I'm working on this for two hours but no luck, always the "Bad substitution" error.
What I want to make (.sh script):
Read from file (names), then I'd like to substitute this name with substring of the given name by the offset and lenght which are the script arguments ($1 = offset, $2 = lenght).
It should work like this (I think) : new_user=${user:$1:$2}
-> where user is read from .txt (in while loop) and $1 and $2 are arguments of this .sc
I've highlighted the important part:
#!/bin/bash
touch postopek.log
while IFS="," read fullName userName passwordLarge
do
pass=$(perl -e 'print crypt(&ARGV[0], "password")' $passwordLarge)
new_up=${fullName:$1:$2} # important line
sudo useradd -m -p $pass -d /home/$new_up -s /bin/bash $new_up
[ $? -eq 0] && echo "Made something bla bla not important..." >> postopek.log
sudo mkdir /home/$new_up/gradivo
sudo cp -r /home/administrator/vaje/* /home/$new_up/gradivo
sudo chown -R $new_up:$new_up /home/$new_up/gradivo
done < /home/administrator/seznam.txt
The sh shell you are running the script with is probably not bash, try running it like
bash ustvari.sh 3 5
or just
/path/to/ustvari.sh 3 5
since your shebang points to bash anyway.
${parameter:offset:length} is not specified by POSIX, thus if your /bin/sh is a shell which does not support the substring syntax, you get the Bad substitution error you encountered, for example:
$ dash
$ echo ${foo:0:1}
dash: 1: Bad substitution
Related
I made this simple bash script to take a full-screen screenshot and save it to the pictures folder
#!/usr/bin/bash
xfce4-screenshooter -f -s /home/rgcodes/Pictures/Screenshot_$scrshotcount.png
let "scrshotcount++"
...which runs into a problem. scrshotcount is a global variable I defined in /etc/environment to be incremented every time the script runs. However, the script fails to increment the variable globally, and causes the script to just overwrite the previous screenshot. Searches on Google and Stack Overflow revealed that the problem isn't straightforward at all (something about child shells being unable to change variables for parents), and finding some other method would be better.
Here's my question. How do we append numbers (in ascending order) to the screenshots the script throws out so that they are saved just like those taken on Windows?(Windows auto-suffixes matching filenames, rather than overwriting them, so all Screenshots have the same name 'Screenshot' and the number of times the screenshot command has been used.)
I am using #erikMD's method as a temporary stop-gap for now.
In addition to the excellent advice about using a date instead of a counter, here's a way to use a counter :/
dir=$HOME/Pictures
# find the current maximum value
current_max=$(
find "$dir" -name Screenshot_\*.png -print0 \
| sort -z -V \
| tail -z -n 1
)
if [[ ! $current_max =~ _([0-9]+)\.png ]]; then
echo "can't find the screenshot with the maximum counter value" >&2
exit 1
fi
# increment it
counter=$(( 1 + ${BASH_REMATCH[1]} ))
# and use it
xfce4-screenshooter -f -s "$dir/Screenshot_${counter}.png"
You'll have to manually create the Screenshot_1.png file.
#rgcodes below is a script that will capture screenshots with a numeric count indicator per your original post. (tested it on Ubuntu 20.04)
Script contents:
#!/bin/bash
set -uo pipefail
# add source file and line number to xtrace output
# i.e. when running: bash -x ./your_script_name.sh
export PS4='+(${BASH_SOURCE}:${LINENO}): ${FUNCNAME[0]:+${FUNCNAME[0]}(): }'
capture_screenshot() {
local output_dir="${1:-/tmp/screenshot}"
local img_name="${2:-Screenshot}"
local img_ext="${3:-png}"
# create output directory (if not already existing)
mkdir -p "${output_dir}"
# get the last image in the sorted ls output
local latest_png=$(tail -n 1 \
<(sort -n -t _ -k 2 \
<(ls ${output_dir}/*.${img_ext} 2> /dev/null)))
# use the latest image to determine img_cnt value for next image
local img_cnt=0
if [[ ${latest_png} =~ _([0-9]+)\.png ]]; then
img_cnt=$((1+${BASH_REMATCH[1]}))
elif [[ ${latest_png} =~ ${img_name}.${img_ext} ]] ; then
img_cnt=1
fi
# build path to output image
local img_path="${output_dir}/${img_name}_${img_cnt}.${img_ext}"
# remove count from output image path if count == 0
if [[ "${img_cnt}" -eq "0" ]] ; then
img_path="${output_dir}/${img_name}.${img_ext}"
fi
xfce4-screenshooter -f -s "${img_path}"
}
capture_screenshot "$#"
The uses the following as defaults, but you can change them to meet your requirements.
output directory for screenshots:
/tmp/screenshot
base screenshot image name:
Screenshot
screenshot file extension:
.png
The script will attempt to create the output directory if it does not already exist (subject to user permission for creation). Below is a sample usage.
Prior to initial script execution, the output directory does not exist:
$ ls screenshot
$
Initial execution (directory is created and Screenshot.png created:
$ ./script.sh
$ ls /tmp/screenshot/
Screenshot.png
Subsequent executions:
$ ./script.sh
$ ls /tmp/screenshot/
Screenshot_1.png Screenshot.png
$ ./script.sh
$ ls /tmp/screenshot/
Screenshot_1.png Screenshot_2.png Screenshot.png
Indeed, as suggested by #j_b in the comments, you should definitely give a try to using a timestamp with the command date +"$format".
FTR, the same idea is implemented here in this project of a gnome-screenshot bash wrapper
(disclaimer: I am the author of this repo).
Example command:
date "+%Y-%m-%d_%H-%M-%S"
↓
2021-07-29_19-13-30
So the overall script could just be something like:
#!/usr/bin/env bash
xfce4-screenshooter -f -s "$HOME/Pictures/Screenshot_$(date "+%Y-%m-%d_%H-%M-%S").png"
(Note that I added missing double-quotes, and modified your shebang, as /usr/bin/env bash is more portable than /bin/bash or /usr/bin/bash.)
I've placed a bash file inside .zshrc and tried all different ways to run it every time I open a new terminal window or source .zshrc but no luck.
FYI: it was working fine on .bashrc
here is .zshrc script:
#Check if ampps is running
bash ~/ampps_runner.sh & disown
Different approach:
#Check if ampps is running
sh ~/ampps_runner.sh & disown
Another approach:
#Check if ampps is running
% ~/ampps_runner.sh & disown
All the above approaches didn't work (meaning it supposes to run an app named ampps but it doesn't in zsh.
Note: It was working fine before switching to zsh from bash. so it does not have permission or syntax problems.
Update: content of ampps_runner.sh
#! /usr/bin/env
echo "########################"
echo "Checking for ampps server to be running:"
check=$(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
This question already has answers here:
': not a valid identifier [duplicate]
(4 answers)
Why would a correct shell script give a wrapped/truncated/corrupted error message? [duplicate]
(1 answer)
Closed 5 years ago.
When running my shell script I get this error:
': not a valid identifiere 17: read: `
Here is my shell script:
#!/usr/bin/env bash
# Mr. Robot Install Wordpress Script
# Script is used for the following:
# add user to server
# change to new user home directory
# download latest version of wordpress
# unzip wordpress
# move all files up a directory level
# move up a directory level
# delete wordpress.zip
# remove wordpress folder
echo "/*****************************************************/"
echo "/************** HELLO MR. ROBOT **********************/"
echo "/*****************************************************/"
echo ".."
echo ".."
echo "Website URL"
echo 'url: \r'
read -p $website
echo 'User: \r'
read -p $newuser
echo 'Pass: \r'
read -p $password
echo "creating account......"
/scripts/wwwacct $website $newuser $password 0 x3 n n n 0 0 0 0 0 0
echo "Changing Directory....."
cd ~/home/$newuser/
echo "Getting Latest Version of Wordpress!"
curl -O http://wordpress.org/latest.tar.gz
echo "Tarball Incoming!!"
tar xvzf latest.tar.gz
echo "removing tar file"
rm latest.tar.gz
echo "moving wordpress folders!"
cp -a ~/home/$newuser/public_html/wordpress/. ~/home/$newuser/public_html/
cd /home/$newuser/public_html/
echo "Part 01 Complete!!"
exit
I've tried to use different versions of the read line with -p or -e. Any help would be appreciated. I've even tried adding it on a separate line with input.
EDIT: Updated file to where it takes inputs, but issue is that the inputs are not being used through the rest of the script. Thus causing errors for directories not being found.
Don't quote the variables names. read needs the name of the variable to assign to, not its value, which is what you get if you have a dollar sign $.
read -p 'website url: ' website
read -p 'Username: ' newuser
read -p 'Password: ' password
It looks like one of the variables holds \r, a carriage return. The error message that bash is trying to print is something like:
bash: ./script: line 17: read: `\r': not a valid identifier
But \r causes the cursor to go back to the beginning of the line, causing ': not a valid identifier to overwrite the beginning of the message.
As mentioned above by John Kugelman, in case you have to check if you Input_file is having carriage returns then you could run following command:
cat -v Input_file
In case you find them then try to remove them from either of following cmmands:
tr -d '\r' < Input_file
OR
awk '{gsub(/\r/,"")} 1' Input_file
Or check if your system(box) has dos2unix utility you could use that also for removing these carriage returns.
echo -n "*.xcodeproj directory: ";
read fileDirectory;
echo -n $fileDirectory;
fileExtension="pbxproj";
find $fileDirectory -name "*.${fileExtension}";
It shows "find: XXXX"(fileDirectory) no such file or directory
However if I replace read fileDirectory by
fileDirectory=$1
It works.
So what's the difference?
$1 is the first argument passed to bash script or to a function inside the script
for example:
mybashfunction /dirtofind
inside the function if you write:
echo "$1"
It should print:
/dirtofind
Edit 1:
You must place the shebang in the beginning of you file
~$ cat a.sh
#!/bin/bash
echo -n "*.xcodeproj directory: ";
read fileDirectory;
echo -n $fileDirectory;
fileExtension="pbxproj";
find "$fileDirectory" -name "*.${fileExtension}";
~$ chmod +x a.sh
~$ ./a.sh
*.xcodeproj directory: /home
/home/home/leonardo/Qt/Tools/QtCreator/share/qtcreator/qbs/share/qbs/examples/cocoa-touch-application/CocoaTouchApplication.xcodeproj/project.pbxproj
/home/leonardo/Qt/Tools/QtCreator/share/qtcreator/qbs/share/qbs/examples/cocoa-application/CocoaApplication.xcodeproj/project.pbxproj
:~$
Works like charm here. Place the shebang
#!/bin/bash
Edit 2
Yes you can use eval. Your script will be like this:
#!/bin/bash
echo -n "*.xcodeproj directory: ";
read fileDirectory;
echo -n $fileDirectory;
fileExtension="pbxproj";
eval fileDirectory=$fileDirectory
find "$fileDirectory" -name "*.${fileExtension}";
read reads data from STDIN (by default), not from positional parameters (arguments).
As you are passing the data as first argument ($1) to the script, read would not catch it; it would catch the input you are providing interactively.
Just to note, you should quote your variable expansions to avoid word splitting and pathname expansion; these are unwanted in most cases.
I'm trying to write a shell script that calls another script that then executes a rsync command.
The second script should run in its own terminal, so I use a gnome-terminal -e "..." command. One of the parameters of this script is a string containing the parameters that should be given to rsync. I put those into single quotes.
Up until here, everything worked fine until one of the rsync parameters was a directory path that contained a space. I tried numerous combinations of ',",\",\' but the script either doesn't run at all or only the first part of the path is taken.
Here's a slightly modified version of the code I'm using
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l '\''/media/MyAndroid/Internal storage'\''' "
Within Backup.sh this command is run
rsync $5 "$path"
where the destination $path is calculated from text in Stamp.
How can I achieve these three levels of nested quotations?
These are some question I looked at just now (I've tried other sources earlier as well)
https://unix.stackexchange.com/questions/23347/wrapping-a-command-that-includes-single-and-double-quotes-for-another-command
how to make nested double quotes survive the bash interpreter?
Using multiple layers of quotes in bash
Nested quotes bash
I was unsuccessful in applying the solutions to my problem.
Here is an example. caller.sh uses gnome-terminal to execute foo.sh, which in turn prints all the arguments and then calls rsync with the first argument.
caller.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh 'long path' arg2 arg3"
foo.sh:
#!/bin/bash
echo $# arguments
for i; do # same as: for i in "$#"; do
echo "$i"
done
rsync "$1" "some other path"
Edit: If $1 contains several parameters to rsync, some of which are long paths, the above won't work, since bash either passes "$1" as one parameter, or $1 as multiple parameters, splitting it without regard to contained quotes.
There is (at least) one workaround, you can trick bash as follows:
caller2.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh '--option1 --option2 \"long path\"' arg2 arg3"
foo2.sh:
#!/bin/bash
rsync_command="rsync $1"
eval "$rsync_command"
This will do the equivalent of typing rsync --option1 --option2 "long path" on the command line.
WARNING: This hack introduces a security vulnerability, $1 can be crafted to execute multiple commands if the user has any influence whatsoever over the string content (e.g. '--option1 --option2 \"long path\"; echo YOU HAVE BEEN OWNED' will run rsync and then execute the echo command).
Did you try escaping the space in the path with "\ " (no quotes)?
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l ''/media/MyAndroid/Internal\ storage''' "