Phone Number - Regular Expression

I'm trying to validate a phone number using regular expressions.
My validation appears as the following:
IF REGEXP_LIKE ( :P10_PHONE, '^$?[[:digit:]]{3}$?[-. ]?[[:digit:]]{3}[-. ]?[[:digit:]]{4}$' ) THEN
RETURN TRUE;
ELSE
RETURN FALSE;
END IF;
The problem I noticed is that it returned true even if I typed in a phone number like this:
(555 123-4567
Is there a way to tell it that if one parenthesis is there that the other one should be there as well? Or should I just do an instr if the regexp_like returns true and double-check it myself?
Thanks,
Chad

You can use the OR operator (|)...
'^($[[:digit:]]{3}$|[[:digit:]]{3})[-. ]?[[:digit:]]{3}[-. ]?[[:digit:]]{4}$'
Rodney

Similar Messages

Phone number Regular expression Help

Hi,
I am trying to validate phone number using regular expression.
The format shoud be either of the two given below. It should not
accept phone number of any other format other than the once given below.
I tried out the following regular expression
pn.matches("[\p{+}][0-9]+ ")
but it accepts alphabets too(which is wrong).
How do i check if the phone number is of the following format(OR condition).
0401234567
+358501234567
Any help would be kindly appriciated.
Thanks
Sanam

There will probably be much more constraints on you phone numbers, but here's a start:String[] numbers = {"0401234567",      // should be ok
                    "040123456",       // wrong, one number too little
                    "+358501234567",   // should be ok
                    "+3585012345670", // wrong, one number too much
                    "+35850123456"};   // wrong, one number too little
String regex = "\\+[0-9]{12}"+         // a + sign, followed by 12 numbers
               "|"+                    // OR
               "0[0-9]{9}";            // a zero, followed by 9 numbers
for(String n : numbers) {
System.out.println("Is "+n+" valid? "+n.matches(regex));
}

Regular expressions in Format Definition add-on

Hello experts,
I have a question about regular expressions. I am a newbie in regular expressions and I could use some help on this one. I tried some 6 hours, but I can't get solve it myself.
Summary of my problem:
In SAP Business One (patch level 42) it is possible to use bank statement processing. A file (full of regular expressions) is to be selected, so it can match certain criteria to the bank statement file. The bank statement file consists of a certain pattern (look at the attached code snippet).
:61:071222D208,00N026
:86:P 12345678BELASTINGDIENST       F8R03782497                $GH
$0000009                         BETALINGSKENM. 123456789123456
0 1234567891234560
:61:071225C758,70N078
:86:0116664495 REGULA B.V. HELPMESTRAAT 243 B 5371 AM HARDCITY HARD
CITY 48772-54314
:61:071225C425,05N078
:86:0329883585 J. MANSSHOT PATTRIOTISLAND 38 1996 PT HELMEN BIJBETA
LING VOOR RELOOP RMP1 SET ORDERNR* 69866 / SPOEDIG LEVEREN
:61:071225C850,00N078
:86:0105327212 POSE TELEFOONSTRAAT 43 6448 SL S-ROTTERDAM MIJN OR
DERNR. 53846 REF. MAIL 21-02
- I am in search of the right type of regular expression that is used by the Format Definition add-on (javascript, .NET, perl, JAVA, python, etc.)
Besides that I need the regular expressions below, so the Format Definition will match the right lines from my bankfile.
- a regular expression that selects lines starting with :61: and line :86: including next lines (if available), so in fact it has to select everything from :86: till :61: again.
- a regular expression that selects the bank account number (position 5-14) from lines starting with :86:
- a regular expression that selects all other info from lines starting with :86: (and following if any), so all positions that follow after the bank account number
I am looking forward to the right solutions, I can give more info if you need any.

Hello Hendri,
Q1:I am in search of the right type of regular expression that is used by the Format Definition add-on (javascript, .NET, perl, JAVA, pythonetc.)
Answer: Format Definition uses .Net regular expression.
You may refer the following examples. If necessary, I can send you a guide about how to use regular expression in Format Defnition. Thanks.
Example 6
Description:
To match a field with an optional field in front. For example, u201C:61:0711211121C216,08N051NONREFu201D or u201C:61:071121C216,08N051NONREFu201D, which comprises of a record identification u201C:61:u201D, a date in the form of YYMMDD, anther optional date MMDD, one or two characters to signify the direction of money flow, a numeric amount value and some other information. The target to be matched is the numeric amount value.
Regular expression:
(?<=:61:\d(\d)?[a-zA-Z]{1,2})((\d(,\d*)?)|(,\d))
Text:
:61:0711211121C216,08N051NONREF
Matches:
1
Tips:
1.     All the fields in front of the target field are described in the look behind assertion embraced by (?<= and ). Especially, the optional field is embraced by parentheses and then a u201C?u201D (question mark). The sub expression for amount is copied from example 1. You can compose your own regular expression for such cases in the form of (?<=REGEX_FOR_FIELDS_IN_FRONT)(REGEX_FOR_TARGET_FIELD), in which REGEX_FOR_FIELDS_IN_FRONT and REGEX_FOR_TARGET_FIELD are respectively the regular expression for the fields in front and the target field. Keep the parentheses therein.
Example 7
Description:
Find all numbers in the free text description, which are possibly document identifications, e.g. for invoices
Regular expression:
(?<=\b)(?<!\.)\d+(?=\b)(?!\.)
Text:
:86:GIRO 6890316
ENERGETICA NATURA BENELU
AFRIKAWEG 14
HULST
3187-A1176
TRANSACTIEDATUM* 03-07-2007
Matches:
6
Tips:
1.     The regular expression given finds all digits between word boundaries except those with a prior dot or following dot; u201C.u201D (dot) is escaped as \.
2.     It may find out some inaccurate matches, like the date in text. If you want to exclude u201C-u201D (hyphen) as prior or following character, resemble the case for u201C.u201D (dot), the regular expression becomes (?<=\b)(?<!\.)(?<!-)\d+(?=\b)(?!\.)(?!-). The matches will be:
:86:GIRO 6890316
ENERGETICA NATURA BENELU
AFRIKAWEG 14
HULST
3187-A1176
TRANSACTIEDATUM* 03-07-2007
You may lose some real values like u201C3187u201D before the u201C-u201D.
Example 8
Description:
Find BP account number in 9 digits with a prior u201CPu201D or u201C0u201D in the first position of free text description
Regular expression:
(?<=^(P|0))\d
Text:
0000006681 FORTIS ASR BETALINGSCENTRUM BV
Matches:
1
Tips:
1.     Use positive look behind assertion (?<=PRIOR_KEYWORD) to express the prior keyword.
2.     u201C^u201D stands for that match starts from the beginning of the text. If the text includes the record identification, you may include it also in the look behind assertion. For example,
:86:0000006681 FORTIS ASR BETALINGSCENTRUM BV
The regular expression becomes
(?<=:86:(P|0))\d
Example 9
Description:
Following example 8, to find the possible BP name after BP account number, which is composed of letter, dot or space.
Regular expression:
(?<=^(P|0)\d)[a-zA-Z. ]*
Text:
0000006681 FORTIS ASR BETALINGSCENTRUM BV
Matches:
1
Tips:
1.     In this case, put BP account number regular expression into the look behind assertion.
Example 10
Description:
Find the possible document identifications in a sub-record of :86: record. Sub-record is like u201C?00u201D, u201C?10u201D etc. A possible document identification sub-record is made up of the following parts:
u2022     keyword u201CREu201D, u201CRGu201D, u201CRu201D, u201CINVu201D, u201CNRu201D, u201CNOu201D, u201CRECHNu201D or u201CRECHNUNGu201D, and
u2022     an optional group made up of following:
     a separator of either a dot, hyphen or slash, and
     an optional space, and
     an optional string starting with keyword u201CNRu201D or u201CNOu201D followed by a separator of either a dot, hyphen or slash, and
     an optional space
u2022     and finally document identification in digits
Regular expression:
(?<=\?\d(RE|RG|R|INV|NR|NO|RECHN|RECHNUNG)((\.|-|/)\s?((NR|NO)(\.|-|/))?\s?)?)\d+
Kind Regards
-Yatsea

Phone number with regular expression

Hi,
is it possible to use regular expression for phone numbers?
I want to create contacts for companies with several direct dial in numbers. My idea is to make one contact for the company itself with the base telephone number and a regular expression like an asterisk and separate contacts for person inside the company containing the direct dial in number.

but a third-party app could not tell me who (anybody from the company or a special direct dial in number) is calling me at this very moment. So sounds like no way anyhow.

Regular Expressions and phone number

HI,
there is a column "Phone_number Varchar2"
Data containing :
123-89556-6852
(123)-857-965
123-(5846)5648
I want to display that
123895566852
123857965
12358465648
Pls help by Regular Expressions.
Edited by: YLN on Feb 17, 2010 3:59 PM

with t
as
select '123-89556-6852' col from dual union all
select '(123)-857-965' col from dual union all
select '123-(5846)5648' col from dual
SELECT regexp_replace(col,'[^[:digit:]]') FROM tRavi Kumar

TA38622 how do you delete a contact name and phone number from this? i have deleted it from my regular contacts, but it remains in a text memory when i begin to type the beginning of a name

How do I delete a contact name and phone number from my texting memory? I have deleted this contact from my regular contact information; however when I text and begin to text a contact beginning with the 1st letter of the name I had deleted, it brings up the old, deleted contact information in my texting history only.

You would have to restore as new.

[Regular Expressions] Saving a variable number of matches

I'm stuck with the following problem and I don't seem to be able to solve without lots of ifs and else's.
I've got a program that you can pass patterns as parameters to. The program receives patterns as one single string.
The string could look like this:
a:i:foo r::bar t:ei:bark
or like this:
a:i:foo
What I'm hinting at is that the string comprises of several parts of the same structure. Each structure can be matched and saved with:
([art]:[ei]{0,2}:.*)
Now I want my regular expression able to match all the occurences without checking the string containing the pattern for something that could indicate the number of structures inside it. The following does not seem to work:
([art]:[ei]{0,2}:.*)+
So now I'm looking for something that would match one or more occurence of the structure and save it for future use.
I'd be really happy if someone could help me out here
Last edited by n0stradamus (2012-05-03 20:27:02)

Procyon wrote:
--> echo "a:i:foo r::bar t:ei:bark" | sed 's/$[art]:[ei]\{0,2\}:[^ ]*$/1/'
1 r::bar t:ei:bark
--> echo "a:i:foo r::bar t:ei:bark" | sed 's/$[art]:[ei]\{0,2\}:[^ ]*$/1/g'
1 1 1
If [^ ]* is not usable (spaces are allowed arbitrarily), you need a non-greedy .* and non-consuming look-ahead of " [art]:"
In python's re module, this is .*?(?=( [art]:|$))
>>> import re
>>> m=re.findall("([art]:[ei]{0,2}:.*?(?=( [art]:|$)))","a:i:foo r::bar t:ei:bark")
>>> print(m)
[('a:i:foo', ' r:'), ('r::bar', ' t:'), ('t:ei:bark', '')]
Exactly what I was looking for! I didn't know that you could specify .* to stop at a certain sequence of characters.
Could you please point me to some materials where I can read up on the topic?
Back to the regex: It works finde in Python, but sadly that is not the language I'm using
The program I need this for is written in C and until now the regex functions from glibc worked fine for me.
Have I missed a function similar to re.findall in glibc?

Checking a number sequence with regular expressions

Hello,
Suppose I have a text in the pattern:
A1=ha,A2=bla,A3=cha,...
I don't know how many sections of "A#=$" (# denotes number, $ denotes text) will be in the text, but I want to verify that the numbers of the A's form the natural ascending number sequence (i.e 1,2,3,...). I prefer to use regular expressions to do this, but if there's another way, I will be glad to hear it too.
Therefore my question is: How can I use regular expressions to check for a sequence of numbers? I know I can search for groups I've caught previously in the expression, but how can I compute the next number in the sequence from the group and search for the result?
Thank you very much!

What I'd do--and I'm not saying this is optimal, just what pops immediately to mind--is have a regex that matches "A(\\d+)=" (assuming the ha, bla, cha can never be "A1" etc.--if they can, you can still do it, but it's more complicated), then you iterate with the Matcher, and each time, you get the Integer.valueOf what you matched. You keep track of the last value, and compare the current to the last. If current is < last (or <= last, depending on your requirements), fail.
Something like this. I don't recall Matcher's methods off the top of my head, so you'll have to fix the details.
Matcher m = Pattern.matcher("A(\\d+)=");
int maxSoFar = Integer.MIN_VALUE;
while (m.matches(input)) {
    int current = Integer.parseInt(m.getMatchedField("$1"));
    if (current <= maxSoFar) {
        // fail
    else {
        maxSoFar = current;
} maxSoFar =

Regular Expression for Invalid Number

Hi everyone,
I am using oracle version as follows:
SQL> select * from v$version;
BANNER
Oracle Database 10g Enterprise Edition Release 10.2.0.4.0 - Prod
PL/SQL Release 10.2.0.4.0 - Production
CORE    10.2.0.4.0      Production
TNS for 32-bit Windows: Version 10.2.0.4.0 - Production
NLSRTL Version 10.2.0.4.0 - Production
I am using regular expression to replace invalid values from a table.
I received oracle error stating "ORA-01722 invalid number"
My query looks like this:
SELECT DISTINCT
MRC_KEY,
PURPOSE_CD,
RESIDENCE_DESC,
to_number(regexp_replace(ICAP_GEN_MADAPTIVE,'[+. ]?0?0?(\d+)[-.]?','\1')) as ICAP_GEN_MADAPTIVE,
From
MRRC_INT
I am not sure what are the invalid values in the table so I can write regexp accordingly.
Any guidance is highly appreciated!
Thanks in advance
J

Or use DML error logging:
create table t1
(col1 number);
exec dbms_errlog.create_error_log ('t1','t1_errors')
insert into t1
with t as
(select '1' col from dual union all
   select '1.1' col from dual union all
   select '.11' col from dual union all
   select '0.11' col from dual union all
   select '-1' col from dual union all
   select '1,1' col from dual union all
   select '11a' col from dual union all
   select '1d' col from dual union all
   select '1e6' col from dual union all
   select '1e6.1' col from dual union all
   select '1e' col from dual
select col
from t
log errors into t1_errors
reject limit 20
col col1 for 999,999,999.99
select * from t1;
           COL1
           1.00
           1.10
            .11
            .11
          -1.00
   1,000,000.00
col col1 for a30
select * from t1_errors;
ORA_ERR_NUMBER$ ORA_ERR_MESG$                  ORA_ERR_ROWID$       OR ORA_ERR_TAG$         COL1
           1722 ORA-01722: invalid number                           I                       1,1
           1722 ORA-01722: invalid number                           I                       11a
           1722 ORA-01722: invalid number                           I                       1d
           1722 ORA-01722: invalid number                           I                       1e6.1
           1722 ORA-01722: invalid number                           I                       1e

Regular expression - parse version number string

Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
    throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
     throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305
Thanks for help or suggestions :)
Best Regards,
Michael

heissm wrote:
Hi,
I try to parse a string using regular expressions but id did not work correctly in all cases.
The string that i want to parse can have one of the following layout:
3.4.5.v20090305 or
3.4.5The first three parts have to be integer values, the last part can be a free string (which is optional).
I use the following code to parse the version number and extract the parts:
Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
Matcher m = versionPattern.matcher(versionString);
if (!m.find()) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// assert that we matched every part
// three groups (without qualifier) or four parts (with qualifier)
int groups = m.groupCount();
if (groups != 4) {
throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
// extract the parts
major = parseInt(m.group(1));
minor = parseInt(m.group(2));
micro = parseInt(m.group(3));
qualifier = m.group(4);The above regular expression works in all cases that i tested, except one.
The follwoing string is accepted as correct input: (but it shouldn't)
3.4.5a.v20090305And i get the result:
major = 3
minor = 4
micro = 5
qualifier = a.v20090305No, that can't be the output. Perhaps you have some old class files you're executing, because with the code you now posted, that can't be the result.
To verify this, execute this:
import java.util.regex.*;
public class Main {
    public static void main(String[] args) {
        String versionString = "3.4.5a.v20090305";
        Pattern versionPattern = Pattern.compile("(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?");
        Matcher m = versionPattern.matcher(versionString);
        if (!m.find()) {
            throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
        // assert that we matched every part
        // three groups (without qualifier) or four parts (with qualifier)
        int groups = m.groupCount();
        if (groups != 4) {
             throw new IllegalArgumentException("Version must be in form <major>.<minor>.<micro>.<qualifier>");
        // extract the parts
        System.out.println("1 -> "+m.group(1));
        System.out.println("2 -> "+m.group(2));
        System.out.println("3 -> "+m.group(3));
        System.out.println("4 -> "+m.group(4));
}and you'll see that the following is the result:
1 -> 3
2 -> 4
3 -> 5
4 -> nullSome remarks about your pattern:
"(\\d+)\\.{1}(\\d+)\\.{1}(\\d+)(?:\\.{1}(\\w+))?"All the "{1}" can be omitted they don't add anything to your regex and are only cluttering it up. And you'll probably also want to "anchor" your regex using the start- and end-of-string-meta-characters, like this:
"^(\\d+)\\.(\\d+)\\.(\\d+)(?:\\.(\\w+))?$"

Regular expression - get longest number from string

I believe it is easy one but I can't get it.Lets say I have a string 'A1_1000' I want to substract the 1000 using regular expression. When I feed Match regular expression I get '1' which is not the longest number. I know other ways of doing that but I want clean solution in one step. Does anybody knows the right regular expression to accomplish that? Thanks!
LV 2011, Win7
Solved!
Go to Solution.

ceties wrote:
This is the best solution I was able to come with. I am just wondering if there is "smoother way" without the cycle.
Since multiple checks are required I would tend to beieve that we do have to loop through the possibilities. in this example
I start check at offset "0" into the string for a number. Provided i find a number I check if it is longer that any previous number I found and if so save the new longer number in the shift register.
Have fun!
Ben
Message Edited by Ben on 04-15-2009 09:23 AM
Ben Rayner
I am currently active on.. MainStream Preppers
Rayner's Ridge is under construction
Attachments:
Find_Longest.PNG ‏33 KB

Regular Expression to Check number with at least one decimal point

Hi,
I would like to use the REGEX_LIKE to check a number with up to two digits and at least one decimal point:
Ex.
10.1
1.1
1
12
This is what I have so far. Any help would be appreciated. Thanks.
if regexp_like(v_expr, '^(\d{0,2})+(\.[0-9]{1})?$') t

Hi,
Whenever you have a question, post a little sample data (CREATE TABLE and INSERT statements, relevant columns only) for all the tables involved, and the results you want from that data.
Explain, using specific examples, how you get those results from that data.
Always say what version of Oracle you're using (e.g. 11.2.0.2.0).
See the forum FAQ: https://forums.oracle.com/message/9362002
SammyStyles wrote:
Hi,
I would like to use the REGEX_LIKE to check a number with up to two digits and at least one decimal point:
Ex.
10.1
1.1
1
12
This is what I have so far. Any help would be appreciated. Thanks.
if regexp_like(v_expr, '^(\d{0,2})+(\.[0-9]{1})?$') t
Do you really mean "up to two digits", that is, 3 or more digits is unacceptable? What if there are 0 digits? (0 is less than 2.)
Do you really mean "at least one decimal point", that is, 2, 3, 4 or more decimal points are okay? Include some examples when you post the sample data and results.
It might be more efficient without regular expressions. For example
WHERE   TRANSLATE ( str              -- nothing except digits and dots
                  , 'A.0123456789'
                  , 'A'
                  )   IS NULL
AND     str           LIKE '%.%'     -- at least 1 dot
AND     LENGTH ( REPLACE ( str       -- up to 2 digits
                )     <= 2

HT201363 How do i delete my iphone 4s from my account on apple ID? I change phones and i am not getting regular txt messages because somehow the messages are still sending through imessage which i no longer have.... I deleted my phone number from the acco

I change phones and i am not getting regular txt messages because somehow the messages are still sending through imessage which i no longer have.... I deleted my phone number from the account though as per instructiions from the phone carrier and is still not working...Any help?

Hello jn_andrade,
It sounds like you are no longer getting text messages becuase your Apple ID is still using your phone number for iMessage. I would use these steps to unlink your phone number from your account:
Unlink a phone number
To remove a phone number from an Apple ID, sign out of FaceTime and Messages on your iPhone:
Settings > Messages > Send & Receive. Tap your Apple ID, then tap Sign Out.
Settings > FaceTime. Tap your Apple ID, then tap Sign Out.
This should remove your phone number from other devices using the same Apple ID with FaceTime and Messages. If the phone number is still available on other devices after you sign out of FaceTime and iMessage on the iPhone, you may need to sign out of iMessage and FaceTime on all your devices, then sign in to FaceTime and Messages again on devices you want to use.
Note: If you no longer have access to the iPhone that is using the number you want to remove, reset your Apple ID password.
From: iOS and OS X: Link your phone number and Apple ID for use with FaceTime and iMessage
http://support.apple.com/kb/ht5538
Thank you for using Apple Support Communities.
Regards,
Sterling

How to prevent users from entering '+' or '0' in front of country code in the phone number field?

Requirement:
How can I prevent guest users from entering '+' sign or '0' in front of country-code in the visitor phone number field during self registration?
Few SMS service providers are not looking for '+' sign or '0' or '00' in front of the international phone numbers to trigger the sms. Providing these values in front of country code during self-registration may fail to deliver the sms to recipient.
Solution:
Using a simple regular expression, you can validate the entered phone number during the guest registration.
The below regular expression will help you to validate the phone number and allows to register only when the phone number is not staring with '+' or '0'.
^[1-9][0-9](\d{7}|\d{8}|\d{9}|\d{10}|\d{11}|\d{12})$
It also performs the below validations.
only numbers are allowed.
first digit of the entered phone number should be 1 to 9, so '+' or '0' is not allowed.
numbers from 0 to 9 are allowed from the second digit.
also validates phone number length, the length of the phone number should be 9 to 14.
Configuration:
To add the above regex in the visitor_phone number filed, please navigate to ClearPass Guest >> Configuration >> (Pages)Guest Self-Registration >> select the self-registration page and go to Edit >> Register Page >> Form >> select the filed visitor_phone and set the Validator to " ISRegexMatch" and enter the above regex in the Validator Argument filed as shown below.
Note: Edit the Validation Error as per your requirement.
Verification
Adding the given regex will validate the phone number and prevent the guest user from registering the phone number starts with '+' or '0'.
Please find below the sample outputs for your reference.
Result when phone number starts with '+' or '0'.
Successful registration.

Is this a Mac Preview issue?

Regular Expressions in num-exp

Hello All,
I had a problem on my SRST gateway with num-exp insterting a repeating pattern into my 7-digit dialing when in fallback mode.
For a brief example, the 7digit internal dialing is 21621.. or 21622..
The num-exp statement of 'num-exp 2... 2162...' was not allowing me to 7-digit dial directly from one IP phone to another while in fallback mode.
When I dialed 2162154 the 2162 would hit the num-exp and be expended to 2162162.
I have a work around that uses a voice translation-rule, applied to the call-manager-fallback config that will translate a 7-digit dialed string to the 4 digit dialed string which then hits the 4-digit to 7-digit num-exp and it is working fine.
However, I was wondering if there is a way to use regular expressions in num-exp so that perhaps I can skip the intermediate step of using the translation-rule. Based off my existing translation-rules that are working properly, I figured something like this might work for num-exp:
'num-exp /^2$[12]..$$/ /2162\1/'
But when I try to issue a num-exp with a regular expression I get the following message.
Incorrect format for Number macro pattern
regular expression must be of the form ^((\+)?([0-9#*A-F.]|(\\\*))+(\$)?)$
I have tried a number of different combinations with no success. I always get the same message. The regular expression that I tried first was:
'num-exp ^2... 2162...'
This is when I first saw the "Incorrect format..." message and figured that is must be possible. Is this just a generic warning similar to when you try to use complex regular expressions with the 'translation-rule' command vs. the 'voice translation-rule' command and in reality you cannot use regular expressions in the num-exp command?
Thank you,
Leo

Hi Chris,
Thank you for taking the time to answer my question. It looks like the answer is no, num-exp does not support regular expressions.
I don't insist on using num-exp for this I was just hoping to kill two birds with one stone and possibly skip the intermediate step of translating the 7-digits dial to 4-digits using a translation-rule just to expand from 4-digit to 7 again. This is only an issue while in SRST if a user tries to dial using 7-digits. We have a 7-digit internal dialing scheme and normally my num-exp is just to expand the 4 digits sent from the telco to our 7-digit internal dialing. The problem is that both our prefix and part of our DID range start with 21 so while in SRST if a user tried to dial a 7-digit DN, say 2162154, after they dialed the 4th digit (2162) that pattern would hit the num-exp and get expanded to 2162162. I was hoping to create a num-exp using a regular expression that would only expand a four digit string that begins with a 2 to seven digits and not any string that begins with a 2. This would 1) expand the four digits sent from the telco and 2) not match a seven digit string that begins with a 2 such as 2162154 which may be dialed by a user.
Again, this is only an issue while in SRST and I have a pretty good work around so I'm fine with not being able to use a regular expression as part of my num-exp config. I just thought it would be a cool application of a regular expression if it was possible.
Thanks again for answering my question.
Leo

Phone Number - Regular Expression

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