Perform action if nothing is returned - bash

I'm trying to echo "It was not found" if the netstat returns no result. But, if it does return a result, then to display the netstat results.
I'm trying to google for what I'm using and can't find much about it.
#!/bin/bash
echo "Which process would you like to run netstat against?"
read psname
echo "Looking for '$psname'"
sleep 2
command=$(netstat -tulpn | grep -e $psname)
[[ ! -z $command ]]
It's to do with the [[ ! -z $command ]]

[[ ! -z $command ]] doesn't 'show' you any output.
Use a if/else setup to show the result;
if [[ ! -z $command ]]; then
echo "$psname was found!"
else
echo "No process named '${psname}' found!"
fi
Or a shorthand variant;
[[ ! -z $command ]] && echo "$psname was found!" || echo "No process named '${psname}' found!"
Note if the user input may contain a space, it's saver to use "" around the grep string;
command=$(netstat -tulpn | grep -e "$psname")
When to wrap quotes around a shell variable?

Do not intercept the output directly. You can instead do something like:
if ! netstat -tulpn | grep -e "$psname"; then
echo "not found" >&2
fi
If grep matches any output, it will print it to stdout. If it mathches nothing, it returns non-zero and the shell will write a message to stderr.

Related

Setting a variable inside a conditional

Is it possible to set a variable from the output of a command inside of a conditional where the conditional is false if nothing gets assigned to the variable.
If I set the variable to a grep with no return and then test:
test=$(echo hello | grep 'helo')
if [[ ! -z $test ]]; then
echo "is set"
else
echo "not set"
fi
Output: not set (this is expected)
But I'm trying to put it all into one statement like this:
test=
if [[ ! -z test=$(echo hello | grep 'helo') ]]; then
echo "is set"
else
echo "not set"
fi
output: "is set" (expected not set)
grep returns success if there is a match, so you can just do:
if test=$(echo hello | grep 'helo')
then
echo "Match: $test"
else
echo "No match"
fi
If you're running something that doesn't differentiate by exit code, you can assign and check in two statements on the same line:
if var=$(cat) && [[ -n $var ]]
then
echo "You successfully piped in some data."
else
echo "Error or eof without data on stdin."
fi
(or ; instead of && if you want to inspect the result even when the command reports failure)
Bit of a hack, using the shell's parameter expansion alternate value syntax, echo -e and some backspaces:
test=$(echo hello | grep 'helo'); echo -e not${test:+\\b\\b\\bis} set
Which outputs is set or not set depending on what grep finds.

shell script handle an error from a command being excuted from echo

Hello I am new to shell script and I need to handle the error coming from a command being excuted inside echo like the following
echo -e "some internal command that I can't share \nq" | nc localhost 10000
I want to say
if [[ there's no error ]]
try
echo "YOUR SUPERSECRET COMMAND" | nc localhost 10000 | grep "Your expected error"
if [[ $? -eq 0 ]]; then
echo "Do something useful with error"
else
echo "Success"
fi
grep return 0 on matching and returns 1 when it doesn't find matching string.
The shell variable $? will give you the exit code. So, you could do:
echo -e "some internal command that I can't share \nq" | nc localhost 10000
rc=$?
if [[ $rc == 0 ]]; then
echo "success"
fi
Or simply,
if echo -e "some internal command that I can't share \nq" | nc localhost 10000; then
echo "success"
fi
Here is a concise way of doing it:
internalcommand &>/dev/null 2>&1 && echo OK || echo FAILED
If internalcommand succeeds OK will be printed to stdout, if it fails FAILED is printed.
Note tested on Bash v4

Bash script ends without doing echo

I am running a script from command line that runs a bunch of Jmeter tests if 4 variables are set.
The script works but I have added parts so the script will end if the server is unknown.
if [ echo "$2" != | grep -iq "^hibagon" ] || [ echo "$2" != | grep -iq "^kameosa" ] ;then
echo "Unkown server stopping tests"
else
echo "Continueing to tests"
when this section of the script runs it will end the script if hibagon or kameosa is not found (not case sensitive).
I want the command line to echo Unknown server stopping tests then end but at the moment it just ends with no echo
That's a strange syntax. Try this:
if echo "$2" | grep -iq "^hibagon\|^kameosa";
then
echo "Continuing to tests"
else
echo "Unkown server, stopping tests"
fi
Or if you're using bash:
if [[ "$2" =~ ^hibagon ]] || [[ "$2" =~ ^kameosa ]]
then
echo "Continuing to tests"
else
echo "Unkown server, stopping tests"
fi
First the test [ echo "$2" != | grep -iq "^hibagon" ] is wrong, then you can use only one (extended)grep with the negation flag -v putting together the two words in one regex ^(hibagon|kameosa). Also the fi was missing. but I suppose it's only a typo here.
if echo "$2" | egrep -ivq "^(hibagon|kameosa)"; then
echo "Unknown server stopping tests"
else
echo "Continuing to tests"
fi
If you like it, even:
if egrep -ivq "^(hibagon|kameosa)" <<< "$2"; then

getops still performs default actions when arguments are provided

I've recently started working with the getopts command in bash. I am confused as to why my script runs the dafult action "cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl" when arguments have been provided. I only want that to run if no arguments were passed to the shell script. I've used getopts:Std in perl where I was able to code somthing like:
unless ($opts{d}) {
do something...}
How would I code something like that in a shell script? Also, how would I code logic such as this:
if ($opts{c}) {
cat ~bin/Temp/mag.txt | ~bin/Scripts/report.pl -c
}
elsif ($opts{d} {
cat ~bin/Temp/mag.txt | ~bin/Scripts/report.pl -d
My code:
#!/bin/sh
while getopts cd name
do
case $name in
c)copt=1;;
d)dopt=1;;
*)echo "Invalid arg";;
esac
done
if [[ ! -z $copt ]] #Specifies what happens if the -c argument was provided
then
echo "CSV file created!"
cat "~/bin/Temp/log.txt" | ~/bin/Scripts/vpnreport/report.pl -c
fi
if [[ ! -z $dopt ]] #Specifies what happens if the -d argument was provided
then
echo "Debug report and files created"
cat ~bin/Temp/mag.txt | ~bin/Scripts/report.pl -d
fi
if [[ ! -z $name ]] #Specifies what happens if no argument was provided
then
echo "Running standard VPN report"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl
fi
shift $(($OPTIND -1))
My Output:
[~/bin/Scripts/report]$ sh getoptstest.sh
Running standard report
[~/bin/Scripts/report]$ sh getoptstest.sh -d
Debug report and files created
Running standard report
[~/bin/Scripts/report]$
The two getopts commands are vasty different from bash to perl and I just can't seem to get the hang of the bash varient even after reading several tutorials. Any help would be greatly appreciated!
On the final run of getopts, your variable (name) will be set to "?".
#!/bin/bash
while getopts abc foo; do :; done
echo "<$foo>"
Output of the above:
$ ./mytest.sh
<?>
$ ./mytest.sh -a
<?>
Insead, use elif, which is like Perl's elsif:
if [[ ! -z $copt ]]
then
# ...
elif [[ ! -z $dopt ]]
then
# ...
else
# ...
fi
Or test if [[ -z $copt && -z $dopt ]], or so forth. Other notes:
See the official if and case documentation in the Bash manual under "Conditional Constructs".
[[ ! -z $name ]] means the same as the more-direct [[ -n $name ]].
Use #!/bin/bash instead of #!/bin/sh, or switch off of [[ in favor of [. The double square bracket (and your use thereof) is specific to bash, and rarely works with sh.
I took Jeff's answer and rewrote my script so it works now:
#!/bin/bash
while getopts cd name
do
case $name in
c)carg=1;;
d)darg=1;;
*)echo "Invalid arg";;
esac
done
#Specifies what happens if the -c argument was provided:
if [[ ! -z $carg ]]
then
if [[ -z $darg ]]
then
echo "CSV created"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl -c
else
echo "Debug CSV created"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl -cd
fi
fi
#Specifies what happens if the -d argurment was provided:
if [[ ! -z $darg ]]
then
echo "Debug report created"
cat ~bin/Temp/log.txt | ~bin/Scripts/report.pl -d
#Specifies what happens if no argument was provided:
else
echo "Standard report created"
cat ~bin/Temp/logs.txt | ~bin/Scripts/report.pl
fi
Thank you again for your assistance!

How to check if a file contains a specific string using Bash

I want to check if a file contains a specific string or not in bash. I used this script, but it doesn't work:
if [[ 'grep 'SomeString' $File' ]];then
# Some Actions
fi
What's wrong in my code?
if grep -q SomeString "$File"; then
Some Actions # SomeString was found
fi
You don't need [[ ]] here. Just run the command directly. Add -q option when you don't need the string displayed when it was found.
The grep command returns 0 or 1 in the exit code depending on
the result of search. 0 if something was found; 1 otherwise.
$ echo hello | grep hi ; echo $?
1
$ echo hello | grep he ; echo $?
hello
0
$ echo hello | grep -q he ; echo $?
0
You can specify commands as an condition of if. If the command returns 0 in its exitcode that means that the condition is true; otherwise false.
$ if /bin/true; then echo that is true; fi
that is true
$ if /bin/false; then echo that is true; fi
$
As you can see you run here the programs directly. No additional [] or [[]].
In case if you want to check whether file does not contain a specific string, you can do it as follows.
if ! grep -q SomeString "$File"; then
Some Actions # SomeString was not found
fi
In addition to other answers, which told you how to do what you wanted, I try to explain what was wrong (which is what you wanted.
In Bash, if is to be followed with a command. If the exit code of this command is equal to 0, then the then part is executed, else the else part if any is executed.
You can do that with any command as explained in other answers: if /bin/true; then ...; fi
[[ is an internal bash command dedicated to some tests, like file existence, variable comparisons. Similarly [ is an external command (it is located typically in /usr/bin/[) that performs roughly the same tests but needs ] as a final argument, which is why ] must be padded with a space on the left, which is not the case with ]].
Here you needn't [[ nor [.
Another thing is the way you quote things. In bash, there is only one case where pairs of quotes do nest, it is "$(command "argument")". But in 'grep 'SomeString' $File' you have only one word, because 'grep ' is a quoted unit, which is concatenated with SomeString and then again concatenated with ' $File'. The variable $File is not even replaced with its value because of the use of single quotes. The proper way to do that is grep 'SomeString' "$File".
Shortest (correct) version:
grep -q "something" file; [ $? -eq 0 ] && echo "yes" || echo "no"
can be also written as
grep -q "something" file; test $? -eq 0 && echo "yes" || echo "no"
but you dont need to explicitly test it in this case, so the same with:
grep -q "something" file && echo "yes" || echo "no"
##To check for a particular string in a file
cd PATH_TO_YOUR_DIRECTORY #Changing directory to your working directory
File=YOUR_FILENAME
if grep -q STRING_YOU_ARE_CHECKING_FOR "$File"; ##note the space after the string you are searching for
then
echo "Hooray!!It's available"
else
echo "Oops!!Not available"
fi
grep -q [PATTERN] [FILE] && echo $?
The exit status is 0 (true) if the pattern was found; otherwise blankstring.
if grep -q [string] [filename]
then
[whatever action]
fi
Example
if grep -q 'my cat is in a tree' /tmp/cat.txt
then
mkdir cat
fi
In case you want to checkif the string matches the whole line and if it is a fixed string, You can do it this way
grep -Fxq [String] [filePath]
example
searchString="Hello World"
file="./test.log"
if grep -Fxq "$searchString" $file
then
echo "String found in $file"
else
echo "String not found in $file"
fi
From the man file:
-F, --fixed-strings
Interpret PATTERN as a list of fixed strings, separated by newlines, any of
which is to be matched.
(-F is specified by POSIX.)
-x, --line-regexp
Select only those matches that exactly match the whole line. (-x is specified by
POSIX.)
-q, --quiet, --silent
Quiet; do not write anything to standard output. Exit immediately with zero
status if any match is
found, even if an error was detected. Also see the -s or --no-messages
option. (-q is specified by
POSIX.)
Try this:
if [[ $(grep "SomeString" $File) ]] ; then
echo "Found"
else
echo "Not Found"
fi
I done this, seems to work fine
if grep $SearchTerm $FileToSearch; then
echo "$SearchTerm found OK"
else
echo "$SearchTerm not found"
fi
grep -q "something" file
[[ !? -eq 0 ]] && echo "yes" || echo "no"

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