Why Unix bash shell script does not run my SAS program?

In my Unix bash shell script I am trying to run my SAS program "sasprog1.sas" and then search within the Unix directory for a file "sasprog1.LOG" and then checking for an ERROR within "sasprog1.LOG" file and then open another file ("file1.txt") and then writing some messages to "file1.txt" file?

Please note: "file1.txt" and "sasprog1.LOG" may or may not currently exist in the Unix directory.

Please see why my Unix bash shell script does not run/execute my SAS program. What am I doing wrong here?
Any examples or comments will be very appriciated!


mydir=/start/dir
plog=/start/log
filename=sasprog1.log
search=ERROR
outfile=file1.txt
sas=/products/SAS/SASFoundation/9.2/sas

$sas <$mydir/sasprog1.sas> $plog/$filename     # is this a proper way to run a SAS program and then create SAS log??
find $plog -type f -name $filename |while read file
  do
    RESULT=$(grep $search $file)         # is this correct?      
       if [[ ! -z $RESULT ]]
         then
            echo "Error(s) in $file: $RESULT" >> $outfile
     fi
  done
exit

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Curently my script does not even run/execute sasprog1.sas
labradorchikAsked:
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DavidPresidentCommented:
Assuming you would ordinarily execute:
/products/SAS/SASFoundation/9.2/sas  > /star/log/sasprog1.log

then you would enter:

$sas $mydir$sasprog1.sas > $plog/$filename


Then the RESULT line nedes to be

RESULT=`grep $search $file`
if [ "$RESULT" == "" ] ; then
    echo error ...
fi



You should also add a PATH line at beginning so it can find grep
PATH=/usr/bin:/bin:/sbin    # make sure the path where grep is installed on your host is in the list

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DavidPresidentCommented:
you can add a "set -x"  at beginning of shell script and it will give a debug dump of exactly what it is doing as it executes.  this will tell you values of all variables as it runs.  Best way to debug a script
labradorchikAuthor Commented:
Thank you very much for your comments, dlethe!!
What do you mean by: "Assuming you would ordinarily execute:
/products/SAS/SASFoundation/9.2/sas  > /star/log/sasprog1.log

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"?

I am assignning and openning SAS software with the following line:
 
sas=/products/SAS/SASFoundation/9.2/sas

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and then will be running SAS program and creating my sasprog1.log file with the following line:
 
$sas $mydir/sasprog1.sas> $plog/$filename   

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So, is this correct?

mydir=/start/dir
plog=/start/log
filename=sasprog1.log
search=ERROR
outfile=file1.txt
sas=/products/SAS/SASFoundation/9.2/sas

$sas $mydir/sasprog1.sas> $plog/$filename    
find $plog -type f -name $filename |while read file
  do
    RESULT= 'grep $search $file'         
       if [["$RESULT" ==""]];
         then
            echo "Error(s) in $file: $RESULT" >> $outfile
     fi
  done
exit

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labradorchikAuthor Commented:
Anyone else would like to comment on the obove Unix script?
DavidPresidentCommented:
no,  you need `   not  the single quote.

many ways to do the if, depending on exact version of shell, but most portable would be

if [ "$RESULT" == "" ]
then

fi
Note the whitespace

I  have no idea if the find $plog ... line will do what you want, that is a function of the output syntax of the sas command.
TintinCommented:
There's no need for the find, assuming you don't have multiple sasprog1.log in sub-directories.

That being the case, you can simply make the script

#!/bin/bash
mydir=/start/dir
plog=/start/log
filename=sasprog1.log
search=ERROR
outfile=file1.txt
sas=/products/SAS/SASFoundation/9.2/sas

$sas $mydir/sasprog1.sas >$plog/$filename

result=$(grep $search $plog/$filename) && echo "Error(s) in $plog/$filename: $result" >>$outfile

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labradorchikAuthor Commented:
Great!! Thank you very much for your comments, dlethe!!
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