Shell Programming and Scripting

I'm new to shell programming, and am having a problem in a (Korn) shell program, which boils down to this:
The program reads a record from an input file and then uses a series of
"cut" commands to break the record into parts and assign the parts to
variables. There are no delimiters in the record. Here's a sequence of
three records from the input file:

S1830031HN106P
S1830032HN113XTA
S1830033MEI A

And here's the section of code that breaks the record into various parts,
with "InputLine" representing the record:

Code:

var1=`echo $InputLine | cut -c 1-5`
var2=`echo $InputLine | cut -c 1`
var3=`echo $InputLine | cut -c 2-5`
var4=`echo $InputLine | cut -c 6-8`
var5=`echo $InputLine | cut -c 9-13`

So the code should assign the characters in columns 9-13 to the var5
variable. The problem is that when the third record above ("S1830033MEI
A") is processed, the var5 should be assigned "MEI ", but ends up getting
assigned "MEI A". In the record, there are four spaces between the "I" and the "A", but the cut command seems to compress the consecutive whitespaces into one space, which then makes the "A" the fifth character in the variable. Is there a way to force the cut command to count each whitespace as a column, or some other simple way to assign exactly those characters in columns 9-13 to var5? I tried using various options with the cut command, but wasn't able to get the results I needed. I appreciate any help anyone can give.


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Hi.

cut does count each whitespace as a character (column, as you put it).

Here's what I get using only cut, based on your input:

Code:

$ cut -c9-13 file1
HN106
HN113
MEI

The problem is with your echo, i.e.

Code:

var5=`echo $InputLine | cut -c 9-13`

Should be:

Code:

var5=`echo "$InputLine" | cut -c 9-13`

to preserve the whitespace.

Thanks a lot, scottn. That's exactly what I needed.