Regular Expression in ADF

I have a requirement to validate one UI field with the following pattern. Can some one advise me about creating the correct pattern for the following requirement.
It can have maximum of 8 characters. It can have only numbers, no - or [A-Z]. If there is -, then it should be only after 5th character. The last two characters must be alphabets.
Example: 12, 1234, 12345, 78696, 12345-AB (Any number between 1 to 99999 and optional -[A-Z] )
Can you please suggest regular expression for the above pattern.
Thanks and Regards,
S R Prasad

Hi Hyangelo,
I have tried the following regular expression suggested by you. I removed the end / as it is not accepting any valid pattern.
<af:validateRegExp pattern="^[0-9]{0,5}(\1[0-9]{5}-[A-Z]{2}){0,1}$" messageDetailNoMatch="Invalid CPt Code..." />
Thanks and Regards,
S R Prasad

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Regular Expression and jUnit in ADF Mobile

Hi Guys,
I tried to implement Regular Expression in ADF Mobile, but in my jUnit that I was configured to unit test purpose in my application.
Shown error below inside jUnit plane.
java.lang.NoClassDefFoundError: oracle/adfmf/util/regex/Pattern
How to solve this issue for jUnit ?

I'll start by answering your Q2 - while in theory you can use the same application for both accessing from a browser on a pc and on a phone - in reality the user experience on the phone is not going to be good because of the limitation of the screen size. Users will need to constantly zoom-in scroll around etc. This is why many people who target application for smart phones prefer to build a separate UI for the mobile device.
Now question 4 which relates to 1 - We haven't tested the ADF Faces components on the browser of blackberry and so there is a chance that some of their functionality won't work there. We did test and certify the ADF Faces components with the safari mobile browser (iOS) - so this is why you see them work there better.
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ADF Email Validation using Regular Expression

Hi,
Wanted to add Email Validation VO search.
It is working if i put
<af:validateRegExp pattern="[A-Z0-9._%+-]+@[A-Z0-9.-]+\.[A-Z]{2,4}"
 messageDetailNoMatch="The value {1} is not a valid email address:"/>However this requires email id to be entered in Capital Letters.
Tried with below option is not working.
 <af:inputText value="#{bindings.xxEmail.inputValue}" label="Email"
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 columns="#{bindings.xxEmail.hints.displayWidth}"
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 </af:inputText>I got above info from
ADF Email Validation using Regular Expression
User don't enter email id Without @ .
Kindly suggest pattern to achive this.
Thanks,
jit
Edited by: appsjit on Jan 25, 2013 7:08 PM

The RegEx to check EMail after RFC2822
[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*@(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?There are still some missing parts in the check as not all suffix combinations are allowed, but this is pretty good.
Timo

ADF Regular expression Jspx page error

Hi Friends
Hi all i have requirement that my inputtext box may include charchters numbers and '&' as a special charchters
so i written the below regular expression in validateRegExp tag
<af:validateRegExp pattern="[a-zA-Z0-9&]" messageDetailNoMatch="Inproper text"/>
but when i enter '&' char in the regex of the tag of my jspx page then it gives me compilier error, so is there any way to write the escape charachters in regular expression like '\ ' in java to avoid above problem.
Thanks in advance
Bipin patil
Edited by: Bipin Patil on Aug 2, 2011 1:39 AM

Hi..
can you try with following
for Ex: aA1& pattern
><af:inputText id="it1">
><af:validateRegExp pattern="[a-z][A-Z][0-9][&]" messageDetailNoMatch="Inproper text"/>
></af:inputText>
for Ex: awerrAFDFDF12343&&& pattern(Every item should shown at least one time..)
><af:inputText id="it11">
><af:validateRegExp pattern="[a-z]+[A-Z]+[0-9]+[&]+" messageDetailNoMatch="Inproper text"/>
></af:inputText>
Hope this will helpful..

Oracle Regular Expressions in Java?

JDEV 10.1.3
ADF BC
ADF Faces
Is there a Java library for dealing with Oracle regular expressions? I would like to be able to test a String against an Oracle regular expression for pattern match.
Thank you,

Hi,
Java supports RegularEpressions: http://www.exampledepot.com/egs/java.util.regex/pkg.html
Frank

How to add regular expression in viewobject bind variables

Hi,
Am using java class to set the adf bind variables
vm.setWhereClause("FIRST_NAME = :fname");
vm.defineNamedWhereClauseParam("fname", null, null);
vm.setNamedWhereClauseParam("fname","Alana");
It works fine and it returns Alana
But what i want is to set a regular expression where it returns all the names that are starting with A.
I tried with this.
vm.setNamedWhereClauseParam("fname","A*");
also
vm.setNamedWhereClauseParam("fname","A%");
but both didn work.
Please help.
Thanks,
Hari

Perfect... Thanks Arun.
I tried including last name too. i.e when either of the one matches (first name or last name)
vm.setWhereClause("upper(FIRST_NAME) LIKE upper(:fname||'%')");
vm.defineNamedWhereClauseParam("fname", null, null);
vm.setNamedWhereClauseParam("fname","A");
vm.setWhereClause("upper(LAST_NAME) LIKE upper(:lname||'%')");
vm.defineNamedWhereClauseParam("lname", null, null);
vm.setNamedWhereClauseParam("lname","A");
But i guess we should not give it like this as it doesnt seem to work.
Hari

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 ?
Baken

First 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.

Help in Regular expression

Hello..
I wanted to write a regular expression to match the foll string..
 
 NEW ORLEANS, Louisiana (CNN) 
-- Two years after Hurricane Katrina devastated coastal areas of Louisiana and Mississippi, residents say much of America has forgotten their plight.
 
I tried doing..
Matcher matcher= Pattern.compile(" ([^<^>]+?)", Pattern.CASE_INSENSITIVE).matcher(story);
Its not working...
is there any other soln?

Theres probably a better way to do this but here's a way that works.
import java.util.regex.*;
public class RegexTester{
public static void main(String[] args){
 String text =
 " " +
 " NEW ORLEANS, Louisiana (CNN) " +
 "-- Two years after Hurricane Katrina devastated coastal areas of Louisiana and Mississippi," +
 "residents say much of America has forgotten their plight." +
 " ";
 String regex = ">((?:\\s*[\\S&&[^<>]]+\\s*)*?)<";
 Pattern p = Pattern.compile(regex);
 Matcher m = p.matcher(text);
 while(m.find()){
 System.out.println("Match: '" + m.group(1) + "'");
}

Help in regular expression matching

I have three expressions like
1) [(y2009)(y2011)]
2) [(y2008M5)(y2011M3)] or [(y2009M5)(y2010M12)]
3) [(y2009M1d20)(y2011M12d31)]
i want regular expression pattern for the above three expressions
I am using :
REGEXP_LIKE(timedomainexpression, '???[:digit:]{4}*[:digit:]{1,2}???[:digit:]{4}*[:digit:]{1,2}??', 'i');
but its giving results for all above expressions while i want different expression for each.
i hav used * after [:digit:]{4}, when i am using ? or . then its giving no results. Please help in this situation ASAP.
Thanks

I dont get your question Can you post your desired output? and also give some sample data.
Please consider the following when you post a question.
1. New features keep coming in every oracle version so please provide Your Oracle DB Version to get the best possible answer.
You can use the following query and do a copy past of the output.
select * from v$version 2. This forum has a very good Search Feature. Please use that before posting your question. Because for most of the questions
that are asked the answer is already there.
3. We dont know your DB structure or How your Data is. So you need to let us know. The best way would be to give some sample data like this.
I have the following table called sales
with sales
as
 select 1 sales_id, 1 prod_id, 1001 inv_num, 120 qty from dual
 union all
 select 2 sales_id, 1 prod_id, 1002 inv_num, 25 qty from dual
select *
from sales 4. Rather than telling what you want in words its more easier when you give your expected output.
For example in the above sales table, I want to know the total quantity and number of invoice for each product.
The output should look like this
Prod_id sum_qty count_inv
1 145 2 5. When ever you get an error message post the entire error message. With the Error Number, The message and the Line number.
6. Next thing is a very important thing to remember. Please post only well formatted code. Unformatted code is very hard to read.
Your code format gets lost when you post it in the Oracle Forum. So in order to preserve it you need to
use the {noformat}{noformat} tags.
The usage of the tag is like this.
<place your code here>\
7. If you are posting a *Performance Related Question*. Please read
 {thread:id=501834} and {thread:id=863295}.
 Following those guide will be very helpful.
8. Please keep in mind that this is a public forum. Here No question is URGENT.
 So use of words like *URGENT* or *ASAP* (As Soon As Possible) are considered to be rude.

Regular expression alphabets

Hi
I want to retrieve the data if the data contains a character or a space or '-' thru select query .
Please help me in writing the combination of 3 with regular expression.
Thanks!!

VT wrote:
Hi,
Try this
SELECT *
FROM <TABLE> WHERE REGEXP_LIKE(<COLUMN>, '[a-z -][A-Z -]');cheers
VTThat won't work as it's expecting at least two characters with the first having to be a-z (lower case) or space or "-" followed by A-Z (upper case) or space or "-".
The correct way is either:
[a-zA-Z -]or
[[:alpha:] -]using the alpha set is often preferable as it can work differently with different character sets/languages rather than restricting to just the a-zA-Z ranges.
Generating a reference for your own database characterset/language can be useful...
SQL> select level-1 as asc_code, decode(chr(level-1), regexp_substr(chr(level-1), '[[:print:]]'), CHR(level-1)) as chr,
2 decode(chr(level-1), regexp_substr(chr(level-1), '[[:graph:]]'), 1) is_graph,
3 decode(chr(level-1), regexp_substr(chr(level-1), '[[:blank:]]'), 1) is_blank,
4 decode(chr(level-1), regexp_substr(chr(level-1), '[[:alnum:]]'), 1) is_alnum,
5 decode(chr(level-1), regexp_substr(chr(level-1), '[[:alpha:]]'), 1) is_alpha,
6 decode(chr(level-1), regexp_substr(chr(level-1), '[[:digit:]]'), 1) is_digit,
7 decode(chr(level-1), regexp_substr(chr(level-1), '[[:cntrl:]]'), 1) is_cntrl,
8 decode(chr(level-1), regexp_substr(chr(level-1), '[[:lower:]]'), 1) is_lower,
9 decode(chr(level-1), regexp_substr(chr(level-1), '[[:upper:]]'), 1) is_upper,
10 decode(chr(level-1), regexp_substr(chr(level-1), '[[:print:]]'), 1) is_print,
11 decode(chr(level-1), regexp_substr(chr(level-1), '[[:punct:]]'), 1) is_punct,
12 decode(chr(level-1), regexp_substr(chr(level-1), '[[:space:]]'), 1) is_space,
13 decode(chr(level-1), regexp_substr(chr(level-1), '[[:xdigit:]]'), 1) is_xdigit
14 from dual
15 connect by level <= 256
16 /
ASC_CODE C IS_GRAPH IS_BLANK IS_ALNUM IS_ALPHA IS_DIGIT IS_CNTRL IS_LOWER IS_UPPER IS_PRINT IS_PUNCT IS_SPACE IS_XDIGIT
 0 1
 1 1
 2 1
 3 1
 4 1
 5 1
 6 1
 7 1
 8 1
 9 1 1
 10 1 1
 11 1 1
 12 1 1
 13 1 1
 14 1
 15 1
 16 1
 17 1
 18 1
 19 1
 20 1
 21 1
 22 1
 23 1
 24 1
 25 1
 26 1
 27 1
 28 1
 29 1
 30 1
 31 1
 32 1 1 1
 33 ! 1 1 1
 34 " 1 1 1
 35 # 1 1 1
 36 $ 1 1 1
 37 % 1 1 1
 38 & 1 1 1
 39 ' 1 1 1
 40 ( 1 1 1
 41 ) 1 1 1
 42 * 1 1 1
 43 + 1 1 1
 44 , 1 1 1
 45 - 1 1 1
 46 . 1 1 1
 47 / 1 1 1
 48 0 1 1 1 1 1
 49 1 1 1 1 1 1
 50 2 1 1 1 1 1
 51 3 1 1 1 1 1
 52 4 1 1 1 1 1
 53 5 1 1 1 1 1
 54 6 1 1 1 1 1
 55 7 1 1 1 1 1
 56 8 1 1 1 1 1
 57 9 1 1 1 1 1
 58 : 1 1 1
 59 ; 1 1 1
 60 < 1 1 1
 61 = 1 1 1
 62 > 1 1 1
 63 ? 1 1 1
 64 @ 1 1 1
 65 A 1 1 1 1 1 1
 66 B 1 1 1 1 1 1
 67 C 1 1 1 1 1 1
 68 D 1 1 1 1 1 1
 69 E 1 1 1 1 1 1
 70 F 1 1 1 1 1 1
 71 G 1 1 1 1 1
 72 H 1 1 1 1 1
 73 I 1 1 1 1 1
 74 J 1 1 1 1 1
 75 K 1 1 1 1 1
 76 L 1 1 1 1 1
 77 M 1 1 1 1 1
 78 N 1 1 1 1 1
 79 O 1 1 1 1 1
 80 P 1 1 1 1 1
 81 Q 1 1 1 1 1
 82 R 1 1 1 1 1
 83 S 1 1 1 1 1
 84 T 1 1 1 1 1
 85 U 1 1 1 1 1
 86 V 1 1 1 1 1
 87 W 1 1 1 1 1
 88 X 1 1 1 1 1
 89 Y 1 1 1 1 1
 90 Z 1 1 1 1 1
 91 [ 1 1 1
 92 \ 1 1 1
 93 ] 1 1 1
 94 ^ 1 1 1
 95 _ 1 1 1
 96 ` 1 1 1
 97 a 1 1 1 1 1 1
 98 b 1 1 1 1 1 1
 99 c 1 1 1 1 1 1
 100 d 1 1 1 1 1 1
 101 e 1 1 1 1 1 1
 102 f 1 1 1 1 1 1
 103 g 1 1 1 1 1
 104 h 1 1 1 1 1
 105 i 1 1 1 1 1
 106 j 1 1 1 1 1
 107 k 1 1 1 1 1
 108 l 1 1 1 1 1
 109 m 1 1 1 1 1
 110 n 1 1 1 1 1
 111 o 1 1 1 1 1
 112 p 1 1 1 1 1
 113 q 1 1 1 1 1
 114 r 1 1 1 1 1
 115 s 1 1 1 1 1
 116 t 1 1 1 1 1
 117 u 1 1 1 1 1
 118 v 1 1 1 1 1
 119 w 1 1 1 1 1
 120 x 1 1 1 1 1
 121 y 1 1 1 1 1
 122 z 1 1 1 1 1
 123 { 1 1 1
 124 | 1 1 1
 125 } 1 1 1
 126 ~ 1 1 1
 127 1
 128 Ç 1 1 1
etc.
{code}

Help in query using regular expression

HI,
I need a help to get the below output using regular expression query. Please help me.
SELECT REGEXP_SUBSTR ('PWRPKG(P/W+P/L+CC)', '[^+]+', 1, lvl) val, lvl
FROM DUAL,(SELECT LEVEL lvl FROM DUAL
CONNECT BY LEVEL <=(SELECT MAX ( LENGTH ('PWRPKG(P/W+P/L+CC)') - LENGTH (REPLACE ('PWRPKG(P/W+P/L+CC)','+',NULL))+ 1) FROM DUAL));
I need the output as
correct result:
==============
val lvl
P/W 1
P/L 2
CC 3
But i tried the above it is not coming the above result. Please help me where i did a mistake.
Thanks in advance

Frank gave you a solution in your other thread. You could simplify it if you are on 11g:
SQL> select * from table_x
2 /
TXT
TECHPKG(INTELLI CC+FRT SONAR)
PWRPKG(P/W+P/L+CC)
select txt,
 regexp_substr(
 txt,
 '(.*\()*([^+)]+)',
 1,
 column_value,
 null,
 2
 ) element,
 column_value element_number
from table_x,
 table(
 cast(
 multiset(
 select level
 from dual
 connect by level <= regexp_count(txt,'\+') + 1
 as sys.OdciNumberList
order by rowid,
 column_value
TXT ELEMENT ELEMENT_NUMBER
TECHPKG(INTELLI CC+FRT SONAR) INTELLI CC 1
TECHPKG(INTELLI CC+FRT SONAR) FRT SONAR 2
PWRPKG(P/W+P/L+CC) P/W 1
PWRPKG(P/W+P/L+CC) P/L 2
PWRPKG(P/W+P/L+CC) CC 3
SQL> SY.

Query help in regular expression

Hi all,
SELECT * FROM emp11
WHERE INSTR(ENAME,'A',1,2) >0;
Please let me know the equivalent query using regular expressions.
i have tried this after going through oracle regular expressions documentation.
SELECT * FROM emp11
WHERE regexp_LIKE(ename,'A{2}')
Any help in this regard would be highly appreciated .
Thanks,
P Prakash

please go here
Introduction to regular expressions ...
Thanks,
P Prakash

Urgent!!! Problem in regular expression for matching braces

Hi,
For the example below, can I write a regular expression to store getting key, value pairs.
example: ((abc def) (ghi jkl) (a ((b c) (d e))) (mno pqr) (a ((abc def))))
in the above example
abc is key & def is value
ghi is key & jkl is value
a is key & ((b c) (d e)) is value
and so on.
can anybody pls help me in resolving this problem using regular expressions...
Thanks in advance

"((key1 value1) (key2 value2) (key3 ((key4 value4)
(key5 value5))) (key6 value6) (key7 ((key8 value8)
(key9 value9))))"
I want to write a regular expression in java to parse
the above string and store the result in hash table
as below
key1 value1
key2 value2
key3 ((key4 value4) (key5 value5))
key4 value4
key5 value5
key6 value6
key7 ((key8 value8) (key9 value9))
key8 value8
key9 value9
please let me know, if it is not possible with
regular expressions the effective way of solving itYes, it is possible with a recursive regular expression.
Unfortunately Java does not provide a recursive regular expression construct.
$_ = "((key1 value1) (key2 value2) (key3 ((key4 value4) (key5 value5))) (key6 value6) (key7 ((key8 value8) (key9 value9))))";
my $paren;
   $paren = qr/
           [^()]+ # Not parens
         |
           (??{ $paren }) # Another balanced group (not interpolated yet)
    /x;
my $r = qr/^(.*?)$(\w+?) (\w+?|(??{$paren}))$\s*(.*?)$/;
while ($_) {
     match()
# operates on $_
sub match {
     my @v;
     @v = m/$r/;
     if (defined $v[3]) {
          $_ = $v[2];
          while (/\(/) {
               match();
          print "\"",$v[1],"\" \"",$v[2],"\"";
          $_ = $v[0].$v[3];
     else { $_ = ""; }
C:\usr\schodtt\src\java\forum\n00b\regex>perl recurse.pl
"key1" "value1"
"key2" "value2"
"key4" "value4"
"key5" "value5"
"key3" "((key4 value4) (key5 value5))"
"key6" "value6"
"key8" "value8"
"key9" "value9"
"key7" "((key8 value8) (key9 value9))"
C:\usr\schodtt\src\java\forum\n00b\regex>

Bracket in Regular Expression constant?

I am a bit puzzled by the behavior I am experiencing in LV 2011. I hope to get some light from experts out there.
I am trying to parse a messy ASCII header file and after having split it into individual lines (strings), I use the "Match Regular Expression" function to remove some of the info before the substantial information.
Some of the strings include square brackets ([, ]), which are special characters for the function, therefore, as documented in the help, one needs to precede them with a backslash.
Example:
I want to parse the following line:
#PR [PR_DEV,I,2]
One way (which I am using because of considerations related to the rest of the header) is the the following:
Note that the first string constant is using "Code Display" whereas the second one is using "Normal Display".
Why did I not put a backslash in front of the bracket in the first string, you may ask? Well, I did, but it disappeared after I typed the other characters. And reverting to "Normal Display" did not restore it.
Of course, the first version does not parse the input string correctly, whereas the second one does it fine.
In other words, the custom display string (which is convenient for cryptic codes such as \s* or to distinguish between space and tab...or simply ENTER tabs!) seems to mess up with the \[ combo (likewise with the \] one).
It is not a huge deal. I can use the "Normal Display" mode, but I tend to think that this qualifies as a hidden "feature". And again, it is still a pain in the ... when dealing with special characters such as tabs, etc...
Solved!
Go to Solution.

I think that [ is a special character which needs to be preceded by a backslash, but it is not one of the defined backslash characters (like \s). So, you need to put in two \\ to get one \ while in '\' Codes Display.
You can put in any character by using \xx where the xx is a hex character using only upper case letters for A..F. I converted the strings to byte arrays and tried to see what made the arrays match and the Match work.
Lynn

Need help with regular expression

I'm trying to use the java.util.regex package to extract URLs from html files.
The URLs that I am interested in extracting from the HTML look like the following:
http://forum.java.sun.com -
So, the URL is always preceeded by:

and then followed by a space character and then a hyphen character. I want to be able to put all these URLs in a Vector object. This doesn't seem like it should be too difficult but for some reason I can't get anywhere with it. Any help would be greatly appreciated. Thanks!

hi gupta am not sure of the java syntax but i can tell u about the regular expression...try this....
(http:\/\/[a-zA-Z0-9.]+) [-]
i dont know the java methods to call...just the reg exp...
Sanjay Acharya

Regular Expression in ADF

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