if [-z $var1 ] && [$var2 == "1" ]; then not working as expected [duplicate] - bash

This question already has answers here:
Getting "command not found" error while comparing two strings in Bash
(4 answers)
Closed 3 years ago.
I'm writing a bash script where i need to combine two conditions with && operator
var1=value
var2=1
if [-z $var1 ] && [$var2=="1"]; then
do something
else
do something else
fi
but it always executes the else part.
My research
google gave me bash conditions like this but its not working for me.
if [condition1] && [condition2]; then
do something
fi
Another method i tried is this but it still completely ignored the true part
if [[-z $var1 ]] && [$var2=="1"]; then
do something
else
do something else
fi
Tried with -a operator like this
if [-z $var1 -a $var2=="1" ]; then
tried nested if
if [-z $var1 ]; then
if [$var2=="1"]; then
do something
fi
else
do something else
fi
So i know i am doing something wrong but my goal is i want to check for any value in $var1 and also want $var2 condition to be true and execute the true part.
UPDATE
i tried this
#! /bin/bash
set -o nounset
#set -o errexit
set -o pipefail
set -o xtrace
var1=pop
var2=1
if test -z "$var1" && test "$var2" -eq 1; then
echo Y1
fi
if [ -z "$var1" ] && [ "$var2" -eq 1 ]; then
echo Y2
fi
and this is the output
root#c847b6423295:/# ./test.sh
+ var1=pop
+ var2=1
+ test -z pop
+ '[' -z pop ']'
root#c847b6423295:/#
What am i doing wrong ?

You want to encode "$var1 is not empty and $var2 is equal to 1", you can do:
if test -z "$var1" && test "$var2" -eq 1; then
echo Y
fi
which is equivalent to:
if [ -z "$var1" ] && [ "$var2" -eq 1 ]; then
echo Y
fi
If you want to compare strings, use a single equals sign:
if test "aaa" = "aaa"; then
echo Y
fi

Related

Bash if statement with prepended character [duplicate]

This question already has answers here:
Why do shell script comparisons often use x$VAR = xyes?
(7 answers)
Closed 4 years ago.
I'm trying to understand a script that will stop when executed as root:
#!/usr/bin/env bash
if [ x"$(whoami)" = x"root" ]; then
echo "Error: don't run this script as root"
exit 1
fi
I have tested this and it works as intended even if I remove the x in the if statement. My question is why is the x in x"$(whoami)" and x"root" needed?
basically the [ is a softlink to an external program called test, therefore the condition is passed to it as program arguments, and doing so if you don't surround a $variable with "$quotes" , and the variable happens to be empty it won't be considered as an empty argument, it will be considered as no argument (nothing)
#!/bin/bash -eu
var=bla
if [[ $var == bla ]];then
echo first test ok
fi
var=""
if [[ $var == "" ]];then
echo second test ok
fi
if [ "$var" == "" ];then
echo third test ok
fi
if [ x$var == "x" ];then
echo fourth test ok
fi
echo this will fail:
if [ $var == "" ];then
echo fifth test ok
fi
echo because it is the same as writing:
if [ == "" ];then
echo sixth test is obviously eroneous
fi
echo but also you should quote your variables because this will work:
var="a b"
if [ "$var" == "a b" ];then
echo seventh test ok
fi
echo ... but this one won\'t as test now has four arguments:
if [ $var == "a b" ];then
echo eighth test ok
fi

How to check if multiple variables are defined or not in bash

I want to check, if multiple variable are set or not, if set then only execute the script code, otherwise exit.
something like:
if [ ! $DB=="" && $HOST=="" && $DATE=="" ]; then
echo "you did not set any variable"
exit 1;
else
echo "You are good to go"
fi
You can use -z to test whether a variable is unset or empty:
if [[ -z $DB || -z $HOST || -z $DATE ]]; then
echo 'one or more variables are undefined'
exit 1
fi
echo "You are good to go"
As you have used the bash tag, I've used an extended test [[, which means that I don't need to use quotes around my variables. I'm assuming that you need all three variables to be defined in order to continue. The exit in the if branch means that the else is superfluous.
The standard way to do it in any POSIX-compliant shell would be like this:
if [ -z "$DB" ] || [ -z "$HOST" ] || [ -z "$DATE" ]; then
echo 'one or more variables are undefined'
exit 1
fi
The important differences here are that each variable check goes inside a separate test and that double quotes are used around each parameter expansion.
If you are ok with writing a function for this purpose, it can be pretty convenient.
This solution uses the ${!VAR_NAME} syntax to check whether the variable is empty and has the added benefit of telling you which variable names are empty.
check_vars()
{
var_names=("$#")
for var_name in "${var_names[#]}"; do
[ -z "${!var_name}" ] && echo "$var_name is unset." && var_unset=true
done
[ -n "$var_unset" ] && exit 1
return 0
}
# Usage for this case
check_vars DB HOST DATE
echo "You are good to go"
I wound up using variable-variables to loop through an easily managed HEREDOC list of variable names:
# Ensure non-empty values.
# Loop through HEREDOC, test variable-variable isn't blank.
while read var; do
[ -z "${!var}" ] && { echo "$var is empty or not set. Exiting.."; exit 1; }
done << EOF
KUBE_NAMESPACE
DOCKER_REGISTRY
DOCKER_DEPLOY_USER
DOCKER_DEPLOY_PASSWORD
DOCKER_DEPLOY_EMAIL
EOF
You can check it also by put the variables name in a file
DB=myDB
HOST=myDB
DATE=myDATE
then test them if currently empty or unset
#!/bin/bash
while read -r line; do
var=`echo $line | cut -d '=' -f1`
test=$(echo $var)
if [ -z "$(test)" ]; then
echo 'one or more variables are undefined'
exit 1
fi
done <var.txt
echo "You are good to go"
Nice solution from #joe.still !
improvement is to exit after checking all variables
i=0
while read var; do
[ -z "${!var}" ] && { echo "$var is empty or not set. Exiting.."; let i=i+1; }
done << EOF
KUBE_NAMESPACE
DOCKER_REGISTRY
DOCKER_DEPLOY_USER
DOCKER_DEPLOY_PASSWORD
DOCKER_DEPLOY_EMAIL
EOF
if [ $i -gt 0 ]; then
echo $i
echo "exiting"
exit 1
fi
Good Day Everyone.
I've personally used this method in my bash scripts. Verified works on bash 4.4 and later in Ubuntu, openSUSE, and ClearLinux.
Can RHEL|CentOS|Alma and Arch Based users let me know it it works fine for you?
( [ "$VAR1""$VAR2""$VAR3""$VAR4""$VAR5" ] && echo -e " Warning: StackIsNotClear" ) || { echo -e " GoodNews: StackIsClear"; }

How to validate shell arguments?

I am trying to write a simple sh script that must be invoked with 2 arguments:
sh myscript.sh --user "some user" --fizz "buzz"
At the top of myscript.sh I have:
#!/bin/sh
# VALIDATION
# 1. Make sure there are 5 positional arguments (that $4 exists).
die () {
echo >&2 "$#"
exit 1
}
[ "$#" -eq 5 ] || die "5 arguments required, $# provided"
# 2. Make sure $1 is "-u" or "--user".
# 3. Make sure $3 is "-f" or "--fizz".
If validation fails, I'd like to print a simple usage message and then exit the script.
I think I have #1 correct (checking # of positional arguments), but have no clue how to implement #2 and #3. Ideas?
# 2. Make sure $1 is "-u" or "--user".
if ! [ "$1" = -u -o "$1" = --user ]; then
# Test failed. Send a message perhaps.
exit 1
fi
# 3. Make sure $3 is "-f" or "--fizz".
if ! [ "$3" = -f -o "$3" = --fizz ]; then
# Test failed. Send a message perhaps.
exit 1
fi
Other forms for testing a variable for two possible possible values:
[ ! "$var" = value1 -a ! "$var" = value2 ]
[ ! "$var" = value1 ] && [ ! "$var" = value2 ]
! [ "$var" = value1 && ! [ "$var" = value2 ]
For Bash and similarly syntaxed shells:
! [[ $var = value1 || $var = value2 ]]
[[ ! $var = value1 || ! $var = value2 ]]
Besides using negated conditions with if blocks, you can also have positive conditions with ||
true_condition || {
# Failed. Send a message perhaps.
exit 1
}
true_condition || exit 1
Of course && on the other hand would apply with negated conditions.
Using case statements:
case "$var" in
value1|value2)
# Valid.
;;
*)
# Failed.
exit 1
;;
esac
Manually:
if [ -z "$1" ];then
fi
if [ -z "$2" ];then
fi
if [ -z "$3" ];then
fi
...
Or check getopt
while getopts "uf" OPTION
do
case $OPTION in
u)
echo "-u"
;;
f)
echo "-f"
;;
esac
done

Bash Boolean testing

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.

bash - Possible to 'override' the test ([[)-builtin?

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

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