This question already has answers here:
How do I suppress shell script error messages?
(6 answers)
Detect if executable file is on user's PATH [duplicate]
(7 answers)
Closed 1 year ago.
I'm writing a shell script, and I need to check for some dependencies being installed before executing anything. I found I can use which <package> to see if it is installed or not. The problem is that when that dependency is not found, it throws the following error into console's output:
which: no abc in (/home/pace/.emacs.d/bin:/usr/local/bin:/home/pace/.emacs.d/bin:/usr/local/bin:/home/pace/.local/bin:/usr/local/sbin:/usr/local/bin:/usr/bin:...)
I want to avoid having such output, as I already have error messages shown when something fails. How can I avoid which from writing anything?
function is_installed() {
if [[ ! $(which $1) ]]
then
echo "[ERROR]: $1 $2"
exit 1
fi
}
Well, there might be better ways to do what you're trying to do (I'm not certain of the "best" way), but you can redirect stderr and stdout to hide the results from the output:
function is_installed() {
if [[ ! $(which $1 > /dev/null 2>&1 ) ]]
then
echo "[ERROR]: $1 $2"
exit 1
fi
}
(recent versions of bash support using >& /dev/null too to do both at once, but the above is slightly more portable)
EDIT -- try this instead
function is_installed() {
which $1 > /dev/null 2>&1
if [ $? = 1 ] ; then
echo "[ERROR]: $1 $2"
exit 1
fi
}
Related
This question already has answers here:
How do I set a variable to the output of a command in Bash?
(15 answers)
Closed 3 years ago.
#!/bin/bash
if [ -f "$1" ]
then
for users in 'cat $1'
do
useradd $users
done
else
echo "input is not a file"
fi
You just have to get the input for the do loop right:
#!/bin/bash
if [ -f "$1" ]
then
for user in $(cat "$1")
do
useradd "$user"
done
else
echo "input is not a file"
fi
Remarks: this works for reading out a file word-by-word and I tried to keep your structure.
For reading out files line by line this is an elegant way: https://stackoverflow.com/a/4642213/2819581
I am working on a script that take 1 string argument and a file. I want it so that if a file is put in that doesn't exist, then it will display the "filename cannot be read" message.
That part does work however it also displays a "grep: grep.txt: No such file or directory" message. Is there any way to stop the grep message from posting and ending the script if the first if statement is true?
#! /bin/sh
if [ ! -f "$2" ]
then
echo "$0" cannot be read 1>&2
fi
if [ $# -eq 2 ]
then
grep "$1" $2
else
echo there is more or less than 2 arguments 1>&2
fi
Exit the script with a non-zero exit code to indicate failure and stop it from continuing on to the grep.
if [ ! -f "$2" ]
then
echo "$0" cannot be read 1>&2
exit 1
fi
You can add /dev/null in grep command it will suppress the error part.
grep "$1" $2 2>/dev/null
The > operator redirects the output usually to a file but it can be to a device. You can also use >> to append.
2> file redirects stderr to file
/dev/null is the null device it takes any input you want and throws it away. It can be used to suppress any output.
You could redirect all errors from grep, for example:
grep "$1" $2 2>/dev/null
(the 2> means redirect standard error, as opposed to standard output with > or 1>).
That introduces a race condition, however: if the file disappears while your script as running, it might still exist when you check that it exists, but be gone by the time grep runs.
You could handle that by checking the exit status...
grep "$1" $2 2>/dev/null
if [[ $? -gt 1 ]]; then
echo "grep failed unexpectedly" >&2
fi
IMHO, in this example it would be better to just let grep print the error.
This question already has answers here:
Checking from shell script if a directory contains files
(30 answers)
How do I check if a folder has contents? [duplicate]
(3 answers)
Closed 6 years ago.
I have a directory. It is empty. If i perform ls -lrt , it shows total 0
How do I specify an If condition to perform something, only if the directory is empty.
I mean to ask how to capture that 0 value.
From here. This should help you run your statements within the if else loop. I saved the DIR in the variable
#!/bin/bash
FILE=""
DIR="/empty_dir"
# init
# look for empty dir
if [ "$(ls -A $DIR)" ]; then
echo "Take action $DIR is not Empty"
else
echo "$DIR is Empty"
fi
# rest of the logic
Remove the -A option :
$ mkdir /tmp/aaa
$ ls /tmp/aaa
$ a=\`ls /tmp/aaa`
$ [[ -z $a ]]
$ echo $?
0
I want to check whether nodejs is installed on the system or not. I am getting this error:
Error : command not found.
How can i fix it?
#!/bin/bash
if [ nodejs -v ]; then
echo "nodejs found"
else
echo "nodejs not found"
fi
You can use the command bash builtin:
if command -v nodejs >/dev/null 2>&1 ; then
echo "nodejs found"
echo "version: $(nodejs -v)"
else
echo "nodejs not found"
fi
The name of the command is node, not nodejs
which returns the path to the command to stdout, if it exists
if [ $(which node 2>/dev/null) ]; then
echo "nodejs found"
else
echo "nodejs not found"
fi
This is not what the OP asked for (nearly 3 years ago!), but for anyone who wants to check multiple dependencies:
#!/bin/bash
echo -n "Checking dependencies... "
for name in youtube-dl yad ffmpeg
do
[[ $(which $name 2>/dev/null) ]] || { echo -en "\n$name needs to be installed. Use 'sudo apt-get install $name'";deps=1; }
done
[[ $deps -ne 1 ]] && echo "OK" || { echo -en "\nInstall the above and rerun this script\n";exit 1; }
Here's how it works. First, we print a line saying that we are checking dependencies. The second line starts a "for name in..." loop, in which we put the dependencies we want to check, in this example we will check for youtube-dl, yad and ffmpeg. The loop commences (with "do") and the next line checks for the existence of each command using the bash command "which." If the dependency is already installed, no action is taken and we skip to the next command in the loop. If it does need to be installed, a message is printed and a variable "deps" is set to 1 (deps = dependencies) and then we continue to the next command to check. After all the commands are checked, the final line checks to see if any dependencies are required by checking the deps variable. If it is not set, it appends "OK" to the line where it originally said "Checking dependencies.... " and continues (assuming this is the first part of a script). If it is set, it prints a message asking to install the dependencies and to rerun the script. It then exits the script.
The echo commands look complicated but they are necessary to give a clean output on the terminal. Here is a screenshot showing that the dependencies are not met on the first run, but they are on the second.
PS If you save this as an script, you will need to be in the same directory as the script and type ./{name_of_your_script} and it will need to be executable.
You may check the existence of a program or function by
type nodejs &>/dev/null || echo "node js not installed"
However, there is a more sophisticated explanation available here.
I was thinking about this and came up with a few versions, then went on the internet to see what others have to say and ended up here. Albeit an old thread, I'll reply with my thoughts.
First to answer the OP's original question: How can i fix it?
if node -v &>/dev/null; then
echo "nodejs found"
else
echo "nodejs not found"
fi
If you are simply checking if node works, this would do it. But it isn't a very generic way to do it.
Another way is to use command in a loop and collect the missing dependencies (in this example looking for the commands kind and kubectl).
for app in kind kubectl; do command -v "${app}" &>/dev/null || not_available+=("${app}"); done
(( ${#not_available[#]} > 0 )) && echo "Please install missing dependencies: ${not_available[*]}" 1>&2 && exit 1
Or less concisely expressed:
unset not_available # script safety, however not necessary.
for app in kind kubectl; do
if ! command -v "${app}" &>/dev/null; then
not_available+=("${app}")
fi
done
if (( ${#not_available[#]} > 0 )); then
echo "Please install missing dependencies: ${not_available[#]}" 1>&2
exit 1
fi
Then I figured I'd want a way to do the same without a loop, so came up with this:
not_installed=$(command -V kind kubectl 2>&1 | awk -F': +' '$NF == "not found" {printf "%s ", $(NF-1)}')
[[ -n ${not_installed} ]] && echo "Please install missing dependencies: ${not_installed}" 1>&2 && exit 1
The command -V can take any number of entries and posts the result back to stdout and stderr (though I redirect both to stdout for the next command to parse).
awk sets the field separator to <colon><one or more space>, expressed as : +. If the last field contains, "not found", print the second to last field, being the name of the command which is not installed.
Lastly, if the variable contains any data, then report back which dependencies that are missing to stderr and exit the script!
You can do dependency checks in a million ways, but here are a few alternatives which are more generally applicable and not too lengthy while still being easy to follow. :]
If all you want is to check to see if a command exists, use which command. It returns the patch if the command is called, and nothing if it is not found
if [ "$(which openssl)" = "" ] ;then
echo "This script requires openssl, please resolve and try again."
exit 1
fi
I want to create a bash file (.sh) which does the following:
I call the script like ./download.sh www.blabla.com/bla.jpg
the script has to echo then if the file has downloaded or not...
How can I do this? I know I can use errorlevel but I'm new to linux so...
Thanks in advance!
Typically applications in Linux will set the value of the environment variable $? on failure. You can examine this return code and see if it gets you any error for wget.
#!/bin/bash
wget $1 2>/dev/null
export RC=$?
if [ "$RC" = "0" ]; then
echo $1 OK
else
echo $1 FAILED
fi
You could name this script download.sh. Change the permissions to 755 with chmod 755. Call it with the name of the file you wish to download. ./download.sh www.google.com
You could try something like:
#!/bin/sh
[ -n $1 ] || {
echo "Usage: $0 [url to file to get]" >&2
exit 1
}
wget $1
[ $? ] && {
echo "Could not download $1" | mail -s "Uh Oh" you#yourdomain.com
echo "Aww snap ..." >&2
exit 1
}
# If we're here, it downloaded successfully, and will exit with a normal status
When making a script that will (likely) be called by other scripts, it is important to do the following:
Ensure argument sanity
Send e-mail, write to a log, or do something else so someone knows what went wrong
The >&2 simply redirects the output of error messages to stderror, which allows a calling script to do something like this:
foo-downloader >/dev/null 2>/some/log/file.txt
Since it is a short wrapper, no reason to forsake a bit of sanity :)
This also allows you to selectively direct the output of wget to /dev/null, you might actually want to see it when testing, especially if you get an e-mail saying it failed :)
wget executes in non-interactive way. This means that wget work in the background and you can't catch de return code with $?.
One solution it's to handle the "--server-response" property, searching http 200 status code
Example:
wget --server-response -q -o wgetOut http://www.someurl.com
sleep 5
_wgetHttpCode=`cat wgetOut | gawk '/HTTP/{ print $2 }'`
if [ "$_wgetHttpCode" != "200" ]; then
echo "[Error] `cat wgetOut`"
fi
Note: wget need some time to finish his work, for that reason I put "sleep 5". This is not the best way to do but worked ok for test the solution.