#!/bin/bash
echo -n "Hurry up and type something! > "
if read -t 10 response ; then
echo "Greate, you made it in time!"
else
echo "sorry, you are too slow!"
fi
I have written above code in terminal and got error "read: Illegal option -t".
Bash supports -t, so it looks like you're trying to execute it with sh or some other shell, which is odd, since you have the correct shebang.
Make sure you run it with ./script or path_to_script/script. If you just run it in the terminal, first start bash.
I had the same problem and then I figured out that I was using #!/bin/sh instead of #!/bin/bash. After changing the shebang everything worked as desired.
bash supports the -t option for the read builtin since version bash-2.04 (see ChangeLog), so either you are using an ancient version of bash (<= 2.03) or are not really running your script under bash.
Run bash --version to check the version and double-check that your shebang really looks like #!/bin/bash in your script.
Related
I am attempting to store the result of an echo command as a variable to be used in a shell script. Debian 4.19.0-6-amd64
The command works in terminal: echo $HOSTNAME returns debian-base, the correct hostname.
I attempt to run it in a shell script, such as:
#!/usr/bin/bash
CURRENT_HOSTNAME=`echo $HOSTNAME`
echo $CURRENT_HOSTNAME
I have tried expansion:
CURRENT_HOSTNAME=$(echo $HOSTNAME)
And just to cover some more bases, I tried things like:
CURRENT_HOSTNAME=$HOSTNAME
# or
CURRENT_HOSTNAME="$HOSTNAME"
# also, in case a problem with reserved names:
test=$HOSTNAME
test="$HOSTNAME"
Works great in the terminal! Output is as follows:
root#debian-base:/scripts# echo $HOSTNAME
debian-base
root#debian-base:/scripts# TEST_HOSTNAME=$HOSTNAME
root#debian-base:/scripts# echo $TEST_HOSTNAME
debian-base
root#debian-base:/scripts# TEST_TWO_HOSTNAME=$(echo $HOSTNAME)
root#debian-base:/scripts# echo $TEST_TWO_HOSTNAME
debian-base
As soon as I run the script (as above):
root#debian-base:/scripts# sh test.sh
root#debian-base:/scripts#
What am I doing wrong?
You are using bash as your terminal. Bash has the variable $HOSTNAME set. You run your script with sh. sh does not have a $HOSTNAME.
Options:
bash test.sh
Or run it as a program:
chmod +x test.sh
./test.sh
But I think you need to change your first line to:
#!/bin/bash
As I don't think bash is installed in /usr/bin in most cases. But you need to try. To figure out where bash is installed use which bash
Another option is to use the hostname binary:
CURRENT_HOSTNAME=$(hostname)
echo $CURRENT_HOSTNAME
Which works in both bash and sh.
You can start sh by just running sh. You will see it has a bash-like terminal. You can try to do echo $HOSTNAME. It will not show, because it's not there. You can use set to see all the variables that are there (as sh does not have tab completion it's harder to figure out).
#!/bin/bash
if [ ! -f readexportfile ]; then
echo "readexportfile does not exist"
exit 0
fi
The above is part of my script. When the current shell is /bin/csh my script fails with the following error:
If: Expression Syntax
Then: Command not found
If I run bash and then run my script, it runs fine(as expected).
So the question is: If there is any way that myscript can change the current shell and then interpretate rest of the code.
PS: If i keep bash in my script, it changes the current shell and rest of the code in script doesn't get executed.
The other replies are correct, however, to answer your question, this should do the trick:
[[ $(basename $SHELL) = 'bash' ]] || exec /bin/bash
The exec builtin replaces the current shell with the given command (in this case, /bin/bash).
You can use SHEBANG(#!) to overcome your issue.
In your code you are already using she-bang but make sure it is first and foremost line.
$ cat test.sh
#!/bin/bash
if [ ! -f readexportfile ]; then
echo "readexportfile does not exist"
exit 0
else
echo "No File"
fi
$ ./test.sh
readexportfile does not exist
$ echo $SHELL
/bin/tcsh
In the above code even though I am using CSH that code executed as we mentioned shebang in the code. In case if there is no shebang then it will take the help of shell in which you are already logged in.
In you case you also check the location of bash interpreter using
$ which bash
or
$ cat /etc/shells |grep bash
When I run the following script, echo does not display anything and I don't know why. It works if I just type it into the terminal, but not from the shell script. Need some insight please. I might be tired but I'm very certain this should work:
#!/bin/sh
for n in `seq 1 10`
do
r=$RANDOM
t=$RANDOM
s=$RANDOM
f=$RANDOM
echo "$r $t $s $f"
done
echo "Done"
Your terminal probably runs a different shell than /bin/sh. For example, on Ubuntu, /bin/sh runs /bin/dash, but $RANDOM does not work there. You have to run /bin/bash or /bin/ksh to make it work.
When run from a terminal, you probably use bash, not sh.
Seems sh doesn't support $RANDOM and thus all variables you assign in your script will be assigned the empty string. Try changing the first line of your script to #!/bin/bash (or whereever bash is installed).
Here is the offending part of my script:
read -d '' TEXT <<'EOF'
Some Multiline
text that
I would like
in
a
var
EOF
echo "$TEXT" > ~/some/file.txt
and the error:
read: 175: Illegal option -d
I use this read -d all over the place and it works fine. Not sure why its not happy now. I'm running the script on Ubuntu 10.10
Fixes? Workarounds?
If you run sh and then try that command, you get:
read: 1: Illegal option -d
If you do it while still in bash, it works fine.
I therefore deduce that your script is not running under bash.
Make sure that your script begins with the line:
#!/usr/bin/env bash
(or equivalent) so that the correct shell is running the script.
Alternatively, if you cannot do that (because the script is not a bash one), just be aware that -d is a bash feature and may not be available in other shells. In that case, you will need to find another way.
The -d option to read is a feature unique to bash, not part of the POSIX standard (which only specifies -r and -p options to read). When you run your script with sh on Ubuntu, it's getting run with dash, which is a POSIX shell, and not bash. If you want the script to run under bash then you should run it with bash, or give it a #!/bin/bash shebang. Otherwise, it should be expected to run under any POSIX sh.
I have a bash script I've written to automate something tedious, so I got the command looking right in echo, but when I run it, it doesn't work. This is what I'm doing:
CMD='custom_script update --flag=value --comment="testing"'
echo -e "Running $CMD"
$CMD
The echo shows:
custom_script update --flag=value --comment="testing"
which is correct, but that is not what is actually run with the $CMD line (I know because if I copy and paste the output from echo, it works, but the error message after running in the script suggests the quoting is off).
I think I can figure this out if I can see the command run by $CMD, but I don't know how to do that.
Run it like
bash -x script.sh
or modify the shebang like
#!/bin/bash -x
Looks like
eval $CMD
is what I needed.