hi there
I am trying to get a value from a remote machine into a local variable. To get this value i want to use awk but im having trouble getting it to run, am i escaping in the right places here and using the right quotes (i must have tried a million combinations :()
# VAR=`ssh server1... (5 Replies)
In awk script,
#!/bin/sh
awk 'BEGIN{i=0;}{i=i+5;}END{print i}' in.txt
vr=0;
vr=$i;
echo "$vr"
How can i assign that value of i in $vr(variable) of shell script? (7 Replies)
Hi,
I have a large flat file from host without delimiter. I'm transforming this file to a csv file using statements like
# Row 03: Customer / field position 3059 +20
WOFABNAM=substr( $0, 3059, 20 );
and deleting the trailing whitespaces before and after with that
sub( /^ +/, "",... (4 Replies)
Hi.
How to change string variable in awk?
for example, I parse with awk script text file named some_name_with_extension.txt
I want to print only some_name in my script
....
varCompName = FILENAME
print varCompName
How to put not all symbols from FILENAME to variable?
thank you
This... (4 Replies)
The following is part of a larger shell script
grep -v "Col1" my_test.log | grep -v "-" | awk '$5 == "Y" {print $1}'
instead of printing, can I set set $1 to a variable that the rest of the shell script can read?
if $5 == Y, I want to call another shell script and pass $1 as a... (2 Replies)
hi i want to find the size of a folder and assign it to a variable and then compare if it is greater than 1 gb.
i am doin this script, but it is throwing error....
#!/bin/ksh
cd . | du -s | size = awk '{print $1}'
if size >= 112000
then
echo size high
fi
ERROR : (4 Replies)
Hello All,
I have csv file, where one of fields is date (yyyy/mm/dd 00:00:00). Using awk I am trying to find all records with date newer/older than specific date. My idea was to compare unix timestamps of both dates:
start=`date +%s -d "$DateStart"`
awk -v start="$start" -v current=`date +%s... (34 Replies)
Hi Experts,
I am trying to get system output to capture inside awk , but not working:
Please advise if this is possible :
I am trying something like this but not working, the output is coming wrong:
echo "" | awk '{d=system ("date") ; print "Current date is:" , d }'
Thanks, (5 Replies)
I have a data file d0 that looks like this:
$cat d0
server1 running -n-cv- 8G 3.1% 1435d 15h
server2 running -n---- 8G 39% 660d 22h
server3 running -n--v- 8G 2.5% 1173d 6h
server4 running -n---- 8G 1.1% 1048d 20h... (2 Replies)
Discussion started by: jake0391S
2 Replies
LEARN ABOUT DEBIAN
shtool-path
SHTOOL-PATH.TMP(1) GNU Portable Shell Tool SHTOOL-PATH.TMP(1)NAME
shtool-path - GNU shtool command dealing with shell path variables
SYNOPSIS
shtool path [-s|--suppress] [-r|--reverse] [-d|--dirname] [-b|--basename] [-m|--magic] [-p|--path path] str [str ...]
DESCRIPTION
This command deals with shell $PATH variables. It can find a program through one or more filenames given by one or more str arguments. It
prints the absolute filesystem path to the program displayed on "stdout" plus an exit code of 0 if it was really found.
OPTIONS
The following command line options are available.
-s, --suppress
Supress output. Useful to only test whether a program exists with the help of the return code.
-r, --reverse
Transform a forward path to a subdirectory into a reverse path.
-d, --dirname
Output the directory name of str.
-b, --basename
Output the base name of str.
-m, --magic
Enable advanced magic search for ""perl"" and ""cpp"".
-p, --path path
Search in path. Default is to search in $PATH.
EXAMPLE
# shell script
awk=`shtool path -p "${PATH}:." gawk nawk awk`
perl=`shtool path -m perl`
cpp=`shtool path -m cpp`
revpath=`shtool path -r path/to/subdir`
HISTORY
The GNU shtool path command was originally written by Ralf S. Engelschall <rse@engelschall.com> in 1998 for Apache. It was later taken
over into GNU shtool.
SEE ALSO shtool(1), which(1).
18-Jul-2008 shtool 2.0.8 SHTOOL-PATH.TMP(1)