UNIX for Dummies Questions & Answers

Dear Friends,

Am very much new to UNIX and this is my first task in UNIX. Can you pls help me with the below problem:

i want to get the latest file from unix to mainframes. I did the following

Code:

remote ls -t $AMR/data01/extract/monthly/source/- 
AMR_D*""_REC_STAT.dat - 
> v1 
remote head -1 v1 > stat

But when I receive stat as

Code:

set receive mode char 
receive - 
"!dcp ""$stat"" - - 
2>/tmp/stat.stderr" - 
DD:O1STAT

I receive nothing.
But when I hardcode the file instead of $stat i was able to receive 535 bytes.
If I give only stat instead of $stat, I get the file with full path as output.
Am I missing something here?

Can you pls help?
Appreciate your response on this.

Regards
Amar

Moderator's Comments:

Please use [code] and [/code] tags when posting code, data or logs etc. to preserve formatting and enhance readability, thanks.

Please tell us what type of mainframe OS you are using. Apart from those few Unix instructions, I can't see anything familiar.

Not sure if it helps you, but to list the latest file in a directory, you could issue

Code:

ls -t1| head -1

Since you issued the ls without the 1, you could have multiple files in that output, since it will not list 1 file per line.

From your post I gather that you need to run stat on the most recent file in the directory which matches your pattern.

If I interpreted you correctly what you need to do is either use xargs, or substitute the value of the ls | head in the arg list of stat using backticks or command substitution.
eg.

Code:

stat `ls -t $AMR/data01/extract/monthly/source/-AMR_D*""_REC_STAT.dat | head -1`
OR
stat $(ls -t $AMR/data01/extract/monthly/source/-AMR_D*""_REC_STAT.dat | head -1)
OR
ls -t $AMR/data01/extract/monthly/source/-AMR_D*""_REC_STAT.dat | head -1| xargs stat