read(), read(), read()
 int count = 0;
 for (int[] bom : BOMS)
 count = testForBOM(bom, bytes);
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}to get rid of a BOM .

Regular Expression Query for SQL

Hi,
I am working on SQL server and there is a requirement of separating street name and street number.
I tried using regexp_substr but not able to do it perfectly.
Here are some sample names which need cleansing
Full name
street name
street no
Bahnhofplatz 1 / Werkstatt
Bahnhofplatz
1 / Werkstatt
Reinacherstr. 149/151
Reinacherstr.
149/151
Kosuth UCA 7/A
Kosuth UCA
7/A
Route des Jeunes 9 / 6 ème
Route des Jeunes
9 / 6 ème
L'Arcadie / Route de Moudon
L'Arcadie / Route de Moudon
null
Hofstetten 206
Hofstetten
206
Rgds
Amitabh

try this
DECLARE @test TABLE (address VARCHAR(50))
INSERT INTO @test values ('Bahnhofplatz 1 / Werkstatt')
,('Reinacherstr.149 / 151')
,('Kosuth UCA 7 / A')
,('Moudon')
,('L''Arcadie / Route de')
SELECT
Address,
CASE
WHEN patindex('%[0-9]%',address) = 0 then address else substring(address,0,patindex('%[0-9]%',address)) end col1,
CASE WHEN patindex('%[0-9]%',address) >0 then SUBSTRING(address,patindex('%[0-9]%',address),LEN(address)) end col2
from @test
Address col1
col2
Bahnhofplatz 1 / Werkstatt
Bahnhofplatz 1 / Werkstatt
Reinacherstr.149 / 151
Reinacherstr. 149 / 151
Kosuth UCA 7 / A
Kosuth UCA 7 / A
Moudon Moudon
NULL
L'Arcadie / Route de
L'Arcadie / Route de
NULL
--Prashanth

Correct regular expression format for hub transport rule??

I want to create a regex such that it tests for two words with a space between them.
Exchange says /abc(\s)def/ig -- abc def or ABC DEF will pass the test -- Exchange says /mls(\s)secure/ig regex won't work.
How do I write the regex so it works in Exchange 2010??
Thank you, Tom

Hi Tom,
As Amit said, you can test the regex via EMS.
I have some tests in my environment using Exchange 2010. "abc def" -match "/abc(\s)def/ig" is false. Also, "ABC DEF" -match "/abc(\s)def/ig" is false. You can use the "^\w+\s\w+$" regex to test for two words with a space between them.
Hope it helps.
Best regards,
Amy Wang
TechNet Community Support

Regular expression help for matching numbers

Hi,
I want a exact match of either 9 digits or 12 digits, my query should give "No Match Found" as the input value is actually 10 digit
select case when regexp_like(regexp_replace( ' 123 4567 890', ' ' ), '^([0-9]{9})|([0-9]{12})$')
then 'Match Found'
else 'No Match Found'
end as test
from dual;
Need help, as I must be doing something very basic thing, wrong.
Regards,
Ash

Remove 2 brackets:
SQL> select case when regexp_like(regexp_replace( ' 123 4567 890', ' ' ), '^([0-9]{9}|[0-9]{12})$')
2 then 'Match Found'
3 else 'No Match Found'
4 end as test
5 from dual;
TEST
No Match Found
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2 then 'Match Found'
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4 end as test
5 from dual;
TEST
Match Found
SQL>

Regular Expression Required for Checksum variable

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