In the last command, you are missing the last single quote.
But you don't need all the quotes. This works:
The only command you need to execute remotely is the grep'ing of the username. That output then gets locally piped to awk.
How do I execute a command containing a double quote ?
I pass a variable to grep that contains spaces, so I need to quote it,
but it does not work.
#!/usr/bin/ksh
set -x
txt='"next to"'
cmd="grep $txt ~dpearso5/file2"
echo $cmd
$cmd
This is the error I get:
+ grep "next to"... (1 Reply)
Hi
Is there any way we can findout which job/process in unix environment is generating error mails. I am continuously getting it with no subject..
I know the hostname.
And the error in mail - SQL server timed out.
There are hundreds of jobs runing there. How can we find the culprit... (3 Replies)
I'm trying to generate this key but getting an error "file not found"
Here is the command:
# ssh-keygen -t dsa
Generating public/private dsa key pair.
Enter file in which to save the key (//.ssh/id_dsa): /export/home/bartadm/.ssh/id_dsa
Enter passphrase (empty for no passphrase):
Enter... (2 Replies)
Hi,
Trying to change the prompt. I have the following code.
export PS1='
<${USER}@`hostname -s`>$ '
The hostname is not displayed
<abc@`hostname -s`>$ uname -a
AIX xyz 1 6 00F736154C00
<adcwl4h@`hostname -s`>$
If I use double quotes, then the hostname is printed properly but... (3 Replies)
RHEL 6.2/Bash shell
root user will be executing the below script. It switches to oracle user logs in using sqlplus and tries to
run the below UPDATE statement. All the commands after su -c are enclosed in a single quote delimited by semicolon.
The execution has failed because the quotes... (3 Replies)
Hello All,
May i please know how do i ensure my split command would NOT generate incomplete output files like below, the last lines in each file is missing some columns or last line is complete.
split -b 50GB File File_
File_aa
|551|70210203|xxxxxxx|12/22/2010 20:44:58|11/01/2010... (1 Reply)
Please help me to use echo or printf type of command to print some value from variable within double quotes - I want to print the double quote ( " ") also.
I tried
#!/bin/bash
VALUE=some_value
echo '{"value" : "$VALUE"}'
I was expecting the above script would produce ..
{"value" :... (3 Replies)
Discussion started by: atanubanerji
3 Replies
LEARN ABOUT PLAN9
grep
GREP(1) General Commands Manual GREP(1)NAME
grep - search a file for a pattern
SYNOPSIS
grep [ option ... ] pattern [ file ... ]
DESCRIPTION
Grep searches the input files (standard input default) for lines (with newlines excluded) that match the pattern, a regular expression as
defined in regexp(6). Normally, each line matching the pattern is `selected', and each selected line is copied to the standard output.
The options are
-c Print only a count of matching lines.
-h Do not print file name tags (headers) with output lines.
-i Ignore alphabetic case distinctions. The implementation folds into lower case all letters in the pattern and input before interpre-
tation. Matched lines are printed in their original form.
-l (ell) Print the names of files with selected lines; don't print the lines.
-L Print the names of files with no selected lines; the converse of -l.
-n Mark each printed line with its line number counted in its file.
-s Produce no output, but return status.
-v Reverse: print lines that do not match the pattern.
Output lines are tagged by file name when there is more than one input file. (To force this tagging, include /dev/null as a file name
argument.)
Care should be taken when using the shell metacharacters $*[^|()= and newline in pattern; it is safest to enclose the entire expression in
single quotes '...'.
SOURCE
/sys/src/cmd/grep.c
SEE ALSO ed(1), awk(1), sed(1), sam(1), regexp(6)DIAGNOSTICS
Exit status is null if any lines are selected, or non-null when no lines are selected or an error occurs.
GREP(1)