I'm trying to check if both conditions return the expected values.
I want to be sure that both return the expected value before continuing...
My problematic line is: if [ [ $ansmob = "y" ] || [ $flagbook != "1" ] ];
read -r -p "Would you like to add $site.booking.local as well? [y/n] " ansbook
if [ $ansbook = "y" ];
then sed "s/ServerAlias.*/& $site.booking.local/" -i $workdir/$site$dom.conf
flagbook="1"
fi
read -r -p "Would you like to add m.$site.booking.local? [y/n] " ansmob
if [ [ $ansmob = "y" ] || [ $flagbook != "1" ] ];
then sed "s/& $site.booking.local/& $site.booking.local m.$site.booking.local/" -i $workdir/$site$dom.conf
else
sed "s/ServerAlias.* /& m.$site.booking.local/" -i $workdir/$site$dom.conf
flagmobile="1"
fi
Replace
if [ [ $ansmob = "y" ] || [ $flagbook != "1" ] ];
with
if [ "$ansmob" = "y" ] || [ "$flagbook" != "1" ]
with bash's double brackets, you can use && and ||
if [[ $ansmob = "y" || $flagbook -ne 1 ]]
Within double brackets, it's not strictly necessary to quote the variables: this command is smart about evaluating expressions with empty variables.
The binary && operator is the syntax for the AND operation.
Related
I am trying to run a shell script in OS X Terminal. The program terminates whenever you enter an m or a.
Problem is, I can't get the OR statement working.
#!/bin/sh
read File
while [ "$File" != "m" ] || [ "$File" != "a" ]
do
read File
done
When I just do
while [ "$File" != "m" ]
It works perfectly fine. I have tried multiple methods, such as
while test $File != "m" || test $File != "a"
while test $File != "m" -o test $File != "a"
while [ $File != "m" -o $File != "a" ]
None of them seem to work.
The above code I posted does not stop looping when the user inputs an "m" or an
I suggest to replace
while [ "$File" != "m" ] || [ "$File" != "a" ]
by
while [ "$File" != "m" ] && [ "$File" != "a" ]
or
until [ "$File" = "m" ] || [ "$File" = "a" ]
or with a regex (bash):
while [[ ! $File =~ m|a ]]
See from bash: help until
I know how to check for a file in bash using this code
file=$1
if [ -f "$file" ]
then
...
fi
But I want to do something when it's not a file.
file=$3
if [ "$1" == "" ] || [ "$2" == "" ] || [ $file is not a file??? ]
then
echo "use: notEmpty notEmpty file"
fi
Can anyone help me out?
if [ "$1" == "" ] || [ "$2" == "" ] || [ ! -f "$file" ]
The whitespaces after [ and before ] are important.
I have a shellscript that tells me missing ] in the line
if [ $status != "2" && $status != "3" && `echo "$temp1 > $upperLimit" | bc` = "1" ]
and also missing ] in the line
if [ $status = "2" && `cat motionsensordate` \> `date +%s` ]
Why is that?
The single [ doesn't support logical operators inside the brackets. You have to use them outside
if [ "$status" != 2 ] && [ "$status" != 3 ] ...
Use double quotes for variables in single brackets to prevent unary operator expected error when the variable is empty.
Or, switch to double brackets:
if [[ $status != 2 && $status != 3 ... ]]
Also, status different to 2 and 3 can be expressed by a pattern:
if [[ $status != [23] && ... ]]
And if you would like to (in addition to the answers here) group together conditions:
if [[ ( COND1 || COND2 ) && COND3 ]]
then
echo "$cmd"
break
fi
why this do not work:
if [ $_TYPE == 1 ] || [ $_TYPE == 2 ]; then
$_TYPE=$(echo "Outbound");
fi
or
if [ $_TYPE == 1 ] || [ $_TYPE == 2 ]; then
$_TYPE=echo "Outbound";
fi
or
if [ $_TYPE == 1 ] || [ $_TYPE == 2 ]; then
$_TYPE="Outbound";
fi
I'm receiving this error: line 251: 2=Outbound: command not found
In POSIX shells such as Bash, $ is not part of the variable-name, it's just the notation for expanding the variable (to obtain its value); so, for example, echo "$_TYPE" prints the value of the variable _TYPE. You don't use the $ when you're assigning to the variable. So you just need:
if [[ "$_TYPE" = 1 || "$_TYPE" = 2 ]] ; then
_TYPE=Outbound
fi
$ is used to access the value, but if you have to assign a value, the syntax is :
_TYPE="newValue"
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