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No change as per the content and behavior.

When you hit space bar it goes to the next character in VI editor. At the end of line, in Solaris it stops and doesn't go to the next line. But in Linux it goes to the first character of the next...

Hi I have a process generating a file in Solaris. Now we have migrated the process to Linux. When we open the file in vi on solaris and hit space bar, it stops after reaching the end of line. But in...

Thanks.

Its erroring out saying

Hi,
I have an input file as follows :

input.txt

abcdTXXqwe
axdfSYYrew
dasgTXXqwt
gtfsTYYwer
gadfSXXerw
gwerSYYTXX

Now I have to get four output files.

output1.txt should have the...

I felt this is the appropriate forum for this post. I don't see a way to delete the earlier post. Anyway the thread has been closed.

Any clue on killing a process without kill command?

Any help...

I want to Kill a process without using kill command as i don't have privileges to kill the process. I know the pid and i am using Linux 2.6.9 OS.

Sorry, posted the question in other forum.

what did u try?

find "$j" -type d -perm +o=w -print

for me this works fine.
Can you paste the contents of the path.txt?

What is the error that u r getting?

PS:U can place 2>error.log in ur cron script to get the error messages logged to a file.

awk -F, '{if(substr($4,1,3)=="923" || substr($4,1,3)=="922") print $0; else print $1","$2","$3","$4","$5","$6","$7",Invalid subscriber."}' file

What is your operating system?
What does the following give?
ls -lrt /dev/null

Check if this works?
ls -ld /usr /foo 2>/directory_where_u_have_permissions_to_write/err.txt

using || for piping? it should be |

1.) awk '{ gsub(/-/,"",$0); for ( i = NF; i > 0; i-- ) printf("%s ",$i); printf("\n");}' <file>

That looks good to me.
Are you trying to replace WORD 8 8 8 with WORD 10 10 10 ??? Then it check the values of the variables, as in the above u r trying to replace all 10 with 12 after WORD on all...

Run ur script using sh -x and check what is the sed command that is getting executed.

This doesn't work for me :


If using variables in sed then use double quotes. Try this :

sed 's/[ ]*([^)]*)[ ]*/ /g'

Duplicate Post : https://www.unix.com/unix-advanced-expert-users/78538-urgent-help-required-awk-sed-help-find-pattern-delete-till-end-line.html

awk '{if($3<"22:30" && $2 <"26") print $0 > "processed.txt"; else print $0 > "nottoprocess.txt";}' inputfile
Just a temporary solution. Actually u need to work on date comparison.

split -l 23 logfile; paste -d' ' xa* | awk '{printf("\n%s ",$3); for(i=1;i<NF;i=i+3) printf("%s ",$i);}' > outfile.

Probably u can work on the header line...

echo $variable | tr ' ' '\n'

Try this, I am not sure whether it works or not as i don't have oracle database.

#! /usr/bin/ksh
sqlplus user/pwd@database<<EOF
! for usr in `cat usernames.txt`
! do
drop user $usr;
!done...