Shell Programming and Scripting

Gooday
I have an argument string that contains 15 arguments.
The first 14 arguments are easy to handle because they are separated by spaces

Code:

ARG14=`echo ${ARGSTRING} | awk '{print $14}'`

The last argument is a text that may be empty or contain spaces. So any ideas on how I get the last argument into the variable ARG15? ie delete from $ARG14 to the beginning of line will be greately appreciated.

regards

Zagga

Moderator's Comments:

Please use code tags

Something like this?

Code:

ARG15=`echo ${ARGSTRING} | awk '{sub(".*" $14 FS,x)}1'`

Thanks very much Franklin52, your solution works fine while arguments 14 and 15 are populated. Unfortunately I omitted to mention that these fields are optional.
After testing: if Arg14 and Arg15 are null, then Arg15 contains the same value as Arg13.
If Arg15 only is null, the it recuperates the value of Arg14.
I figured I would add tests on each field to set to "NONE" if null, unless you have a better trick up your sleeve?

Regards

Zagga

Do you have 2 spaces (field separators) at the end of the line if the last 2 fields are empty?

NO in fact I have 56 spaces at the end of the record. But the good news is the last two fields are fixed length which is not the case of all the preceding fields. Arg14= 4 cars Arg15=50 cars + 2 FS=56.

So I need to get from the end of $Arg13 to the EoL into a new variable and then I should be able to cut Arg14 and 15 from their positions.

Can you do a cut backwards from the end of line?

Any ideas welcome?

Zagga

Try this:

Code:

Arg14=`echo ${ARGSTRING} | awk '{print substr($14, 1 ,4)}'`

Arg15=`echo ${ARGSTRING} | awk '{print substr($14, 5 ,50)}'`

Sorry Franklin52 that doesn't seem to work:

Case 1:
if both $14 and 15 are null ie full of spaces.
I get Arg14=" " and Arg15 has nothing.

Case 2:
if $14="3333" and $15 is null
I get Arg14="3333" and Arg15 has nothing.

Case 3:
if $14 is null ie " " and $15 contains "12345678901234567890123456789012345678901234567890"
I get Arg14="1234" and Arg15="567890......"

Case 4:
If $14="9999" and $15="one two three four five six seven ....."
I get Arg14="9999" and Arg15=

I think I need to create an intermediary variable and get everything from the end of $13 into it.

I'm not sure I've made myself clear.

Thanks in advance

Zagga

---------- Post updated at 11:51 AM ---------- Previous update was at 11:39 AM ----------

The reply I just posted may have been a little misleading as I quoted ARG14 but not ARG15. so output looked different. As it turns out Case 1 and 2 give expected results. Case 3 and 4 don't.
Zagga