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Top Forums Shell Programming and Scripting ps command showing full code of running script Post 302647499 by rajeevra on Monday 28th of May 2012 08:41:53 AM
Old 05-28-2012
The rsync command I am using is :
rsync -Sz --append --bwlimit= --progress --verbose -e ssh /tmp/Image-15.img root@localhost11::imgxfer/ 2>error.log | /tmp/rsync-progress.sh localhost11 >process.log

when I enter grep -ef | grep rsync I got the following lines which shows the the contents of script.

root 5774 5771 0 09:05 pts/0 00:00:00 sed --unbuffer ? # Get rid .............
root 5775 5771 0 09:05 pts/0 00:00:00 expect -- /opt/IBM/ITM/lx8266/53/bin/53-unbuffer -p awk ? BEGIN {.........
root 5777 5775 0 09:05 pts/7 00:00:00 awk ? BEGIN { FS = " "; startTi..........

The above all programs has been written into a single script "rsync-progress.sh" which is called as :
"$PROGDIR/53-unbuffer" -p tr '\r' '\n' | sed --unbuffer "$sedProgram" | "$PROGDIR/53-unbuffer" -p awk "$awkProgram" >> "$logFile"

Here, "$sedProgram" is being replaced by all the code and displayed on ps -ef. same happens with awk and expect scripts. How could I supress the printing of content of varible in ps -ef?

Last edited by rajeevra; 05-28-2012 at 11:27 AM..
 

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UNBUFFER(1)						      General Commands Manual						       UNBUFFER(1)

NAME
unbuffer - unbuffer output SYNOPSIS
unbuffer program [ args ] INTRODUCTION
unbuffer disables the output buffering that occurs when program output is redirected from non-interactive programs. For example, suppose you are watching the output from a fifo by running it through od and then more. od -c /tmp/fifo | more You will not see anything until a full page of output has been produced. You can disable this automatic buffering as follows: unbuffer od -c /tmp/fifo | more Normally, unbuffer does not read from stdin. This simplifies use of unbuffer in some situations. To use unbuffer in a pipeline, use the -p flag. Example: process1 | unbuffer -p process2 | process3 CAVEATS
unbuffer -p may appear to work incorrectly if a process feeding input to unbuffer exits. Consider: process1 | unbuffer -p process2 | process3 If process1 exits, process2 may not yet have finished. It is impossible for unbuffer to know long to wait for process2 and process2 may not ever finish, for example, if it is a filter. For expediency, unbuffer simply exits when it encounters an EOF from either its input or process2. In order to have a version of unbuffer that worked in all situations, an oracle would be necessary. If you want an application-specific solution, workarounds or hand-coded Expect may be more suitable. For example, the following example shows how to allow grep to finish pro- cessing when the cat before it finishes first. Using cat to feed grep would never require unbuffer in real life. It is merely a place- holder for some imaginary process that may or may not finish. Similarly, the final cat at the end of the pipeline is also a placeholder for another process. $ cat /tmp/abcdef.log | grep abc | cat abcdef xxxabc defxxx $ cat /tmp/abcdef.log | unbuffer grep abc | cat $ (cat /tmp/abcdef.log ; sleep 1) | unbuffer grep abc | cat abcdef xxxabc defxxx $ BUGS
The man page is longer than the program. SEE ALSO
"Exploring Expect: A Tcl-Based Toolkit for Automating Interactive Programs" by Don Libes, O'Reilly and Associates, January 1995. AUTHOR
Don Libes, National Institute of Standards and Technology 1 June 1994 UNBUFFER(1)
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