i really new in linux and just heard about shell scripting couple days ago..
i did exercises on linux in online tutorial but as a beginner, i'm facing problems in developing the script as there are errors that sometimes i dun have any idea on how to solve it.What i'm doing now is not homework, but i had been assigned to develop some system in linux.
i appreciate if there are people that willing to guide me on this matter.
my problem currently is on sub-string matter, where i need to read the line from file/directory and based on the line retrieved,i need to seperate the information in the line and assign as variable..
below are my script:
(the script reside in test.sh)
Code:
!/bin/sh
dir=/home/user/input/test.log #(this is the input file)
msg=`tail -1 $dir`
#(it will take only the last line as log file will be updated every day)
echo "$msg"
thus, the output something like this:
file_name successfully ran on Mon Jul 12 23:15:00 SST 2009.
now, what i need to do is to extract certain information from that line which is the name, date, time and the status
the desired output is:
name date time status
file_name Mon Jul 12 23:15:00 success
so,my next step is to identify the sub-string and assigned as variable first in order to parse the info and output it.
thus,my script is:
Code:
!/bin/sh
dir=/home/user/input/test.log #(this is the input file)
msg=`tail -1 $dir`
#(it will take only the last line as log file will be updated every day)
sub= $msg
name=${msg:0:8}
echo "$msg"
echo "name"
but i received error message which is
Code:
./test.sh[10]: name=${msg:0:8}: The specified substitution is not valid for this command.
guide me please..
Last edited by pludi; 07-14-2010 at 03:55 AM..
Reason: code tags, please...
is not valid POSIX sh syntax. It is valid in bash or ksh93. If you want that you need to change the first line to the desired shell. If you want to use POSIX shell, have a look here to see what is or is not allowed: Shell & Utilities: Detailed ToC
ok...i've edited my script and i got my desired output..
but the position of the string is always changes..Thus the script is no longer efficient.It may output wrong result. i'm thinking of creating a script that can read the exact sub-string that i wanted...let say from the whole line, i want to capture only "my file name"..instead of i have to set the position.
maybe in the log file, the last line has long/many words,so i just wanted to capture certain word in order for me to assign it as variable..
i've tried to search and did some try an error but still doesn't work.. any idea?
below are the scripts that i've edited based on my understanding:
Code:
#!/bin/sh
dir=/home/devusr02/nadiah/input/file_name.log
msg=`tail -1 $dir`
name=${msg%% Failed *}
status=Failed
if [ ${#name} -eq ${#msg} ]
then
# length of string is same as original message
# ie replace matched nothing - assume success string
# status=success
name=${msg%% successfully *}
status=Success
fi
datetime=${msg##*ran on } # remove "ran on " and all proceeding
datetime=${datetime%% SST *} # remove " SST " and all following
time=${datetime##* } # remove last space and all proceeding
date=${datetime% *} # remove last space and all following
echo "$msg"
echo "$name"
echo "$status"
the script output:
if success:
Code:
report_name successfully ran on Tue Jul 11 23:15:00 SST 2010.
report_name
Success
00:15:00
if failed:
Code:
ERROR: report_name is Failed on Tue Jul 11 23:15:00 SST 2010. ErrorCode : 500
ERROR: report_name is
Failed
00:15:00
BUT the output for name should be only "report_name" instead of "ERROR: report_name is" if failed
Moderator's Comments:
Use code tags, please...
---------- Post updated at 04:57 AM ---------- Previous update was at 03:51 AM ----------
one more question..
i wonder if is it possible if the date can be set as : 11 Jul 2010
based on script below,
Code:
datetime=${msg##*ran on } # remove "ran on " and all proceeding
datetime=${datetime%% SST *} # remove " SST " and all following
time=${datetime##* } # remove last space and all proceeding
date=${datetime% *} # remove last space and all following
the output is : "Tue Jul 11"
is there any other easier way to make the date as 11 Jul 2010 instead of i need to do step by step and combine the all the variable for example : echo "$day" "$month" "$year" ??
in addition, can someone guide me on how to eliminate day in front of the month " Tue Jul 11" as the day is changing from monday-sunday..
thanks.
Last edited by Scott; 07-15-2010 at 07:01 AM..
Reason: Code tags
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07-15-2010
