on 07-23-2008 10:39 AM
Hi,
I want to write a validation on Identification no.
The format is like first 3 are numer from 0 to 9 and then there are alphabets any from a to Z.
Help!!
Brad
Hi ,
U can write in the following ways:
1) If there is not a space b/w nos and alphabets
HAS_ALL_CHARS( MID(Id no.,1,3),"0","9") AND HAS_ALL_CHARS( MID(Id no.,3),"a","Z")
2)If space is there and no of alphabets are not constant:
HAS_ALL_CHARS(MID(Id no,1,3),"0","9") AND HAS_ALL_CHARS( MID(Id no,FIND(Id no," ")),"a","Z")
Hope this may help u .
I have not run this expression.Kindly check them if they are working
Rgds
Ankit
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
Hi Braddy,
IF(LEN(Identification no)=8, HAS_ALL_CHARS( LEFT(Identification no,3),0,9) AND (MID(Id no.,4,1) = "-") AND HAS_ALL_CHARS( RIGHT(Identification no,4),"a","Z"),FALSE)
Please award accordingly
Thanks,
Mandeep Saini
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
Hi,
It must work.
IF(LEN(Identification no)=8, HAS_ALL_CHARS( LEFT(Identification no,3),0,9) AND (MID(Id no.,4,1) = "-") AND HAS_ALL_CHARS( RIGHT(Identification no,4),"a","Z")
REWARDS IF USEFUL.....
THANKS,
MANDEEP SAINI
Edited by: Mandeep Saini on Jul 23, 2008 12:30 PM
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
Hi Brad,
U can use the HAS ALLCharacters and MID or Find property of MDM data manager.They are given on page no. 240-245.
I will try top wrie and then give the expression.Just u also try to write.
Rgds
Ankit
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.
User | Count |
---|---|
87 | |
23 | |
11 | |
9 | |
8 | |
5 | |
5 | |
5 | |
5 | |
4 |
You must be a registered user to add a comment. If you've already registered, sign in. Otherwise, register and sign in.