I'm writing a script that checks for 3 different variables and matches on true. However I need it to only return true if ALL conditions match. Right now its returning true if only one or 2 conditions match from my list of operators.
if [ -z "$VAR3" ] || [ "$VAR1" -ge 10 ] || [ "$VAR2" != "WORKING" ]; then
echo "app failed "$VAR1" check-ins"
exit 2
elif [ -n "$VAR3" ] || [ "$VAR1" -le 10 ] || [ "$VAR2" == "WORKING" ]; then
echo "$VAR3" pid active connection is "$VAR2"
exit 0
fi
Should I just replace the || with && ??
Yes, || represents the OR logical operator, it returns true if at least one of the operators is true. && represents the AND logical operator, it returns true if and only if all of the operators are true
if [ -z "$VAR3" ] && [ "$VAR1" -ge 10 ] && [ "$VAR2" != "WORKING" ]; then
echo "app failed "$VAR1" check-ins"
exit 2
elif [ -n "$VAR3" ] && [ "$VAR1" -le 10 ] && [ "$VAR2" == "WORKING" ]; then
echo "$VAR3" pid active connection is "$VAR2"
exit 0
fi
Related
I have to find a way to have my script read from one of these three options:
a file argument
standard input
a previously established environment variable
Here's what I currently have:
#!/bin/bash
key=$1
[ $# -ge 1 -a -f "$2" ] && input="$2" || [ -f "$INPUT" ] && input="$INPUT" || input="-"
echo $input
Only the environment variable refuses to work, the rest works fine.
I've tried using the export INPUT="pathnametofile" before but it doesn't make any difference, I end up with the shell asking me to enter info as if I called on cat.
The problem in your script
Your attemp is not working due to the way the shell processes a Lists of Commands:
‘&&’ and ‘||’ have equal precedence.
AND and OR lists are executed with left associativity.
Your sentence:
[ $# -ge 1 -a -f "$2" ] && input="$2" || [ -f "$INPUT" ] && input="$INPUT" || input="-"
does the same as follows:
[ $# -ge 1 -a -f "$2" ] && input="$2"
[ $? -eq 0 ] || [ -f "$INPUT" ]
[ $? -eq 0 ] && input="$INPUT"
[ $? -eq 0 ] || input="-"
Now yo may see why your unexpected behaviour.
A better attempt grouping commands
{ [ $# -ge 1 -a -f "$2" ] && input="$2"; } || { [ -f "$INPUT" ] && input="$INPUT"; } || input="-"
Now, due to precedence, the first group is not needed at all:
[ $# -ge 1 -a -f "$2" ] && input="$2" || { [ -f "$INPUT" ] && input="$INPUT"; } || input="-"
Furthermore, unless you have set the positional parameters by hand, you can remove the first check (after all, if $2 is emtpy, -f "" fails the same).
[ -f "$2" ] && input="$2" || { [ -f "$INPUT" ] && input="$INPUT"; } || input="-"
An alternative with the if conditional construct
if [ -f "$2" ]; then
input=$2
elif [ -f "$INPUT" ]; then
input=$INPUT
fi
echo "${input:=-}"
untested, but you'll probably have better luck with if commands, and test that the variable is not empty:
if [ $# -ge 1 -a -f "$2" ]; then
input="$2"
elif [ -n "$INPUT" -a -f "$INPUT" ]; then
input="$INPUT"
else
input="-"
fi
Suppose I am writing the following in a bash script:
if [ -z $a ] || [ -z $b ] ; then
usage
fi
It works but I would like to write it with short-circuiting as follows:
[ -z $a ] || [ -z $b ] || usage
Unfortunately it does not work. What am I missing ?
You want to execute usage in case either 1st or 2nd condition are accomplished. For that, you can do:
[ -z $a ] || [ -z $b ] && usage
Test:
$ [ -z "$a" ] || [ -z "$b" ] && echo "yes"
yes
$ b="a"
$ [ -z "$a" ] || [ -z "$b" ] && echo "yes"
yes
$ a="a"
$ [ -z "$a" ] || [ -z "$b" ] && echo "yes"
$
You could make use of the following form:
[[ expression ]]
and say:
[[ -z "$a" || -z "$b" ]] && usage
This would execute usage if either a or b is empty.
Always quote your variables. Saying
[ -z $a ]
if the variable a is set to foo bar would return an error:
bash: [: foo: binary operator expected
When I run this bash script :
if [ [$EUID -ne 0] ]; then
echo "This script must be run as root" 1>&2
exit 1
else
printf " whathever "
exit 0
fi
I have this error :
./myScript: 15: [: Illegal number: [
Do you see any problem ?
You have syntax error in your if condition, use this if condition:
if [ "$EUID" -ne 0 ];
OR using [[ and ]]
if [[ "$EUID" -ne 0 ]];
You have syntax error in your if condition, use this if condition:
if [ "$EUID" -ne 0 ];
OR using [[ and ]]
if [[ "$EUID" -ne 0 ]];
If you use the KSH88+/Bash 3+ internal instruction [[, it's not necessary to use doubles quotes around the variables operands :
[ ~/test]$ [[ $var2 = "string with spaces" ]] && echo "OK" || echo "KO"
OK
Instead of the external command test or his fork [ :
[ ~/test]$ [ $var2 = "string with spaces" ] && echo "OK" || echo "KO"
bash: [: too many arguments
KO
[ ~/test]$ [ "$var2" = "string with spaces" ] && echo "OK" || echo "KO"
OK
Of course, you also have to choose the operators according to the type of operands :
[ ~/test]$ var1="01"
[ ~/test]$ [ "$var1" = "1" ] && echo "OK" || echo "KO"
KO
[ ~/test]$ [ "$var1" -eq "1" ] && echo "OK" || echo "KO"
OK
two suggestions apart from what everyone else has pointed out already.
rather than doing else [bunch of code because we are root] fi, just replace the else with fi. once you've tested for the failure condition you are concerned about and taken appropriate action, no need to continue to be within the body of the conditional.
$EUID is a bashism, if you would like to make this portable to shells such as ksh, replacing it with:
if [ $(id -u) -ne 0 ]; then echo "ur not root bro"; exit 1; fi
would be a good way to do it.
using
sudo bash shell_script.sh
instead of
sudo sh shell_script.sh
solved in my case.
I'm trying to write a script that will check two error flags, and in case one flag (or both) are changed it'll echo-- error happened. My script:
my_error_flag=0
my_error_flag_o=0
do something.....
if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then
echo "$my_error_flag"
else
echo "no flag"
fi
Basically, it should be, something along:
if ((a=1 or b=2) or (a=1 and b=2))
then
display error
else
no error
fi
The error I get is:
line 26: conditional binary operator expected
line 26: syntax error near `]'
line 26: `if [[ "$my_error_flag"=="1" || "$my_error_flag_o"=="2" ] || [ "$my_error_flag"="1" && "$my_error_flag_o"="2" ]]; then'
Are my brackets messed up?
Use -a (for and) and -o (for or) operations.
tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html
Update
Actually you could still use && and || with the -eq operation. So your script would be like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ] || ([ $my_error_flag -eq 1 ] && [ $my_error_flag_o -eq 2 ]); then
echo "$my_error_flag"
else
echo "no flag"
fi
Although in your case you can discard the last two expressions and just stick with one or operation like this:
my_error_flag=1
my_error_flag_o=1
if [ $my_error_flag -eq 1 ] || [ $my_error_flag_o -eq 2 ]; then
echo "$my_error_flag"
else
echo "no flag"
fi
You can use either [[ or (( keyword. When you use [[ keyword, you have to use string operators such as -eq, -lt. I think, (( is most preferred for arithmetic, because you can directly use operators such as ==, < and >.
Using [[ operator
a=$1
b=$2
if [[ a -eq 1 || b -eq 2 ]] || [[ a -eq 3 && b -eq 4 ]]
then
echo "Error"
else
echo "No Error"
fi
Using (( operator
a=$1
b=$2
if (( a == 1 || b == 2 )) || (( a == 3 && b == 4 ))
then
echo "Error"
else
echo "No Error"
fi
Do not use -a or -o operators Since it is not Portable.
Please try following
if ([ $dateR -ge 234 ] && [ $dateR -lt 238 ]) || ([ $dateR -ge 834 ] && [ $dateR -lt 838 ]) || ([ $dateR -ge 1434 ] && [ $dateR -lt 1438 ]) || ([ $dateR -ge 2034 ] && [ $dateR -lt 2038 ]) ;
then
echo "WORKING"
else
echo "Out of range!"
You can get some inspiration by reading an entrypoint.sh script written by the contributors from MySQL that checks whether the specified variables were set.
As the script shows, you can pipe them with -a, e.g.:
if [ -z "$MYSQL_ROOT_PASSWORD" -a -z "$MYSQL_ALLOW_EMPTY_PASSWORD" -a -z "$MYSQL_RANDOM_ROOT_PASSWORD" ]; then
...
fi
How can I consolidate the following if statements into a single line?
if [ $# -eq 4 ]
then
if [ "$4" = "PREV" ]
then
print "yes"
fi
fi
if [ $# -eq 3 ]
then
if [ "$3" = "PREV" ]
then
print "yes"
fi
fi
I am using ksh.
Why does this give an error?
if [ [ $# -eq 4 ] && [ "$4" = "PREV" ] ]
then
print "yes"
fi
Error:
0403-012 A test command parameter is not valid.
Try this:
if [[ $# -eq 4 && "$4" == "PREV" ]]
then
print "yes"
fi
You can also try putting them all together like this:
if [[ $# -eq 4 && "$4" == "PREV" || $# -eq 3 && "$3" == "PREV" ]]
then
print "yes"
fi
Do you just want to check if the last argument is "PREV"? If so, you can also do something like this:
for last; do true; done
if [ "$last" == "PREV" ]
then
print "yes"
fi
'[' is not a grouping token in sh. You can do:
if [ expr ] && [ expr ]; then ...
or
if cmd && cmd; then ...
or
if { cmd && cmd; }; then ...
You can also use parentheses, but the semantics is slightly different as the tests will run in a subshell.
if ( cmd && cmd; ); then ...
Also, note that "if cmd1; then cmd2; fi" is exactly the same as "cmd1 && cmd2", so you could write:
test $# = 4 && test $4 = PREV && echo yes
but if your intention is to check that the last argument is the string PREV, you might consider:
eval test \$$# = PREV && echo yes
Try this :
if [ $# -eq 4 ] && [ "$4" = "PREV" ]
then
print "yes"
fi