Log File Writing and Reading

valluvan · #1 (**permalink**) 11-25-2008

Hi all,
I have the following shell script code which tries to sftp and writes the log into the log file.

Code:

TestConnection ()
{
  echo 'Connection to ' $DESTUSERNAME@$DESTHOSTNAME
  $SETDEBUG
  if [[ -a $SCRIPT ]]; then rm $SCRIPT ; fi
  touch $SCRIPT
  echo "cd" $REMOTEDIR >> $SCRIPT
  echo "quit" >> $SCRIPT
  
  chmod 700 $SCRIPT
  sftp -b $SCRIPT $DESTUSERNAME@$DESTHOSTNAME > $SESSIONLOG
  
  echo '**** FTP log reproduced below:'
  echo '-----------------------------------------------------------'
  cat $SESSIONLOG
  echo '-----------------------------------------------------------'
}

the code executes well and when i manually vi the log file i can see some data written into it...

Whereas the portion of script(cat $SESSIONLOG) does nothing.It prints nothing.Does anyone know what could be the problem?.

I am new to unix and iam thinking is it because iam trying to cat from a file which might be still Open?If yes how do i get around it?

Please help.

Regards

mk1216 · #2 (**permalink**) 11-25-2008

try "echo $SESSIONLOG" before "cat" command and see if the variable has a proper value .

zaxxon · #3 (**permalink**) 11-25-2008

If $SESSIONLOG was empty or had nonsense inside, cat would tell something like "file not found".
You should also write stderr out to your log like:

Code:

sftp -b $SCRIPT $DESTUSERNAME@$DESTHOSTNAME > $SESSIONLOG 2>&1

If it is still empty, maybe put a -v for verboste to the sftp.

valluvan · #4 (**permalink**) 11-25-2008

Quote:

Originally Posted by zaxxon

If $SESSIONLOG was empty or had nonsense inside, cat would tell something like "file not found".
You should also write stderr out to your log like:

Code:

sftp -b $SCRIPT $DESTUSERNAME@$DESTHOSTNAME > $SESSIONLOG 2>&1

If it is still empty, maybe put a -v for verboste to the sftp.

Thats correct Zaxxon.
Adding
sftp -b $SCRIPT $DESTUSERNAME@$DESTHOSTNAME > $SESSIONLOG 2>&1
did the trick....thanks a lot guys

valluvan · #5 (**permalink**) 11-25-2008

One more question:

I want to compare the value of the variable against a string and i am using the below code,but doe's'nt seem to do it...is there anything wrong with it??

Code:

if echo "$PRECHECK" | grep 'FALSE'; 
    then
      echo "File Ftped Successfully: " $f
      mv $f "FTPED_"$f
    fi

Thanks

mk1216 · #6 (**permalink**) 11-25-2008

if [ ! -z $PRECHECK ] && [ $PRECHECK == "FALSE" ]
then
echo "File Ftped Successfully: " $f
mv $f "FTPED_"$f
fi

valluvan · #7 (**permalink**) 11-25-2008

Quote:

Originally Posted by mk1216

if [ ! -z $PRECHECK ] && [ $PRECHECK == "FALSE" ]
then
echo "File Ftped Successfully: " $f
mv $f "FTPED_"$f
fi

Thank you very much.It worked...
I assume [ ! -z $PRECHECK ] means does the variable precheck exist??

One more thing is that I need to email the file log contents to a user.
I am using the code below:

Code:

cat $SESSIONLOG|mailx -s "Alert created on $DATERUN : Transaction date is $DATERUN : Interface failure from RCUK to BottomLine " $EMAILADD

Is it correct? I don't seem to get any emails....

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