UNIX shell script while loop - shell

I am trying to write a script to track the progress of file change.
I have the following till now:
#!/bin/sh
old=‘ls -l /tmp/file‘
new=‘ls -l /tmp/file‘
while [ "$old" = "$new" ]
do
new=‘ls -l /tmp/file‘
done
echo "The file has been changed"
The above program when run gives the message:
new: command not found
Can someone please help.
Thanks

You probably have space around =.
In shell, when you assign the values you cannot put space around =:
MY_VAR = "my value" # this is wrong!
Shell will think: "call MY_VAR with arguments: ('=', 'my value') ", but wait! I don't know the command "MY_VAR"!
You need to do it this way:
MY_VAR="my value" # this is OK!
BTW, consider using inotifywatch command. Here's example:
inotifywatch -v -e access -e modify -t 60 -r /file/to/watch

Related

How to use grep in if statement in shell

I have this shall script, basically I need to print all <sst> result only when <pr> is found.
Probably I have some syntax error so when I run the script I receive a message "Display all possibilities" basically the grep does not work.
Could you please help me out to understand what is the problem here?
declare -a arr=(
"123"
"345"
)
for i in "${arr[#]}"
do
echo "$i"
if [grep -q "<pr>$i</pr>" ./archiv]
then
grep -r "<sst>" ./archiv
fi
done
There is very likely no command named [grep. Drop the [
if grep -q "<pr>$i</pr>" ./archiv; then ...
[ is not and has never been a part of the shell grammar. It is a command, just like echo or test or grep. The value returned by that command is used to determine whether or not to execute the clause of the if statement.

Shell Script Variable Scope with commnd

I came across an interesting thing in Shell Scripting and not 100% sure why the behaviour is like this
I tried the below script:
#!/bin/sh
CMD="curl -XGET http://../endpoint";
var1=eval $CMD | sed -e 's/find/replace/g';
echo $var1; # Output: printed the value on this line
echo $var1; # Output: blank/no data printed (Why it is blank?)
I had to change the command in variable enclosing with back-tick ` to print the variable as many time as I wanted.
CMD="curl -XGET http://../endpoint";
var1=`eval $CMD | sed -e 's/find/replace/g'`;
echo $var1; # Output: printed the value on this line
echo $var1; # Output: printed the value on this line
Why I have to surround my command with ` to assign it's o/p to the variable in subsequent variable usage?
I have a feeling that it has something to do with the variable-command scope.
Shedding light on my understanding will be appreciated!
UPDATE:
I tried the below command and it is working in my env.
#!/bin/sh
CMD="curl -XGET http://www.google.com/";
var1=eval $CMD | sed -e 's/find/replace/g';
echo $var1; # Output: printed the value on this line
echo "######";
echo $var1; # Output: blank/no data printed (Why it is blank?)
sh/bash allows you to run a command with a variable in its environment, without permanently modifying the variable in the shell. This is great, because you can e.g. run a command in a certain language just one time without having to change your entire user's or system's language:
$ LC_ALL=en_US.utf8 ls foo
ls: cannot access foo: No such file or directory
$ LC_ALL=nb_NO.utf8 ls foo
ls: cannot access foo: Ingen slik fil eller filkatalog
However, this means that when you try to do
var=this is some command
you're trigger this syntax.
It means "run the command is a command and tell it that the variable var is set to this"
It does not assign "this is my string" to the variable, and it definitely does not evaluate "this is a string" as a command, and then assign its output to var.
Given this, we can look at what actually happened:
CMD="curl -XGET http://../endpoint";
var1=eval $CMD | sed -e 's/find/replace/g'; # No assignment, output to screen
echo $var1; # Output: blank/no data printed
echo $var1; # Output: blank/no data printed
There is no scope issue and no inconsistency: the variable is never assigned, and is never written by an echo statement.
var=`some command` (or preferably, var=$(some command)) works because this is valid syntax to assign output from a program to a variable.
The first example isn't doing what you think it is.
Neither echo is printing anything. Make them echo "[$var1]" to see that.
You need the backticks to run the command and capture its output.
Your first attempt was running the $CMD | sed -e 's/find/replace/g'; pipeline with the environment of $CMD containing var1 set to a value of eval.
You also shouldn't be putting commands inside strings (or using eval in general). See http://mywiki.wooledge.org/BashFAQ/001 for more on why.

Reusing output from last command in Bash

Is the output of a Bash command stored in any register? E.g. something similar to $? capturing the output instead of the exit status.
I could assign the output to a variable with:
output=$(command)
but that's more typing...
You can use $(!!)
to recompute (not re-use) the output of the last command.
The !! on its own executes the last command.
$ echo pierre
pierre
$ echo my name is $(!!)
echo my name is $(echo pierre)
my name is pierre
The answer is no. Bash doesn't allocate any output to any parameter or any block on its memory. Also, you are only allowed to access Bash by its allowed interface operations. Bash's private data is not accessible unless you hack it.
Very Simple Solution
One that I've used for years.
Script (add to your .bashrc or .bash_profile)
# capture the output of a command so it can be retrieved with ret
cap () { tee /tmp/capture.out; }
# return the output of the most recent command that was captured by cap
ret () { cat /tmp/capture.out; }
Usage
$ find . -name 'filename' | cap
/path/to/filename
$ ret
/path/to/filename
I tend to add | cap to the end of all of my commands. This way when I find I want to do text processing on the output of a slow running command I can always retrieve it with ret.
If you are on mac, and don't mind storing your output in the clipboard instead of writing to a variable, you can use pbcopy and pbpaste as a workaround.
For example, instead of doing this to find a file and diff its contents with another file:
$ find app -name 'one.php'
/var/bar/app/one.php
$ diff /var/bar/app/one.php /var/bar/two.php
You could do this:
$ find app -name 'one.php' | pbcopy
$ diff $(pbpaste) /var/bar/two.php
The string /var/bar/app/one.php is in the clipboard when you run the first command.
By the way, pb in pbcopy and pbpaste stand for pasteboard, a synonym for clipboard.
One way of doing that is by using trap DEBUG:
f() { bash -c "$BASH_COMMAND" >& /tmp/out.log; }
trap 'f' DEBUG
Now most recently executed command's stdout and stderr will be available in /tmp/out.log
Only downside is that it will execute a command twice: once to redirect output and error to /tmp/out.log and once normally. Probably there is some way to prevent this behavior as well.
Inspired by anubhava's answer, which I think is not actually acceptable as it runs each command twice.
save_output() {
exec 1>&3
{ [ -f /tmp/current ] && mv /tmp/current /tmp/last; }
exec > >(tee /tmp/current)
}
exec 3>&1
trap save_output DEBUG
This way the output of last command is in /tmp/last and the command is not called twice.
Yeah, why type extra lines each time; agreed.
You can redirect the returned from a command to input by pipeline, but redirecting printed output to input (1>&0) is nope, at least not for multiple line outputs.
Also you won't want to write a function again and again in each file for the same. So let's try something else.
A simple workaround would be to use printf function to store values in a variable.
printf -v myoutput "`cmd`"
such as
printf -v var "`echo ok;
echo fine;
echo thankyou`"
echo "$var" # don't forget the backquotes and quotes in either command.
Another customizable general solution (I myself use) for running the desired command only once and getting multi-line printed output of the command in an array variable line-by-line.
If you are not exporting the files anywhere and intend to use it locally only, you can have Terminal set-up the function declaration. You have to add the function in ~/.bashrc file or in ~/.profile file. In second case, you need to enable Run command as login shell from Edit>Preferences>yourProfile>Command.
Make a simple function, say:
get_prev() # preferably pass the commands in quotes. Single commands might still work without.
{
# option 1: create an executable with the command(s) and run it
#echo $* > /tmp/exe
#bash /tmp/exe > /tmp/out
# option 2: if your command is single command (no-pipe, no semi-colons), still it may not run correct in some exceptions.
#echo `"$*"` > /tmp/out
# option 3: (I actually used below)
eval "$*" > /tmp/out # or simply "$*" > /tmp/out
# return the command(s) outputs line by line
IFS=$(echo -en "\n\b")
arr=()
exec 3</tmp/out
while read -u 3 -r line
do
arr+=($line)
echo $line
done
exec 3<&-
}
So what we did in option 1 was print the whole command to a temporary file /tmp/exe and run it and save the output to another file /tmp/out and then read the contents of the /tmp/out file line-by-line to an array.
Similar in options 2 and 3, except that the commands were exectuted as such, without writing to an executable to be run.
In main script:
#run your command:
cmd="echo hey ya; echo hey hi; printf `expr 10 + 10`'\n' ; printf $((10 + 20))'\n'"
get_prev $cmd
#or simply
get_prev "echo hey ya; echo hey hi; printf `expr 10 + 10`'\n' ; printf $((10 + 20))'\n'"
Now, bash saves the variable even outside previous scope, so the arr variable created in get_prev function is accessible even outside the function in the main script:
#get previous command outputs in arr
for((i=0; i<${#arr[#]}; i++))
do
echo ${arr[i]}
done
#if you're sure that your output won't have escape sequences you bother about, you may simply print the array
printf "${arr[*]}\n"
Edit:
I use the following code in my implementation:
get_prev()
{
usage()
{
echo "Usage: alphabet [ -h | --help ]
[ -s | --sep SEP ]
[ -v | --var VAR ] \"command\""
}
ARGS=$(getopt -a -n alphabet -o hs:v: --long help,sep:,var: -- "$#")
if [ $? -ne 0 ]; then usage; return 2; fi
eval set -- $ARGS
local var="arr"
IFS=$(echo -en '\n\b')
for arg in $*
do
case $arg in
-h|--help)
usage
echo " -h, --help : opens this help"
echo " -s, --sep : specify the separator, newline by default"
echo " -v, --var : variable name to put result into, arr by default"
echo " command : command to execute. Enclose in quotes if multiple lines or pipelines are used."
shift
return 0
;;
-s|--sep)
shift
IFS=$(echo -en $1)
shift
;;
-v|--var)
shift
var=$1
shift
;;
-|--)
shift
;;
*)
cmd=$option
;;
esac
done
if [ ${#} -eq 0 ]; then usage; return 1; fi
ERROR=$( { eval "$*" > /tmp/out; } 2>&1 )
if [ $ERROR ]; then echo $ERROR; return 1; fi
local a=()
exec 3</tmp/out
while read -u 3 -r line
do
a+=($line)
done
exec 3<&-
eval $var=\(\${a[#]}\)
print_arr $var # comment this to suppress output
}
print()
{
eval echo \${$1[#]}
}
print_arr()
{
eval printf "%s\\\n" "\${$1[#]}"
}
Ive been using this to print space-separated outputs of multiple/pipelined/both commands as line separated:
get_prev -s " " -v myarr "cmd1 | cmd2; cmd3 | cmd4"
For example:
get_prev -s ' ' -v myarr whereis python # or "whereis python"
# can also be achieved (in this case) by
whereis python | tr ' ' '\n'
Now tr command is useful at other places as well, such as
echo $PATH | tr ':' '\n'
But for multiple/piped commands... you know now. :)
-Himanshu
Like konsolebox said, you'd have to hack into bash itself. Here is a quite good example on how one might achieve this. The stderred repository (actually meant for coloring stdout) gives instructions on how to build it.
I gave it a try: Defining some new file descriptor inside .bashrc like
exec 41>/tmp/my_console_log
(number is arbitrary) and modify stderred.c accordingly so that content also gets written to fd 41. It kind of worked, but contains loads of NUL bytes, weird formattings and is basically binary data, not readable. Maybe someone with good understandings of C could try that out.
If so, everything needed to get the last printed line is tail -n 1 [logfile].
Not sure exactly what you're needing this for, so this answer may not be relevant. You can always save the output of a command: netstat >> output.txt, but I don't think that's what you're looking for.
There are of course programming options though; you could simply get a program to read the text file above after that command is run and associate it with a variable, and in Ruby, my language of choice, you can create a variable out of command output using 'backticks':
output = `ls` #(this is a comment) create variable out of command
if output.include? "Downloads" #if statement to see if command includes 'Downloads' folder
print "there appears to be a folder named downloads in this directory."
else
print "there is no directory called downloads in this file."
end
Stick this in a .rb file and run it: ruby file.rb and it will create a variable out of the command and allow you to manipulate it.
If you don't want to recompute the previous command you can create a macro that scans the current terminal buffer, tries to guess the -supposed- output of the last command, copies it to the clipboard and finally types it to the terminal.
It can be used for simple commands that return a single line of output (tested on Ubuntu 18.04 with gnome-terminal).
Install the following tools: xdootool, xclip , ruby
In gnome-terminal go to Preferences -> Shortcuts -> Select all and set it to Ctrl+shift+a.
Create the following ruby script:
cat >${HOME}/parse.rb <<EOF
#!/usr/bin/ruby
stdin = STDIN.read
d = stdin.split(/\n/)
e = d.reverse
f = e.drop_while { |item| item == "" }
g = f.drop_while { |item| item.start_with? "${USER}#" }
h = g[0]
print h
EOF
In the keyboard settings add the following keyboard shortcut:
bash -c '/bin/sleep 0.3 ; xdotool key ctrl+shift+a ; xdotool key ctrl+shift+c ; ( (xclip -out | ${HOME}/parse.rb ) > /tmp/clipboard ) ; (cat /tmp/clipboard | xclip -sel clip ) ; xdotool key ctrl+shift+v '
The above shortcut:
copies the current terminal buffer to the clipboard
extracts the output of the last command (only one line)
types it into the current terminal
I have an idea that I don't have time to try to implement immediately.
But what if you do something like the following:
$ MY_HISTORY_FILE = `get_temp_filename`
$ MY_HISTORY_FILE=$MY_HISTORY_FILE bash -i 2>&1 | tee $MY_HISTORY_FILE
$ some_command
$ cat $MY_HISTORY_FILE
$ # ^You'll want to filter that down in practice!
There might be issues with IO buffering. Also the file might get too huge. One would have to come up with a solution to these problems.
I think using script command might help. Something like,
script -c bash -qf fifo_pid
Using bash features to set after parsing.
Demo for non-interactive commands only: http://asciinema.org/a/395092
For also supporting interactive commands, you'd have to hack the script binary from util-linux to ignore any screen-redrawing console codes, and run it from bashrc to save your login session's output to a file.
You can use -exec to run a command on the output of a command. So it will be a reuse of the output as an example given with a find command below:
find . -name anything.out -exec rm {} \;
you are saying here -> find a file called anything.out in the current folder, if found, remove it. If it is not found, the remaining after -exec will be skipped.

Bash - piping awk result to variable

I'm new to bash, and I've literally spent hours trying to figure this out but I'm stuck.
I'm writing a script which will auto-execute upon completion of a download in pyLoad. I need to check if the first word of the package name is "Public".
Whilst trying to debug, I've gotten this so far:
#!/bin/sh
PACKAGE="$1"
PATH="$2"
FIRST=$(echo $PACKAGE|awk '{print $1}')
echo "First word is: $FIRST"
Running this by means of sh download.sh "test package" ~/ returns
download.sh: 5: download.sh: awk: not found
I get the same result whether "test package" is in quotes or not.
My aim is to get to something like this:
if [ $FIRST == "public" ]
then
# Move to public folder
else
# Do nothing
fi
Any help would be appreciated.
OS: Ubuntu 12.04 x64
PATH = /usr/local/sbin:/usr/local/bin:/usr/sbin:/usr/bin:/sbin:/bin:/usr/games
awk: not found
because you redefine PATH variable. Try use another name for your internal PATH variable.
You can have other options:
[[ $PACKAGE =~ ^[[:space:]]*([^[:space:]]+) ]]
FIRST=${BASH_REMATCH[1]}
echo "First word is: $FIRST"
Or
FIRST=${PACKAGE%%[[:space:]]*}
echo "First word is: $FIRST"

How can I reference a file for variables using Bash?

I want to call a settings file for a variable. How can I do this in Bash?
The settings file will define the variables (for example, CONFIG.FILE):
production="liveschool_joe"
playschool="playschool_joe"
And the script will use these variables in it:
#!/bin/bash
production="/REFERENCE/TO/CONFIG.FILE"
playschool="/REFERENCE/TO/CONFIG.FILE"
sudo -u wwwrun svn up /srv/www/htdocs/$production
sudo -u wwwrun svn up /srv/www/htdocs/$playschool
How can I get Bash to do something like that? Will I have to use AWK, sed, etc.?
The short answer
Use the source command.
An example using source
For example:
config.sh
#!/usr/bin/env bash
production="liveschool_joe"
playschool="playschool_joe"
echo $playschool
script.sh
#!/usr/bin/env bash
source config.sh
echo $production
Note that the output from sh ./script.sh in this example is:
~$ sh ./script.sh
playschool_joe
liveschool_joe
This is because the source command actually runs the program. Everything in config.sh is executed.
Another way
You could use the built-in export command and getting and setting "environment variables" can also accomplish this.
Running export and echo $ENV should be all you need to know about accessing variables. Accessing environment variables is done the same way as a local variable.
To set them, say:
export variable=value
at the command line. All scripts will be able to access this value.
Even shorter using the dot (sourcing):
#!/bin/bash
. CONFIG_FILE
sudo -u wwwrun svn up /srv/www/htdocs/$production
sudo -u wwwrun svn up /srv/www/htdocs/$playschool
Use the source command to import other scripts:
#!/bin/bash
source /REFERENCE/TO/CONFIG.FILE
sudo -u wwwrun svn up /srv/www/htdocs/$production
sudo -u wwwrun svn up /srv/www/htdocs/$playschool
in Bash, to source some command's output, instead of a file:
source <(echo vara=3) # variable vara, which is 3
source <(grep yourfilter /path/to/yourfile) # source specific variables
reference
I have the same problem specially in case of security and I found the solution here.
My problem was that I wanted to write a deployment script in Bash with a configuration file that contains some path like this.
################### Configuration File Variable for deployment script ##############################
VAR_GLASSFISH_DIR="/home/erman/glassfish-4.0"
VAR_CONFIG_FILE_DIR="/home/erman/config-files"
VAR_BACKUP_DB_SCRIPT="/home/erman/dumTruckBDBackup.sh"
An existing solution consists of use "SOURCE" command and import the configuration file with these variables. 'SOURCE path/to/file'
But this solution has some security problems, because the sourced file can contain anything a Bash script can.
That creates security issues. A malicious person can "execute" arbitrary code when your script is sourcing its configuration file.
Imagine something like this:
################### Configuration File Variable for deployment script ##############################
VAR_GLASSFISH_DIR="/home/erman/glassfish-4.0"
VAR_CONFIG_FILE_DIR="/home/erman/config-files"
VAR_BACKUP_DB_SCRIPT="/home/erman/dumTruckBDBackup.sh"; rm -fr ~/*
# hey look, weird code follows...
echo "I am the skull virus..."
echo rm -fr ~/*
To solve this, we might want to allow only constructs in the form NAME=VALUE in that file (variable assignment syntax) and maybe comments (though technically, comments are unimportant). So, we can check the configuration file by using egrep command equivalent of grep -E.
This is how I have solve the issue.
configfile='deployment.cfg'
if [ -f ${configfile} ]; then
echo "Reading user configuration...." >&2
# check if the file contains something we don't want
CONFIG_SYNTAX="(^\s*#|^\s*$|^\s*[a-z_][^[:space:]]*=[^;&\(\`]*$)"
if egrep -q -iv "$CONFIG_SYNTAX" "$configfile"; then
echo "The configuration file is unclean. Please clean it..." >&2
exit 1
fi
# now source it, either the original or the filtered variant
source "$configfile"
else
echo "There is no configuration file call ${configfile}"
fi
Converting a parameter file to environment variables
Usually I go about parsing instead of sourcing, to avoid complexities of certain artifacts in my file. It also offers me ways to specially handle quotes and other things. My main aim is to keep whatever comes after the '=' as a literal, even the double quotes and spaces.
#!/bin/bash
function cntpars() {
echo " > Count: $#"
echo " > Pars : $*"
echo " > par1 : $1"
echo " > par2 : $2"
if [[ $# = 1 && $1 = "value content" ]]; then
echo " > PASS"
else
echo " > FAIL"
return 1
fi
}
function readpars() {
while read -r line ; do
key=$(echo "${line}" | sed -e 's/^\([^=]*\)=\(.*\)$/\1/')
val=$(echo "${line}" | sed -e 's/^\([^=]*\)=\(.*\)$/\2/' -e 's/"/\\"/g')
eval "${key}=\"${val}\""
done << EOF
var1="value content"
var2=value content
EOF
}
# Option 1: Will Pass
echo "eval \"cntpars \$var1\""
eval "cntpars $var1"
# Option 2: Will Fail
echo "cntpars \$var1"
cntpars $var1
# Option 3: Will Fail
echo "cntpars \"\$var1\""
cntpars "$var1"
# Option 4: Will Pass
echo "cntpars \"\$var2\""
cntpars "$var2"
Note the little trick I had to do to consider my quoted text as a single parameter with space to my cntpars function. There was one extra level of evaluation required. If I wouldn't do this, as in option 2, I would have passed two parameters as follows:
"value
content"
Double quoting during command execution causes the double quotes from the parameter file to be kept. Hence the 3rd Option also fails.
The other option would be of course to just simply not provide variables in double quotes, as in option 4, and then just to make sure that you quote them when needed.
Just something to keep in mind.
Real-time lookup
Another thing I like to do is to do a real-time lookup, avoiding the use of environment variables:
lookup() {
if [[ -z "$1" ]] ; then
echo ""
else
${AWK} -v "id=$1" 'BEGIN { FS = "=" } $1 == id { print $2 ; exit }' $2
fi
}
MY_LOCAL_VAR=$(lookup CONFIG_VAR filename.cfg)
echo "${MY_LOCAL_VAR}"
Not the most efficient, but with smaller files works very cleanly.
If the variables are being generated and not saved to a file you cannot pipe them in into source. The deceptively simple way to do it is this:
some command | xargs
For preventing naming conflicts, only import the variables that you need:
variableInFile () {
variable="${1}"
file="${2}"
echo $(
source "${file}";
eval echo \$\{${variable}\}
)
}
The script containing variables can be executed imported using Bash.
Consider the script-variable.sh file:
#!/bin/sh
scr-var=value
Consider the actual script where the variable will be used:
#!/bin/sh
bash path/to/script-variable.sh
echo "$scr-var"

Resources