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Top Forums UNIX for Beginners Questions & Answers Issue with emailing format from UNIX Post 302982056 by mrn6430 on Thursday 22nd of September 2016 08:07:43 PM
Old 09-22-2016
Wrench Issue with emailing format from UNIX

Need help as to why when I try email the following it does not come out right:


Code:
DateOt=`date +"%a %b %d %T %Y"`
TimeOt=`date +%H:%M:%S`
interval=`cat $SecDiff| awk '{print $3}'`
PortInReq2=`cat /shared/ftpdir/tools/quelog/logs/quelog.NPS_TO_SMG.log | grep PQI | wc -l`
PortInRes2=`cat /shared/ftpdir/tools/quelog/logs/quelog.SMG_TO_NPS.log | grep PRI | wc -l`
PortOtReq2=`cat /shared/ftpdir/tools/quelog/logs/quelog.SMG_TO_NPS.log | grep PQ2 | wc -l`
PortOtRes2=`cat /shared/ftpdir/tools/quelog/logs/quelog.NPS_TO_SMG.log | grep PR2 | wc -l`
echo "$DateOt " >> $Rptfile
echo "@$TimeOt==> PortIn Requests count = $PortInReq2 " >> $Rptfile
echo "@$TimeOt==> PortIn Response count = $PortInRes2 " >> $Rptfile
echo "@$TimeOt==> PortOut Requests count= $PortOtReq2 " >> $Rptfile
echo "@$TimeOt==> PortOut Respense count= $PortOtRes2 " >> $Rptfile
$scriptpath/sendmail.ksh "SMG PortIn/Out counts Report." $alarmtos $Rptfile

The report file looks fine:

Quote:
@19:00:00==> PortIn Requests count = 13399
@19:00:00==> PortIn Response count = 12945
@19:00:00==> PortOut Requests count= 16022
@19:00:00==> PortOut Respense count= 16744
when emailed it is all in one line !!! as follows:

Quote:
Thu Sep 22 18:59:05 2016
@18:59:05==> PortIn Requests count = 13369 @18:59:05==> PortIn Response count = 12917 @18:59:05==> PortOut Requests count= 16007 @18:59:05==> PortOut Response count= 16726
 

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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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