It turns out that left or right justification need not matter for the string
id. So, if the oldid is a string variable, this set of syntax will declare a
new string variable, and then compute the new string number from the
existing string:
STRING newid(A6).
COMPUTE newid=STRING(NUMBER(oldid,F6.0),N6).
This somewhat convoluted COMPUTE creates an intermediate numeric value using
the NUMBER function as input to the STRING function. The NUMBER function
reads the string value as a number with a standard F format (to handle
leading or trailing blanks). The STRING function then converts that number
to a string using the N format to create a display representation to
convert.
---Original Message-----
From: ViAnn Beadle [mailto:
[hidden email]]
Sent: Monday, January 07, 2008 6:43 PM
To: '
[hidden email]'
Subject: RE: Adding Leading Zeros To An ID
The reply assumes that the ID variable is a number. Is it a string or a
number? If it is a string, is it displayed left-justified or right-justified
within the data editor? That will determine how to modify the string value.
-----Original Message-----
From: SPSSX(r) Discussion [mailto:
[hidden email]] On Behalf Of
Oliver, Richard
Sent: Monday, January 07, 2008 6:14 PM
To:
[hidden email]
Subject: Re: Adding Leading Zeros To An ID
formats ID (n6).
________________________________
From: SPSSX(r) Discussion on behalf of Don Asay
Sent: Mon 1/7/2008 5:54 PM
To:
[hidden email]
Subject: Adding Leading Zeros To An ID
I have an ID variable that should be 6 charachters long some ranging from
000001 to 999999. Some of the records lost their leading zeros so instead of
having 000015 the computer stored 15, this is not the case for all of the
IDs with leading zeros some are stored properly as 000023 and the very next
one wil be 24 without the leading zeros. I want to create syntax to add the
leading zeros, but I can't figure out how to conditionally prepend the zeros
to just the cases that don't already have them. Any help would be
appreciated.
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