I am trying to make the statement pass as true, if and only if the user input through stdin is within the guidelines of
[a-z_][a-z0-9_-]*
so if the user were to input a % or $ in their argument passed then it would return false. How could i go about that?
This reads from stdin and reports on true or false status (and exits if it is false):
grep -q '^[a-z_][a-z0-9_-]*$' && echo true || { echo false; exit 1 ; }
If grep finds a match to your regex, it sets its exit code to true (0) in which case the "and" (&&) clause is executed and "true" is returned. If grep fails to find a match, the "or" (||) clause is executed and "false" is returned. The -q flag to grep tells grep to be quiet.
If one were to use this in a script, one would probably want to capture the user input into a shell variable and then test it. That might look like:
read -p "Enter a name: " var
echo "$var" | grep -q '^[a-z_][a-z0-9_-]*$' && echo true || { echo false; exit 1 ; }
To make it easy to add more statements to execute if the result is "true", we can write it out in a longer form with a place marked to add more statements:
read -p "Enter a name: " var
if echo "$var" | grep -q '^[a-z_][a-z0-9_-]*$'
then
echo true
# other statements to execute if true go here
else
echo false
exit 1
fi
The answer depends on what you mean by return. If by return you mean literal false, well, you have a small problem: UNIX scripts can only return an integer in the range 0-255. Normally in UNIX when a program returns it returns an exit status that indicates (among other things) the success or failure of the program. So you could just write:
grep -q ''^[a-z_][a-z0-9_-]*' || exit
At the top of your script. Since a shell script exits with the last value of $? anyway, that would only exit if the grep fails, and would exit with the same exit code as the grep statement itself. Technically this would mean returning 1, but in UNIX that would be akin to false. If the grep succeeds, the script would continue, not returning anything until completion or another error condition occurs.
If what you really want is to print the string "false", then see John1024's answer.
Related
So I'm having a bit of a problem when running this command in bash;
echo "$usr_age" | grep "^[0-9]*$" > $null || echo "Please only use numbers in the Age field." || exit 1
When running it, if the "$usr_age" variable has anything that is not a number, it warns the user, but it doesn't exit the script. I also tried changing the last || to && but if I do so it will just exit the script even if the variable is all numbers.
Note: the "$null" variable is just "/dev/null"
Thank you.
Use a grouping operator to combine the echo and the exit. a || b runs b only if a fails, whereas you want to run exit whether or not echo succeeds.
grep -q "^[0-9]*$" <<<"$usr_age" || { echo "Only use numbers in the Age field."; exit 1; }
By the way -- grep, as an external command, is quite slow to start up compared to using a shell builtin. Consider instead bash's built-in regex support:
[[ $usr_age =~ ^[0-9]*$ ]] || { echo "Only use numbers in the Age field."; exit 1; }
Assuming thoses functions :
return_0() {
return 0
}
return_1() {
return 1
}
Then the following code :
if return_0; then
echo "we're in" # this will be displayed
fi
if return_1; then
echo "we aren't" # this won't be displayed
fi
if return_0 -a return_1; then
echo "and here we're in again" # will be displayed - Why ?
fi
Why I am getting into the last ifstatement ?
Aren't we supposed to be out of the condition with those 0 and 1 ?
-a is one of the options of the test command (which is also implemented by [ and [[). So you can't just use -a by itself. You probably want to use &&, which is a control operator token for an AND list.
if return_0 && return_1; then ...
You can use -a to tell test to "and" two different test expressions, like
if test -r /file -a -x /file; then
echo 'file is readable and executable'
fi
But this is equivalent to
if [ -r /file -a -x /file ]; then ...
which may be more readable because the brackets make the test part of the expression clearer.
See the Bash Reference Manual for further information on...
&&, see lists
if statements and the various test commands and keywords, see conditional constructs
When you execute
if return_0 -a return_1; then
echo "and here we're in again" # will be displayed - Why ?
fi
You execute the line return_0 -a return_1. This actually means that you pass -a and return_1 as arguments to return_0. If you want to have an and operation, you should make use of the && syntax.
if return_0 && return_1; then
echo "and here we're in again" # will be displayed - Why ?
fi
The useful information to understand this is:
AND and OR lists are sequences of one of more pipelines separated by the && and || control operators, respectively. AND and OR lists are executed with left associativity. An AND list has the form
command1 && command2
command2 is executed if, and only if, command1 returns an exit status of zero.
An OR list has the form
command1 || command2
command2 is executed if and only if command1 returns a non-zero exit status. The return status of AND and OR lists is the exit status of the last command executed in the list.
According to this accepted answer using the set -e builtin should suffice for a bash script to exit on the first error. Yet, the following script:
#!/usr/bin/env bash
set -e
echo "a"
echo "b"
echo "about to fail" && /bin/false && echo "foo"
echo "c"
echo "d"
prints:
$ ./foo.sh
a
b
about to fail
c
d
removing the echo "foo" does stop the script; but why?
To simplify EtanReisner's detailed answer, set -e only exits on an 'uncaught' error. In your case:
echo "about to fail" && /bin/false && echo "foo"
The failing code, /bin/false, is followed by && which tests its exit code. Since && tests the exit code, the assumption is that the programmer knew what he was doing and anticipated that this command might fail. Ergo, the script does not exit.
By contrast, consider:
echo "about to fail" && /bin/false
The program does not test or branch on the exit code of /bin/false. So, when /bin/false fails, set -e will cause the script to exit.
Alternative that exits when /bin/false fails
Consider:
set -e
echo "about to fail" && /bin/false ; echo "foo"
This version will exit if /bin/false fails. As in the case where && was used, the final statement echo "foo" would therefore only be executed if /bin/false were to succeed.
Because that answer is not sufficiently specific enough.
It should say (bolded text is my addition):
# Any subsequent simple commands which fail will cause the shell script to exit immediately
Since the man page reads thusly:
-e Exit immediately if a simple command (see SHELL GRAMMAR
above) exits with a non-zero status. The shell does not
exit if the command that fails is part of the command
list immediately following a while or until keyword,
part of the test in an if statement, part of a && or ││
list, or if the command’s return value is being inverted
via !. A trap on ERR, if set, is executed before the
shell exits.
And SHELL GRAMMAR expands thusly:
SHELL GRAMMAR
Simple Commands
A simple command is a sequence of optional variable assignments fol-
lowed by blank-separated words and redirections, and terminated by a
control operator. The first word specifies the command to be executed,
and is passed as argument zero. The remaining words are passed as
arguments to the invoked command.
The return value of a simple command is its exit status, or 128+n if
the command is terminated by signal n.
I came across set -e for Bash scripts but had problems understanding what happens regarding the evaluation of the command following the last && or || in a && or || list. I know of the following quote from http://man7.org/linux/man-pages/man1/bash.1.html about set -e:
The shell does not exit if the command that fails is (...) part of
any command executed in a && or || list except the command
following the final && or || (...)
To test this, I wrote a small Bash script:
#!/bin/bash
bash -c "set -e ; true ; echo -n A"
bash -c "set -e ; false ; echo -n B"
bash -c "set -e ; true && true ; echo -n C"
bash -c "set -e ; true && false ; echo -n D"
bash -c "set -e ; false && true ; echo -n E"
bash -c "set -e ; false && false ; echo -n F"
bash -c "set -e ; true || true ; echo -n G"
bash -c "set -e ; true || false ; echo -n H"
bash -c "set -e ; false || true ; echo -n I"
bash -c "set -e ; false || false ; echo -n J"
echo ""
It prints:
ACEFGHI
About A:
true does not have a non-zero status. Therefore, the shell doesn't exit.
About B:
false does have a non-zero status and is not part of a && or || list. Therefore, the shell exits.
About C:
This is a && or || list. We will have to look at the command following the last && or ||. The command is true which does not have a non-zero status. So it doesn't matter if the command is evaluated or not - the shell doesn't exit either way.
About D:
This is a && or || list. We will have to look at the command following the last && or ||. This time, the command is false which does have a non-zero status. So we have to check if that false is being evaluated - which is indeed the case since && is following the true. Therefore, the shell exits.
About E:
Same reasoning as with C: true is the command following the last && or ||. Therefore, the shell doesn't exit.
About F:
Similar to D: This is a && or || list. We will have to look at the command following the last && or ||. Again the command is false which does have a non-zero status. But this time it doesn't matter, because the first command is false as well. Since it's a && list, the second false won't be evaluated. Therefore, the shell doesn't exit.
About G:
Same reasoning as with C or E: true is the command following the last && or ||. Therefore, the shell doesn't exit.
About H:
This is a && or || list. We will have to look at the command following the last && or ||. This command is false which does have a non-zero status, but it won't be evaluated since || is preceded by true. Therefore, the shell doesn't exit.
About I:
Same reasoning as with C, E or G: true is the command following the last && or ||. Therefore, the shell doesn't exit.
About J:
This is a && or || list. We will have to look at the command following the last && or ||. This command is false which does have a non-zero status. Since || is preceded by false the second false will be evaluated. Therefore, the shell does exit.
You should be able to apply these test cases to your case: true && false && true. Since the command following the last && or || is true which doesn't have a non-zero status, it doesn't matter what precedes && or ||, the shell won't exit either way.
I started using set -e in my bash scripts,
and discovered that short form of conditional expression breaks the script execution.
For example the following line should check that $var is not empty:
[ -z "$var" ] && die "result is empty"
But causes silent exit from script when $var has non-zero length.
I used this form of conditional expression in many places...
What should I do to make it run correctly? Rewrite everything with "if" construction (which would be ugly)? Or abandon "set -e"?
Edit: Everybody is asking for the code. Here is full [non]working example:
#!/bin/bash
set -e
function check_me()
{
ws="smth"
[ -z "$ws" ] && echo " fail" && exit 1
}
echo "checking wrong thing"
check_me
echo "check finished"
I'd expect it to print both echoes before and after function call.
But it silently fails in the check_me function. Output is:
checking wrong thing
Use
[ -n "$var" ] || die "result is empty"
This way, the return value of the entire statement is true if $var is non-empty, so the ERR trap is not triggered.
I'm afraid you will have to rewrite everything so no false statements occur.
The definition of set -e is clear:
-e Exit immediately if a simple command (see SHELL GRAMMAR above) exits with a non-zero status. The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test in an if statement, part of a && or || list, or if the command's return value is being inverted via !. A trap on ERR, if set, is executed before the shell exits.
You are using the "optimization" system of Bash: because a false statement will cause an AND (&&) statement never to be true, bash knows it doesn't have to execute the second part of the line. However, this is a clever "abuse" of the system, not intended behaviour and therefore incompatible with set -e. You will have to rewrite everything so it is using proper ifs.
You should write your script such that no command ever exits with non-zero status.
In your command [ -z "$var" ] can be true, in which case you call die, or false in which case -e does it's thing.
Either write it with if, as you say, or use something like this:
[ -z "$var" ] && die "result is empty" || true
I'd recommend if though.
What the bash help isn't very clear on is that only the last statement in an && or || chain is subject to causing an exit under set -e. foo && bar will exit if bar returns false, but not if foo returns false.
So your script should work... but it doesn't. Why?
It's not because of the failed -z test. It's because that failure makes the function return a non-zero status:
#!/bin/bash
set -e
function check_me()
{
ws="smth"
[ -z "$ws" ] && echo " fail" && exit 1
# The line above fails, setting $? to 1
# The function now returns, returning 1!
}
echo "checking wrong thing"
check_me # function returns 1, causing exit here
echo "check finished"
So there are multiple ways to fix this. You could add ||true to the conditional inside the function, or to the line that calls check_me. But as others have pointed out, using ||true has its own problems.
In this specific scenario, where the desired postcondition of check_me is "either this thing is valid or the script has exited", the straightforward thing to do is to write it like that, i.e. [[ -n "$ws" ]] || die "whatever".
But using && conditions will actually work fine with set -e in general, as long as you don't use such a conditional as the last thing in a function. You need to add an explicit true or return 0 or even : as a statement following such a conditional, unless you intend the function to return false when the condition fails.
Bash: I want to run a command and pipe the results through some filter, but if the command fails, I want to return the command's error value, not the boring return value of the filter:
E.g.:
if !(cool_command | output_filter); then handle_the_error; fi
Or:
set -e
cool_command | output_filter
In either case it's the return value of cool_command that I care about -- for the 'if' condition in the first case, or to exit the script in the second case.
Is there some clean idiom for doing this?
Use the PIPESTATUS builtin variable.
From man bash:
PIPESTATUS
An array variable (see Arrays
below) containing a list of exit
status values from the processes in
the most-recently-executed foreground
pipeline (which may contain only a
single command).
If you didn't need to display the error output of the command, you could do something like
if ! echo | mysql $dbcreds mysql; then
error "Could not connect to MySQL. Did you forget to add '--db-user=' or '--db-password='?"
die "Check your credentials or ensure server is running with /etc/init.d/mysqld status"
fi
In the example, error and die are defined functions. elsewhere in the script. $dbcreds is also defined, though this is built from command line options. If there is no error generated by the command, nothing is returned. If an error occurs, text will be returned by this particular command.
Correct me if I'm wrong, but I get the impression you're really looking to do something a little more convoluted than
[ `id -u` -eq '0' ] || die "Must be run as root!"
where you actually grab the user ID prior to the if statement, and then perform the test. Doing it this way, you could then display the result if you choose. This would be
UID=`id -u`
if [ $UID -eq '0' ]; then
echo "User is root"
else
echo "User is not root"
exit 1 ##set an exit code higher than 0 if you're exiting because of an error
fi
The following script uses a fifo to filter the output in a separate process. This has the following advantages over the other answers. First, it is not bash specific. In particular it does not rely on PIPESTATUS. Second, output is not stalled until the command has completed.
$ cat >test_filter.sh <<EOF
#!/bin/sh
cmd()
{
echo $1
echo $2 >&2
return $3
}
filter()
{
while read line
do
echo "... $line"
done
}
tmpdir=$(mktemp -d)
fifo="$tmpdir"/out
mkfifo "$fifo"
filter <"$fifo" &
pid=$!
cmd a b 10 >"$fifo" 2>&1
ret=$?
wait $pid
echo exit code: $ret
rm -f "$fifo"
rmdir "$tmpdir"
EOF
$ sh ./test_filter.sh
... a
... b
exit code: 10