How to do a logical OR operation for integer comparison in shell scripting?
Solution 1
This should work:
#!/bin/bash
if [ "$#" -eq 0 ] || [ "$#" -gt 1 ] ; then
echo "hello"
fi
I'm not sure if this is different in other shells but if you wish to use <, >, you need to put them inside double parenthesis like so:
if (("$#" > 1))
...
Solution 2
This code works for me:
#!/bin/sh
argc=$#
echo $argc
if [ $argc -eq 0 -o $argc -eq 1 ]; then
echo "foo"
else
echo "bar"
fi
I don't think sh supports "==". Use "=" to compare strings and -eq to compare ints.
man test
for more details.
Solution 3
If you are using the bash exit code status $? as variable, it's better to do this:
if [ $? -eq 4 -o $? -eq 8 ] ; then
echo "..."
fi
Because if you do:
if [ $? -eq 4 ] || [ $? -eq 8 ] ; then
The left part of the OR alters the $? variable, so the right part of the OR doesn't have the original $? value.
Solution 4
Sometimes you need to use double brackets, otherwise you get an error like too many arguments
if [[ $OUTMERGE == *"fatal"* ]] || [[ $OUTMERGE == *"Aborting"* ]]
then
fi
Solution 5
If a bash script
If [[ $input -gt number || $input -lt number ]]
then
echo .........
else
echo .........
fi
exit
Strawberry
I want to learn industry practices and apply them to my projects to make myself successful
Updated on July 16, 2022Comments
-
Strawberry almost 2 years
I am trying to do a simple condition check, but it doesn't seem to work.
If
$#
is equal to0
or is greater than1
then say hello.I have tried the following syntax with no success:
if [ "$#" == 0 -o "$#" > 1 ] ; then echo "hello" fi if [ "$#" == 0 ] || [ "$#" > 1 ] ; then echo "hello" fi