Different results when running commands in braces within a bash script - linux

I was editing a script and as the script was getting a bit long I decided to enclose the main part of the script in braces and divert the output to a log file instead of having individual log redirects for commands. Then I noticed that a command block that checks for a running copy of the script gives 2 different results depending if it is enclosed in braces.
I run the script as:
$ /bin/bash scriptname.bash
My question is why the same command block returns 2 different results and if it is possible to have the command block work inside the braces.
Below is the command block:
#!/bin/bash
#set -x # Uncomment to debug this shell script
#
##########################################################
# DEFINE FILES AND VARIABLES HERE
##########################################################
THIS_SCRIPT=$(basename $0)
TIMESTAMP=$(date +%Y-%m-%d_%H%M%S)
LOGFILE=process_check_$TIMESTAMP.log
##########################################################
# BEGINNING OF MAIN
##########################################################
{
printf "%s\n" "Checking for currently runnning versions of this script"
MYPID=$$ # Capture this scripts PID
MYOTHERPROCESSES=$(ps -ef | \grep $THIS_SCRIPT | \grep -v $MYPID | \grep -v grep | awk '{print $2}')
if [[ "$MYOTHERPROCESSES" != "" ]]
then
printf "%s\n" "ERROR: Another version of this script is running...exiting!"
exit 2
else
printf "%s\n" "No other versions running...proceeding"
fi
printf "%s\n" "Doing some script stuff..."
exit 0
} | tee -a $LOGFILE 2>&1
# End of script

This is not due to the braces, this is due to the pipe.
When you combine commands with a pipe like command | tee, each side of the pipe is executed in a separate sub-process. Shell commands are therefore executed in a sub-shell. That's this sub-shell that you detect.
PS: avoid constructs like ps | grep -v grep, use pidof or pgrep instead

Related

Can't run bash file inside ZSH

I've placed a bash file inside .zshrc and tried all different ways to run it every time I open a new terminal window or source .zshrc but no luck.
FYI: it was working fine on .bashrc
here is .zshrc script:
#Check if ampps is running
bash ~/ampps_runner.sh & disown
Different approach:
#Check if ampps is running
sh ~/ampps_runner.sh & disown
Another approach:
#Check if ampps is running
% ~/ampps_runner.sh & disown
All the above approaches didn't work (meaning it supposes to run an app named ampps but it doesn't in zsh.
Note: It was working fine before switching to zsh from bash. so it does not have permission or syntax problems.
Update: content of ampps_runner.sh
#! /usr/bin/env
echo "########################"
echo "Checking for ampps server to be running:"
check=$(pgrep -f "/usr/local/ampps" )
#[ -z "$check" ] && echo "Empty: Yes" || echo "Empty: No"
if [ -z "$check" ]; then
echo "It's not running!"
cd /usr/local/ampps
echo password | sudo -S ./Ampps
else
echo "It's running ..."
fi
(1) I believe ~/.ampps_runner.sh is a bash script, so, its first line should be
#!/bin/bash
or
#!/usr/bin/bash
not
#! /usr/bin/env
(2) Then, the call in zsh script (~/.zshrc) should be:
~/ampps_runner.sh
(3) Note: ~/.ampps_runner.sh should be executable. Change it to executable:
$ chmod +x ~/ampps_runner.sh
The easiest way to run bash temporarily from a zsh terminal is to
exec bash
or just
bash
Then you can run commands you previously could only run in bash. An example
help exec
To exit
exit
Now you are back in your original shell
If you want to know your default shell
echo $SHELL
or
set | grep SHELL=
If you want to reliably know your current shell
ps -p $$
Or if you want just the shell name you might use
ps -p $$ | awk "NR==2" | awk '{ print $4 }' | tr -d '-'
And you might just put that last one in a function for later, just know that it is only available if it was sourced in a current shell.
whichShell(){
local defaultShell=$(echo $SHELL | tr -d '/bin/')
echo "Default: $defaultShell"
local currentShell=$(ps -p $$ | awk "NR==2" | awk '{ print $4 }' | tr -d '-')
echo "Current: $currentShell"
}
Call the method to see your results
whichShell

Bash output to screen and logfile differently

I have been trying to get a bash script to output different things on the terminal and logfile but am unsure of what command to use.
For example,
#!/bin/bash
freespace=$(df -h / | grep -E "/" | awk '{print $4}')
greentext="\033[32m"
bold="\033[1m"
normal="\033[0m"
logdate=$(date +"%Y%m%d")
logfile="$logdate"_report.log
exec > >(tee -i $logfile)
echo -e $bold"Quick system report for "$greentext"$HOSTNAME"$normal
printf "\tSystem type:\t%s\n" $MACHTYPE
printf "\tBash Version:\t%s\n" $BASH_VERSION
printf "\tFree Space:\t%s\n" $freespace
printf "\tFiles in dir:\t%s\n" $(ls | wc -l)
printf "\tGenerated on:\t%s\n" $(date +"%m/%d/%y") # US date format
echo -e $greentext"A summary of this info has been saved to $logfile"$normal
I want to omit the last output (echo "A summary...") in the logfile while displaying it in the terminal. Is there a command to do so? It would be great if a general solution can be provided instead of a specific one because I want to apply this to other scripts.
EDIT 1 (after applying >&6):
Files in dir: 7
A summary of this info has been saved to 20160915_report.log
Generated on: 09/15/16
One option:
exec 6>&1 # save the existing stdout
exec > >(tee -i $logfile) # like you had it
#... all your outputs
echo -e $greentext"A summary of this info has been saved to $logfile"$normal >&6
# writes to the original stdout, saved in file descriptor 6 ------------^^^
The >&6 sends echo's output to the saved file descriptor 6 (the terminal, if you're running this from an interactive shell) rather than to the output path set up by tee (which is on file descriptor 1). Tested on bash 4.3.46.
References: "Using exec" and "I/O Redirection"
Edit As OP found, the >&6 message is not guaranteed to appear after the lines printed by tee off stdout. One option is to use script, e.g., as in the answers to this question, instead of tee, and then print the final message outside of the script. Per the docs, the stdbuf answers to that question won't work with tee.
Try a dirty hack:
#... all your outputs
echo >&6 # <-- New line
echo -e $greentext ... >&6
Or, equally hackish, (Note that, per OP, this worked)
#... all your outputs
sleep 0.25s # or whatever time you want <-- New line
echo -e ... >&6

Bash Script output is always 'ps' when piping to grep from ps regardless of PID results

given an array of pids and the code:
for i in ${listedPids[#]}
do
runningCheck="ps -u $USER | grep $i"
grepRes=(${runningCheck})
if [[ -n $grepRes ]]
then
echo $grepRes
echo $runningCheck
... code not related to the issue
fi
done
Regardless if those pids are active or not; I keep getting 'ps' from echo $grepRes while the output of echo $runningCheck shows up with the correct user name and pid. What am I missing?
Replace
"ps -u $USER | grep $i"
by
$(ps -u $USER | grep $i)
Command Substitution: Bash performs the expansion by executing your command and replacing the command substitution with the standard output of the
command, with any trailing newlines deleted.
I simplified your script and here's what it should look like.
for i in "${listedPids[#]}"
do
grepRes=$(ps --no-heading -p $i)
if [[ -n "$grepRes" ]]
then
echo "$grepRes"
... code not related to the issue
fi
done
An even shorter code could be written using while loop.
ps --noheading -p "${listedPids[#]}" | while read grepRes
do
echo "$grepRes"
... code not related to the issue
done
As alvits and l0b0 pointed out, I made a few syntax errors: grepRes=(${runningCheck}) when I just wanted to execute that line and not turn it to a list, and the fact pipes and redirects don't work in variables. In the end pgrep did the job as I just needed to continue looping till all the background processes ended.
Maybe you could try eval.
runningCheck1="ps -u $USER"
runningCheck2=" | grep $i"
echo $runningCheck1$runningCheck
eval $runningCheck1$runningCheck2

Setting variable in bash -c

I try to set variable which get interface ip-address from ifconfig and read it later. But when I execute a echo command, variable still empty. Please look at my code:
/usr/bin/bash -c "HOST_IPS=$(/usr/bin/ifconfig | /usr/bin/awk 'BEGIN {cnt=0} {if($0 ~ /inet / && cnt==1) {print $2} cnt++}'); echo $HOST_IPS"
But /bin/echo work fine with same command:
/usr/bin/bash -c "echo $(/usr/bin/ifconfig | /usr/bin/awk 'BEGIN {cnt=0} {if($0 ~ /inet / && cnt==1) {print $2} cnt++}')"
You have to escape the $ sign in the final echo command, or the variable $HOST_IPS will be substituted into the command string before the subshell is spawned:
/usr/bin/bash -c "HOST_IPS=$(/usr/bin/ifconfig | /usr/bin/awk 'BEGIN {cnt=0} {if($0 ~ /inet / && cnt==1) {print $2} cnt++}'); echo \$HOST_IPS"
For more immediate visibility:
# v-- insert backslash here
/usr/bin/bash -c "HOST_IPS=$(same as before); echo \$HOST_IPS"
Contrary to #gniourf_gniourf's comment, it is not actually necessary to escape the other dollar signs. However, as written, the command substitution is not performed by the subshell (!); its result is substituted into the command string that is passed to the subshell. The calls
mypid() { echo $$; }
bash -c "pid=$(mypid); echo \$pid; mypid"
demonstrate the way it works: it will once print the PID of the parent shell and once complain that mypid is not a known command because the subshell does not know the function.
Since running the ifconfig | awk command in the parent shell is unlikely to be a problem, you can probably leave the command substitution part unchanged. If it is important that the command be run by the subshell, you'll have to escape all the things there as well.
With /usr/bin/bash you start a subshell. When the shell is finished, all settings in the shell are lost.
The following sub is set in the subshell and lost before being echoed:
/usr/bin/bash -c sub=1; echo $sub
Do you want to set a variable in a subshell, use stdout for transporting the value:
sub=$(/usr/bin/bash -c "echo 1"); echo $sub
From your question and example, your task at hand doesn't require you to leave your current environment, hence you don't need to start a new one and be concerned with data lost.
HOST_IPS=();
while read -r;
do [[ $REPLY =~ "inet "([^/]*) ]] && HOST_IPS+=(${BASH_REMATCH[1]});
done < <( ip -f inet address show )
If you wish to maintain a list of interface IP-addresses you can process the output of ip (or ifconfig) in your current shell without calling awk.

Saving a command into a variable instead of running it

I'm trying to get the output of the ps command to output to a file, then to use that file to populate a radiolist. So far I'm having problems.
eval "ps -o pid,command">/tmp/process$$
more /tmp/process$$
sed -e '1d' /tmp/process$$ > /tmp/process2$$
while IFS= read -r pid command
do
msgboxlist="$msgboxlist" $($pid) $($command) "off"
done</tmp/process2$$
height=`wc -l "/tmp/process$$" | awk '{print $1}'`
width=`wc --max-line-length "/tmp/process$$" | awk '{print $1}'`
echo $height $width
dialog \
--title "Directory Listing" \
--radiolist "Select process to terminate" "$msgboxlist" $(($height+7)) $(($width+4))
So far not only does the while read not split the columns into 2 variables ($pid is the whole line and $command is blank) but when I try to run this the script is trying to run the line as a command. For example:
+ read -r pid command
++ 7934 bash -x assessment.ba
assessment.ba: line 322: 7934: command not found
+ msgboxlist=
+ off
assessment.ba: line 322: off: command not found
Basically I have no idea where I'm supposed to be putting quotes, double quotes and backslashes. It's driving me wild.
tl;dr Saving a command into a variable without running it, how?
You're trying to execute $pid and $command as commands:
msgboxlist="$msgboxlist" $($pid) $($command) "off"
Try:
msgboxlist="$msgboxlist $pid $command off"
Or use an array:
msgboxlist=() # do this before the while loop
msgboxlist+=($pid $command "off")
# when you need to use the whole list:
echo "${msgboxlist[#]}"
Your script can be refactored by removing some unnecessary calls like this:
ps -o pid=,command= > /tmp/process$$
msgboxlist=""
while read -r pid command
do
msgboxlist="$msgboxlist $pid $command off"
done < /tmp/process2$$
height=$(awk 'END {print NR}' "/tmp/process$$")
width=$(awk '{if (l<length($0)) l=length($0)} END{print l}' "/tmp/process$$")
dialog --title "Directory Listing" \
--radiolist "Select process to terminate" "$msgboxlist" $(($height+7)) $(($width+4))
I have to admit, I'm not 100% clear on what you're doing; but I think you want to change this:
msgboxlist="$msgboxlist" $($pid) $($command) "off"
to this:
msgboxlist+=("$pid" "$command" off)
which will add the PID, the command, and "off" as three new elements to the array named msgboxlist. You'd then change "$msgboxlist" to "${msgboxlist[#]}" in the dialog command, to include all of those elements as arguments to the command.
Use double quotes when you want variables to be expanded. Use single quotes to disable variable expansion.
Here's an example of a command saved for later execution.
file="readme.txt"
cmd="ls $file" # $file is expanded to readme.txt
echo "$cmd" # ls readme.txt
$cmd # lists readme.txt
Edit adressing the read:
Using read generally reads an entire line. Consider this instead (tested):
ps o pid=,command= | while read line ; do
set $line
pid=$1
command=$2
echo $pid $command
done
Also note the different usage of 'ps o pid=,command=' to skip displaying headers.

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