I have the following bash script:
dpkg-query --show --showformat='${Status}\n' "$i" 2> \
/dev/null | grep "install ok installed" &> /dev/null
if [[ $? -eq 0 ]]; then
l_var_is_desktop="true"
fi
and the ShellCheck utility (https://www.shellcheck.net/) is giving me the following output:
$ shellcheck myscript
Line 17:
if [[ $? -eq 0 ]]; then
^-- SC2181: Check exit code directly with e.g. 'if mycmd;', not indirectly with $?.
$
The link to this warning is the following: https://github.com/koalaman/shellcheck/wiki/SC2181
What is the best way for modifying this. The command is really too long to put into one line. I would like to avoid using ShellCheck ignore rules.
I've tried creating a local variable and storing the output of the command, but this breaks other rules.
The command doesn't really get much longer by putting it directly in if, you're just adding 3 characters.
if dpkg-query --show --showformat='${Status}\n' "$i" 2> \
/dev/null | grep "install ok installed" &> /dev/null
then
l_var_is_desktop="true"
fi
Related
I have been testing bash and found the following code:
#!/bin/bash
[[ `which pacman` ]] && echo pacman
[[ `which apt-get` ]] && echo apt-get
It will test what package manager is installed and echo it.
On some systems, a failed which command prints the error to stderr. I want to suppress all output from the which command.
So I came up with the following code:
#!/bin/bash
[[ `which pacman >/dev/null 2>&1` ]] && echo pacman
[[ `which apt-get >/dev/null 2>&1` ]] && echo apt-get
But this doesn't output anything. When I run each command on the command line like this:
which pacman >/dev/null 2>&1 && echo $?
On a system with pacman, it prints 0 which it should. The && also proves that the previous command succeeded.
So why doesn't the redirection code work like it does on the command line? What can I add to the script to make it work?
This is really confusing to me, as I have never had this type of problem before. Usually, any command that works on the command line should also work in a bash script, shouldn't it?
[[ ... ]] without a specified test runs [[ -n ... ]]. In this case, it tests whether the captured output is non-empty. But if you redirect everything to /dev/null, the output is indeed empty!
You don't need to capture the output. which should already return a non-zero status when it cannot find the file to execute.
which pacman &> /dev/null && echo pacman
I've got the following code in my shell script:
SERVER=`ps -ef | grep -v grep | grep -c sof2ded`
if ["$SERVER" != "0"]; then
echo "Already Running, exiting"
exit
else
echo "Starting up the server..."
cd /home/sof2/
/home/sof2/crons/start.sh > /dev/null 2>&1
fi
I did chmod a+x status.sh
Now I try to run the script but it's returning this error:
./status.sh: line 5: [1: command not found
Starting up the server...
Any help would be greatly appreciated.
Could you please try changing a few things in your script as follows and let me know if that helps you?(changed back-tick to $ and changed [ to [[ in code)
SERVER=$(ps -ef | grep -v grep | grep -c sof2ded)
if [[ "$SERVER" -eq 0 ]]; then
echo "Already Running, exiting"
exit
else
echo "Starting up the server..."
cd /home/sof2/
/home/sof2/crons/start.sh > /dev/null 2>&1
fi
The problem is with the test command. "But", I hear you say, "I am not using the test command". Yes you are, it is also known as [.
if statement syntax is if command. The brackets are not part of if syntax.
Commands have arguments separated (tokenized) by whitespace, so:
[ "$SERVER" != "0" ]
The whitespace is needed because the command is [ and then there are 4 arguments passed to it (the last one must be ]).
A more robust way of comparing numerics is to use double parentheses,
(( SERVER == 0 ))
Notice that you don't need the $ or the quotes around SERVER. Also the spacing is less important, but useful for readability.
[[ is used for comparing text patterns.
As a comment, backticks ` ` are considered deprecated because they are difficult to read, they are replaced with $( ... ).
I have following code in my build script:
if [ -z "$1" ]; then
make -j10 $1 2>&1 | tee log.txt && notify-send -u critical -t 7 "BUILD DONE"
else
make -j10 $1 2>&1 | tee log.txt | grep -i --color "Error" && notify-send -u critical -t 7 "BUILD DONE"
fi
I tried to optimize it to:
local GREP=""
[[ ! -z "$1" ]] && GREP="| grep -i --color Error" && echo "Grepping for ERRORS"
make -j10 $1 2>&1 | tee log.txt "$GREP" && notify-send -u critical -t 7 "BUILD DONE"
But error thrown in make line if $1 isn't empty. I just can't figure out how to pass command with grep pipe through the variable.
Like others have already pointed out, you cannot, in general, expect a command in a variable to work. This is a FAQ.
What you can do is execute commands conditionally. Like this, for example:
( make -j10 $1 2>&1 && notify-send -u critical -t 7 "BUILD DONE" ) |
tee log.txt |
if [ -z "$1" ]; then
grep -i --color "Error"
else
cat
fi
This has the additional unexpected benefit that the notify-send is actually conditioned on the exit code of make (which is probably what you intended) rather than tee (which I would expect to succeed unless you run out of disk or something).
(Or if you want the notification regardless of the success status, change && to just ; -- I think this probably makes more sense.)
This is one of those rare Useful Uses of cat (although I still feel the urge to try to get rid of it!)
You can't put pipes in command variables:
$ foo='| cat'
$ echo bar $foo
bar | cat
The linked article explains how to do such things very well.
As mentioned in #l0b0's answer, the | will not be interpreted as you are hoping.
If you wanted to cut down on repetition, you could do something like this:
if [ $(make -j10 "$1" 2>&1 > log.txt) ]; then
[ "$1" ] && grep -i --color "error" log.txt
notify-send -u critical -t 7 "BUILD DONE"
fi
The inside of the test is common to both branches. Instead of using tee so that the output can be piped, you can just indirect the output to log.txt. If "$1" isn't empty, grep for any errors in log.txt. Either way, do the notify-send.
Good day. I have a series of commands that I wanted to execute via a function so that I could get the exit code and perform console output accordingly. With that being said, I have two issues here:
1) I can't seem to direct stderr to /dev/null.
2) The first echo line is not displayed until the $1 is executed. It's not really noticeable until I run commands that take a while to process, such as searching the hard drive for a file. Additionally, it's obvious that this is the case, because the output looks like:
sh-3.2# ./runScript.sh
sh-3.2# com.apple.auditd: Already loaded
sh-3.2# Attempting... Enable Security Auditing ...Success
In other words, the stderr was displayed before "Attempting... $2"
Here is the function I am trying to use:
#!/bin/bash
function saveChange {
echo -ne "Attempting... $2"
exec $1
if [ "$?" -ne 0 ]; then
echo -ne " ...Failure\n\r"
else
echo -ne " ...Success\n\r"
fi
}
saveChange "$(launchctl load -w /System/Library/LaunchDaemons/com.apple.auditd.plist)" "Enable Security Auditing"
Any help or advice is appreciated.
this is how you redirect stderr to /dev/null
command 2> /dev/null
e.g.
ls -l 2> /dev/null
Your second part (i.e. ordering of echo) -- It may be because of this you have while invoking the script. $(launchctl load -w /System/Library/LaunchDaemons/com.apple.auditd.plist)
The first echo line is displayed later because it is being execute second. $(...) will execute the code. Try the following:
#!/bin/bash
function saveChange {
echo -ne "Attempting... $2"
err=$($1 2>&1)
if [ -z "$err" ]; then
echo -ne " ...Success\n\r"
else
echo -ne " ...Failured\n\r"
exit 1
fi
}
saveChange "launchctl load -w /System/Library/LaunchDaemons/com.apple.auditd.plist" "Enable Security Auditing"
EDIT: Noticed that launchctl does not actually set $? on failure so capturing the STDERR to detect the error instead.
I'm looking for a way to create a switch for this bash script so that I have the option of either printing (echo) it to stdout or executing the command for debugging purposes. As you can see below, I am just doing this manually by commenting out one statement over the other to achieve this.
Code:
#!/usr/local/bin/bash
if [ $# != 2 ]; then
echo "Usage: testcurl.sh <localfile> <projectname>" >&2
echo "sample:testcurl.sh /share1/data/20110818.dat projectZ" >&2
exit 1
fi
echo /usr/bin/curl -c $PROXY --certkey $CERT --header "Test:'${AUTH}'" -T $localfile $fsProxyURL
#/usr/bin/curl -c $PROXY --certkey $CERT --header "Test:'${AUTH}'" -T $localfile $fsProxyURL
I'm simply looking for an elegant/better way to create like a switch from the command line. Print or execute.
One possible trick, though it will only work for simple commands (e.g., no pipes or redirection (a)) is to use a prefix variable like:
pax> cat qq.sh
${PAXPREFIX} ls /tmp
${PAXPREFIX} printf "%05d\n" 72
${PAXPREFIX} echo 3
What this will do is to insert you specific variable (PAXPREFIX in this case) before the commands. If the variable is empty, it will not affect the command, as follows:
pax> ./qq.sh
my_porn.gz copy_of_the_internet.gz
00072
3
However, if it's set to echo, it will prefix each line with that echo string.
pax> PAXPREFIX=echo ./qq.sh
ls /tmp
printf %05d\n 72
echo 3
(a) The reason why it will only work for simple commands can be seen if you have something like:
${PAXPREFIX} ls -1 | tr '[a-z]' '[A-Z]'
When PAXPREFIX is empty, it will simply give you the list of your filenames in uppercase. When it's set to echo, it will result in:
echo ls -1 | tr '[a-z]' '[A-Z]'
giving:
LS -1
(not quite what you'd expect).
In fact, you can see a problem with even the simple case above, where %05d\n is no longer surrounded by quotes.
If you want a more robust solution, I'd opt for:
if [[ ${PAXDEBUG:-0} -eq 1 ]] ; then
echo /usr/bin/curl -c $PROXY --certkey $CERT --header ...
else
/usr/bin/curl -c $PROXY --certkey $CERT --header ...
fi
and use PAXDEBUG=1 myscript.sh to run it in debug mode. This is similar to what you have now but with the advantage that you don't need to edit the file to switch between normal and debug modes.
For debugging output from the shell itself, you can run it with bash -x or put set -x in your script to turn it on at a specific point (and, of course, turn it off with set +x).
#!/usr/local/bin/bash
if [[ "$1" == "--dryrun" ]]; then
echoquoted() {
printf "%q " "$#"
echo
}
maybeecho=echoquoted
shift
else
maybeecho=""
fi
if [ $# != 2 ]; then
echo "Usage: testcurl.sh <localfile> <projectname>" >&2
echo "sample:testcurl.sh /share1/data/20110818.dat projectZ" >&2
exit 1
fi
$maybeecho /usr/bin/curl "$1" -o "$2"
Try something like this:
show=echo
$show /usr/bin/curl ...
Then set/unset $show accordingly.
This does not directly answer your specific question, but I guess you're trying to see what command gets executed for debugging. If you replace #!/usr/local/bin/bash with #!/usr/local/bin/bash -x bash will run and echo the commands in your script.
I do not know of a way for "print vs execute" but I know of a way for "print and execute", and it is using "bash -x". See this link for example.