I have written some code that uses the /bin/true by accident and I want to force the usage of the builtin of true and false.. any Idea what is the correct way of doing so?
#!/usr/bin/env bash
set -eEuo pipefail
CONFIRM=false
if [[ $1 == "Y" ]]; then
CONFIRM=true
fi
"${CONFIRM}" || echo "Deletion will not happen, please run with '${0} Y' to confirm"
if $CONFIRM; then
echo "I deleted the stuff!"
fi
To me, this looks like a job for if ... else:
#!/usr/bin/env bash
set -eEuo pipefail
if [[ $# -eq 1 && $1 == Y ]]; then
echo "I deleted the stuff!"
else
echo "Deletion will not happen, please run with '${0} Y' to confirm"
fi
If you need to save the result:
#!/usr/bin/env bash
set -eEuo pipefail
# with the set-mode above, you can set `conform` like this:
[[ $# -eq 1 && $1 == Y ]] && confirm=$? || confirm=$?
# otherwise, you could just set it after the above test
# truth test
[[ $confirm -eq $() ]] || echo "Deletion will not happen, please run with '${0} Y' to confirm"
# truth test again
if [[ $confirm -eq $() ]]; then
echo "I deleted the stuff!"
fi
Related
I wanna write a script which check variables of this script.
I have tried some, but it isn't working. The idea is:
If on of the parameters is a number, print that it is number
If on of the parameters is a character, print that it is character
If 'man parameter' is executable, print that it is might be a function
Script I have tried:
#!/bin/bash
echo Hello $LOGNAME'!'
test $# -eq 0 && echo 'Try again, no parameters were entered' || echo 'Num of parameters is '$#
re='^[0-9]+$'
for i in $*
do
if ![["$i" =~ $re]];then
echo 'Parameter '$i' is alphabetical'
else
if [["$i" =~ $re]];then
echo 'Parameter '$i' is digital'
else
if [ $i];then
echo $i' might be a function. Try to use man of --help'
fi
fi
fi
done
#!/bin/bash
echo "Hello ${LOGNAME}!"
[ "$#" -eq 0 ] && { echo 'Try again, no parameters were entered'; exit 1; }
echo 'Num of parameters is '$#
re='^[0-9]+$'
for i in "$#"
do
if ! [[ "$i" =~ $re ]];then
echo "Parameter '$i' is alphabetical"
man "$i" > /dev/null 2> /dev/null
if [ "$?" -eq 0 ];then
echo "$i might be a function. Try to use man of --help"
fi
else
echo "Parameter '$i' is digital"
fi
done;
When you write a test you need spaces around your brackets.
You can easily find and fix those bugs if you use shellcheck
#!/bin/bash
# Code to generate script usage
if [[ "$#" -ne 1 ]] && [[ "$#" -ne 2 ]]; then
flag=1;
elif ! [[ "$1" == "abcd" || "$1" == "dcba" ]]; then
echo "Invalid"
flag=1;
fi
while [ $# -gt 1 ]
do
case $2 in
'streams')
;;
*)
echo "unrecognised optional arg $2"; flag=1;
;;
esac
shift
done
if [ "$flag" == "1" ]; then
echo "Usage:"
exit
fi
function main {
arg1=$1
streams=$2
if [ "${streams}" == "streams" ]; then
echo entering here
else
echo entering there
fi
}
parent_dir=`pwd`
find $parent_dir -name "*" -type d | while read d; do
cd $denter code here
main $1 $2
done
Why the code does not enter "entering here" when script run with arguments "abcd" and "streams" ?
I feel that function having two arguments is causing the problem, code was working fine with one argument
Several things you might want to fix in your code, before attempts are made to find the specific problem. It is possible that it will disappear after modifying your script accordingly. If the problem is still alive, I'll edit my answer with a solution. If you decide to apply the following changes, please update your code in the question.
Consistent usage of either [[ or [. [[ is a Bash keyword similar to (but more powerful than) the [ command.
See
Bash FAQ 31
Tests And Conditionals
Unless you're writing for POSIX sh, I recommend [[.
Use (( for arithmetic expressions. ((...)) is an arithmetic command, which returns an exit status of 0 if the expression is nonzero, or 1 if the expression is zero. Also used as a synonym for let, if assignments are needed. See Arithmetic Expression.
Use the variable PWD instead of pwd. PWD is a builtin variable in all POSIX shells that contains the current working directory. pwd(1) is a POSIX utility that prints the name of the current working directory to stdout. Unless you're writing for some non-POSIX system, there is no reason to waste time executing pwd(1) rather than just using PWD.
The function keyword is not portable. I suggest you to avoid using it and simply write function_name() { your code here; } # Usage
$parent_dir is not double-quoted. "Double quote" every literal that contains spaces/metacharacters and every expansion: "$var", "$(command "$var")", "${array[#]}", "a & b". See
Quotes
Arguments
ShellCheck your code before uploading.
Replace the while condition logic with an if condition, so that shift is no longer required. Shift was the devil I was facing I found.
#!/bin/bash
# Code to generate script usage
if [[ "$#" -ne 1 ]] && [[ "$#" -ne 2 ]]; then
flag=1;
elif ! [[ "$1" == "abcd" || "$1" == "dcba" ]]; then
echo "Invalid"
flag=1;
fi
#while [[ $# -gt 1 ]]
#do
# case $2 in
# 'streams')
# ;;
# *)
# echo "unrecognised optional arg $2"; flag=1;
# ;;
# esac
# shift
#done
if [[ $2 == "streams" ]]; then
:
elif [[ (-z $2) ]]; then
:
else
echo "unrecognised optional arg $2"; flag=1;
fi
if [[ "$flag" == "1" ]]; then
echo "Usage:"
exit
fi
function main {
streams=$2
if [[ "${streams}" == "streams" ]]; then
echo entering here
else
echo entering there
fi
}
parent_dir=`pwd`
find $parent_dir -name "*" -type d | while read d; do
cd $d
main $1 $2
done
For some reason, I can't figure out how to test truth in bash:
#!/bin/bash
FORCE_DELETE=""
BE_VERBOSE=""
OPTIND=1
while getopts ":fv" FLAG "$#" ; do
if [[ "$FLAG" == "f" ]] ; then
FORCE_DELETE="true"
fi
if [[ "$VALUE" == "v" ]] ; then
BE_VERBOSE="true"
fi
if [[ "$FLAG" == "?" ]] ; then
echo "Usage: $0 [-fv] file ..."
exit 1
fi
done
shift `expr $OPTIND - 1`
if [[ "$FORCE_DELETE" == "true" && "BE_VERBOSE" == "true" ]] ; then
echo "FORCE_DELETE AND BE_VERBOSE $#"
elif [[ "$FORCE_DELETE" == "true" ]] ; then
echo "FORCE_DELETE $#"
elif [[ "$BE_VERBOSE" == "true" ]] ; then
echo "BE_VERBOSE $#"
else
echo "$#"
fi
exit 0
Transcript:
$ test a b
a b
$ test -f a b
FORCE_DELETE a b
$ test -v a b
a b
$ test -fv a b
FORCE_DELETE a b
Why does my bash script respond to the -f flag but not the -v flag?
Most likely a typo :
[[ "$VALUE" == "v" ]],
this should be
[[ "$FLAG" == "v" ]]
You specifically ask about testing true/false. These are built in to the language rather than using strings, and you don't need the [[ test. Here is how I would write this:
#!/bin/bash
force_delete=false # Don't use UPPERCASE
be_verbose=false # they could collide with reserved variables
# OPTIND does not need to be initialised
while getopts :fv flag
do
# appears one of your if statements is incorrect
# a case is often used with getopts
case $flag in
f) force_delete=true
;;
v) be_verbose=true
;;
\?) echo "Usage: $0 [-fv] file ..."
exit 1
;;
esac
done
shift $((OPTIND-1)) # don't create a child process for simple arithmetic
if $force_delete && $be_verbose
then
echo "force_delete AND be_verbose $#"
elif $force_delete
then
echo "force_delete $#"
elif $be_verbose
then
echo "be_verbose $#"
else
echo "$#"
fi
# Bash exits 0 by default
I am attempting to run a block of code if one flag is set to true and the other is set to false. ie
var1=true
var2=false
if [[ $var1 && ! $var2 ]]; then var2="something"; fi
Since that did not evaluate the way that I expected I wrote several other test cases and I am having a hard time understanding how they are being evaluated.
aa=true
bb=false
cc="python"
if [[ "$aa" ]]; then echo "Test0" ; fi
if [[ "$bb" ]]; then echo "Test0.1" ; fi
if [[ !"$aa" ]]; then echo "Test0.2" ; fi
if [[ ! "$aa" ]]; then echo "Test0.3" ; fi
if [[ "$aa" && ! "$bb" ]]; then echo "Test1" ; fi
if [[ "$aa" && ! "$aa" ]]; then echo "Test2" ; fi
if [[ "$aa" ]] && ! [[ "$bb" ]]; then echo "test3" ; fi
if [[ "$aa" ]] && ! [[ "$cc" ]]; then echo "test4" ; fi
if [[ $aa && ! $bb ]]; then echo "Test5" ; fi
if [[ $aa && ! $aa ]]; then echo "Test6" ; fi
if [[ $aa ]] && ! [[ $bb ]]; then echo "test7" ; fi
if [[ $aa ]] && ! [[ $cc ]]; then echo "test8" ; fi
When I run the preceding codeblock the only output I get is
Test0
Test0.1
Test0.2
however, my expectation is that I would get
Test0
Test1
Test3
Test5
Test7
I have tried to understand the best way to run similar tests, however most examples I have found are set up in the format of
if [[ "$aa" == true ]];
which is not quite what I want to do. So my question is what is the best way to make comparisons like this, and why do several of the test cases that I would expect to pass simply not?
Thank you!
Without any operators, [[ only checks if the variable is empty. If it is, then it is considered false, otherwise it is considered true. The contents of the variables do not matter.
Your understanding of booleans in shell context is incorrect.
var1=true
var2=false
Both the above variables are true since those are non-empty strings.
You could instead make use of arithmetic context:
$ a=1
$ b=0
$ ((a==1 && b==0)) && echo y
y
$ ((a==0 && b==0)) && echo y
$
$ ((a && !(b))) && echo y; # This seems to be analogous to what you were attempting
y
The shell does not have Boolean variables, per se. However, there are commands named true and false whose exit statuses are 0 and 1, respectively, and so can be used similarly to Boolean values.
var1=true
var2=false
if $var1 && ! $var2; then var2="something"; fi
The difference is that instead of testing if var1 is set to a true value, you expand it to the name of a command, which runs and succeeds. Likewise, var2 is expanded to a command name which runs and fails, but because it is prefixed with ! the exit status is inverted to indicate success.
(Note that unlike most programming languages, an exit status of 0 indicates success because while most commands have 1 way to succeed, there are many different ways they could fail, so different non-zero values can be assigned different meanings.)
true and false are evaluated as strings ;)
[[ $var ]] is an equivalent of [[ -n $var ]] that check if $var is empty or not.
Then, no need to quote your variables inside [[. See this reminder.
Finally, here is an explication of the difference between && inside brackets and outside.
The closest you can come seems to be use functions instead of variables because you can use their return status in conditionals.
$ var1() { return 0; }
$ var2() { return 1; } # !0 = failure ~ false
and we can test this way
$ var1 && echo "it's true" || echo "it's false"
it's true
$ var2 && echo "it's true" || echo "it's false"
it's false
or this way
$ if var1; then echo "it's true"; else echo "it's false"; fi
it's true
$ if var2; then echo "it's true"; else echo "it's false"; fi
it's false
Hope this helps.
Is it possible to override Bash's test builtin? So that
[[ $1 = 'a' ]]
not just does the test but also outputs which result was expected when it fails? Something like
echo "Expected $1 to be a.'
EDIT
I know this is bad :-).
The test expression compound command does real short-circuiting that affects all expansions.
$ set -x
$ [[ 0 -gt x=1+1 || ++x -eq $(tee /dev/fd/3 <<<$x) && $(echo 'nope' >&3) ]] 3>&1
+ [[ 0 -gt x=1+1 ]]
++ tee /dev/fd/2
2
+ [[ ++x -eq 2 ]]
So yes you could do anything in a single test expression. In reality it's quite rare to have a test produce a side-effect, and almost never used to produce output.
Also yes, reserved words can be overridden. Bash is more lenient with ksh-style function definitions than POSIX style (which still allows some invalid names).
function [[ { [ "${#:1:${##}-1}" ]; }; \[[ -a -o -a -o -a ]] || echo lulz
Yet another forky bomb.
if function function if function if if \function & then \if & fi && \if & then \function & fi && then \function fi
Something like this?
if [[ $1 == 'a' ]]; then
echo "all right";
else
echo 'Expected $1 to be "a"'
fi
Anyway, what's the point of the test if you only expect one answer? Or do you mean that for debugging purposes?
[[ 'a' = 'a' ]] || echo "failed"
[[ 'b' = 'a' ]] || echo "failed"
failed