Delete word from the file

Ram_Kumar_BE · August 28, 2015, 7:32am

My file has contents like below. It is named as output_file.txt

/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-27_tmptcmsaslva2_19142.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-27_tmptcmsaslva2_19142.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-24_tmptcmsaslva2_18208.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-24_tmptcmsaslva2_19142.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-24_tmptcmsaslva2_29115.log

Now I would like to remove, /usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/ from all the records. I need only the file names.

When I tried the commands like below, it is throwing an error. Kindly need some help.

-bash-4.1$ sed 's//usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs///g' output_file.txt > output.txt
sed: -e expression #1, char 8: unknown option to `s'
-bash-4.1$

[/CODE]

sea · August 28, 2015, 7:50am

Why did you open a new thread instead of answering RudiC's question in your other thread: http://www.unix.com/shell-programming-and-scripting/260424-loop-unexpected-token-do-2.html\#post302953450

Thank you

EDIT:
Thanks for cleaning your keyboard, its easier to read now again.

---------- Post updated at 13:50 ---------- Previous update was at 13:48 ----------

And about your issue, try:

while read item
do  echo "${item##*/}"
done<output_file.txt

hth

amit2015 · August 28, 2015, 8:13am

$ cat f1

/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-27_tmptcmsaslva2_19142.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-27_tmptcmsaslva2_19142.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-24_tmptcmsaslva2_18208.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-24_tmptcmsaslva2_19142.log
/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_2015-08-24_tmptcmsaslva2_29115.log

$ cat f1.sh

#!/bin/sh

awk -F"/" '{ print $NF }' f1 > a.out

while read line
do
        program_name=`basename $line .log`
        echo $program_name
done < a.out
rm a.out

$ ./f1.sh

SASApp_STPServer_2015-08-27_tmptcmsaslva2_19142
SASApp_STPServer_2015-08-27_tmptcmsaslva2_19142
SASApp_STPServer_2015-08-24_tmptcmsaslva2_18208
SASApp_STPServer_2015-08-24_tmptcmsaslva2_19142
SASApp_STPServer_2015-08-24_tmptcmsaslva2_29115

Ram_Kumar_BE · August 28, 2015, 8:20am

When I run the script below,

#!/bin/sh
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
echo "${/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/##*/}"
done < output_file.txt

Got error as

line 4: ${/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/##*/}: bad substitution

My requirement was to have only the file name (instead of filename with directories) in output_file.txt

---------- Post updated at 07:20 AM ---------- Previous update was at 07:16 AM ----------

My script location and output file location is /usr/sas/tir/test/loganalysis.

I need to search for a file in /usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/.

Now how to modify your script to include the directories?

RavinderSingh13 · August 28, 2015, 8:21am

ram kumar_be:

When I run the script below,

#!/bin/sh
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
echo "${/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/##*/}"
done < output_file.txt

Got error as

line 4: ${/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/##*/}: bad substitution

My requirement was to have only the file name (instead of filename with directories) in output_file.txt

Hello Ram,

You should put $i as follows in code.

#!/bin/sh
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
       echo "${i##*/}"
done < output_file.txt

Thanks,
R. Singh

Ram_Kumar_BE · August 28, 2015, 8:43am

I got the required output in command prompt. However I would like redirect the result to file (eg.output_file.txt).

I would also request you to explain the echo statement

echo "${i##*/}"

as well.

RavinderSingh13 · August 28, 2015, 9:19am

Hello Ram,

It is an example of parameter expansion, you can go through following.

${parameter#word}${parameter##word}
The word is expanded to produce a pattern just as in filename expansion (see Filename Expansion). If the pattern matches the beginning of the expanded value of parameter, then the result of the expansion is the expanded value of parameter with the shortest matching pattern (the �#� case) or the longest matching pattern (the �##� case) deleted. If parameter is �@� or ��, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with �@� or ��, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.

${parameter%word}${parameter%%word}
The word is expanded to produce a pattern just as in filename expansion. If the pattern matches a trailing portion of the expanded value of parameter, then the result of the expansion is the value of parameter with the shortest matching pattern (the �%� case) or the longest matching pattern (the �%%� case) deleted. If parameter is �@� or ��, the pattern removal operation is applied to each positional parameter in turn, and the expansion is the resultant list. If parameter is an array variable subscripted with �@� or ��, the pattern removal operation is applied to each member of the array in turn, and the expansion is the resultant list.

Thanks,
R. Singh

Corona688 · August 28, 2015, 11:15am

To get output in a file, put it in a file.

while something
do
...
done < file > outfile

Ram_Kumar_BE · August 31, 2015, 5:04am

I've one last question.

With this code I can get a .txt file as required. However, I would like to get a fresh file everyday instead of the appending the the same .txt file.

#!/bin/sh
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
       echo "${i##*/}" < output_file.txt
done

e.g. If I run today I may get 3 records in output_file.txt. When I run tomorrow I should not get the same records along with current day's records in output_file.txt rather I should get only current day's records in output_file.txt

RavinderSingh13 · August 31, 2015, 5:18am

Hello Ram,

There are two ways to complete this request.
I- Keep getting a output file which has particular day's date in it's name itself, so each day files will be unique and different names.

#!/bin/sh
DATE=`date +%y%m%d`
 for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
       echo "${i##*/}" >>  "output_file_"$DATE".txt"
done

II- Each day check for file named output_file.txt and rename it to that's day's date and create a output file with name output_file.txt.

#!/bin/sh
DATE=`date +%y%m%d`
if [[ -f output_file.txt ]]
then
      mv output_file.txt "output_file_"$DATE".txt"
fi
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
       echo "${i##*/}" >>  output_file.txt
done

Thanks,
R. Singh

Ram_Kumar_BE · August 31, 2015, 6:04am

There is other way which I found now:). Just I put a remove command at the beginning of the script like below.

#!/bin/sh
rm output_file.txt
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "1 day ago")"_tmptcmsaslva2_"*".log"
do
    echo "${i##*/}" >> output_file.txt
done

Now I'm looking to capture the script execution in log files (.log)

RudiC · August 31, 2015, 6:05am

ram kumar_be:

.
.
.

#!/bin/sh
for i in "/usr/sas/sas_config/Lev1/SASApp/StoredProcessServer/Logs/SASApp_STPServer_"$(date +%Y-%m-%d --date "4 day ago")"_tmptcmsaslva2_"*".log"
do  
   echo "${i##*/}" < output_file.txt
done

.
.
.

With this code snippet, you won't get any output_file.txt . Replacing < with > , you'd get an output_file.txt with exactly the last .log file name encountered.
Putting the > output_file.txt right after the done , as proposed by Corona688 in post#8, you'll have a new output file every time you run the snippet.

Ram_Kumar_BE · August 31, 2015, 6:14am

Script is working fine. Earlier there was a line feed issue.

Now I was looking to get a script execution log file.