I would like to customize the shell for a Makefile, but I am running into trouble. Here is a MWE. I have a Makefile,
SHELL=./my_shell.sh
all: abc def
abc:
touch abc
def:
touch def
clean:
rm -f abc def
and a simple custom script, my_shell.sh.
#!/bin/bash
eval $*
I made sure to run chmod +x my_shell.sh beforehand. Then, when I typed make, I got an error.
make[3]: Entering directory `<CONFIDENTIAL>'
touch abc
./my_shell.sh: line 2: eval: -c: invalid option
eval: usage: eval [arg ...]
make[3]: *** [abc] Error 2
make[3]: Leaving directory `<CONFIDENTIAL>`
I have been struggling to get rid of the -c option. Setting .SHELLFLAGS='' doesn't seem to work.
First, if you want to set a make variable to empty use
.SHELLFLAGS =
By adding the quotes you've actually set the variable to the literal string '' (make is not the shell and does not do shell quote stripping).
Second, the .SHELLFLAGS variable was added in GNU make 3.82 so if your version is older than that, it won't work. If you have 3.82 or newer then it does work (I just tried it).
Lastly, as Jean-François points out you will problems if your command has escaped spaces. However, his solution of using "$#" cannot work as-is, because then the entire command is seen as a single string.
A more reliable implementation of the script would be:
#!/bin/bash
exec /bin/bash "$#"
And then do not modify .SHELLFLAGS. Or else, do set .SHELLFLAGS to empty and make your script:
#!/bin/bash
exec /bin/bash -c "$#"
SHELL expects a shell command like bash or csh. make invokes the shell command using -c as first argument (command option).
Since you pass all arguments directly to the eval commands you get that error.
Use shift to skip the first -c argument that you are not using.
#!/bin/bash
shift
eval $*
And maybe you'll need "$#" instead of $* if the arguments have spaces.
Related
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
This bash script gives me Bad substitution error on ubuntu. Any help will be highly appreciated.
The default shell (/bin/sh) under Ubuntu points to dash, not bash.
me#pc:~$ readlink -f $(which sh)
/bin/dash
So if you chmod +x your_script_file.sh and then run it with ./your_script_file.sh, or if you run it with bash your_script_file.sh, it should work fine.
Running it with sh your_script_file.sh will not work because the hashbang line will be ignored and the script will be interpreted by dash, which does not support that string substitution syntax.
I had the same problem. Make sure your script didnt have
#!/bin/sh
at the top of your script. Instead, you should add
#!/bin/bash
For others that arrive here, this exact message will also appear when using the env variable syntax for commands, for example ${which sh} instead of the correct $(which sh)
Your script syntax is valid bash and good.
Possible causes for the failure:
Your bash is not really bash but ksh or some other shell which doesn't understand bash's parameter substitution. Because your script looks fine and works with bash.
Do ls -l /bin/bash and check it's really bash and not sym-linked to some other shell.
If you do have bash on your system, then you may be executing your script the wrong way like: ksh script.sh or sh script.sh (and your default shell is not bash). Since you have proper shebang, if you have bash ./script.sh or bash ./script.sh should be fine.
Try running the script explicitly using bash command rather than just executing it as executable.
Also, make sure you don't have an empty string for the first line of your script.
i.e. make sure #!/bin/bash is the very first line of your script.
Not relevant to your example, but you can also get the Bad substitution error in Bash for any substitution syntax that Bash does not recognize. This could be:
Stray whitespace. E.g. bash -c '${x }'
A typo. E.g. bash -c '${x;-}'
A feature that was added in a later Bash version. E.g. bash -c '${x#Q}' before Bash 4.4.
If you have multiple substitutions in the same expression, Bash may not be very helpful in pinpointing the problematic expression. E.g.:
$ bash -c '"${x } multiline string
$y"'
bash: line 1: ${x } multiline string
$y: bad substitution
Both - bash or dash - work, but the syntax needs to be:
FILENAME=/my/complex/path/name.ext
NEWNAME=${FILENAME%ext}new
I was adding a dollar sign twice in an expression with curly braces in bash:
cp -r $PROJECT_NAME ${$PROJECT_NAME}2
instead of
cp -r $PROJECT_NAME ${PROJECT_NAME}2
I have found that this issue is either caused by the marked answer or you have a line or space before the bash declaration
Looks like "+x" causes problems:
root#raspi1:~# cat > /tmp/btest
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
root#raspi1:~# chmod +x /tmp/btest
root#raspi1:~# /tmp/btest
root#raspi1:~# sh -x /tmp/btest
+ jobname=job_201312161447_0003
/tmp/btest: 4: /tmp/btest: Bad substitution
in my case (under ubuntu 18.04), I have mixed $( ${} ) that works fine:
BACKUPED_NB=$(ls ${HOST_BACKUP_DIR}*${CONTAINER_NAME}.backup.sql.gz | wc --lines)
full example here.
I used #!bin/bash as well tried all approaches like no line before or after #!bin/bash.
Then also tried using +x but still didn't work.
Finally i tried running the script ./script.sh it worked fine.
#!/bin/bash
jobname="job_201312161447_0003"
jobname_post=${jobname:17}
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# sh jaru.sh
jaru.sh: 3: jaru.sh: Bad substitution
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# ./jaru.sh
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts#
This question already has answers here:
Why does shell ignore quoting characters in arguments passed to it through variables? [duplicate]
(3 answers)
Closed 6 years ago.
I'm trying to write a database call from within a bash script and I'm having problems with a sub-shell stripping my quotes away.
This is the bones of what I am doing.
#---------------------------------------------
#! /bin/bash
export COMMAND='psql ${DB_NAME} -F , -t --no-align -c "${SQL}" -o ${EXPORT_FILE} 2>&1'
PSQL_RETURN=`${COMMAND}`
#---------------------------------------------
If I use an 'echo' to print out the ${COMMAND} variable the output looks fine:
echo ${COMMAND}
screen output:-
#---------------
psql drupal7 -F , -t --no-align -c "SELECT DISTINCT hostname FROM accesslog;" -o /DRUPAL/INTERFACES/EXPORTS/ip_list.dat 2>&1
#---------------
Also if I cut and paste this screen output it executes just fine.
However, when I try to execute the command as a variable within a sub-shell call, it gives an error message.
The error is from the psql client to the effect that the quotes have been removed from around the ${SQL} string.
The error suggests psql is trying to interpret the terms in the sql string as parameters.
So it seems the string and quotes are composed correctly but the quotes around the ${SQL} variable/string are being interpreted by the sub-shell during the execution call from the main script.
I've tried to escape them using various methods: \", \\", \\\", "", \"" '"', \'"\', ... ...
As you can see from my 'try it all' approach I am no expert and it's driving me mad.
Any help would be greatly appreciated.
Charlie101
Instead of storing command in a string var better to use BASH array here:
cmd=(psql ${DB_NAME} -F , -t --no-align -c "${SQL}" -o "${EXPORT_FILE}")
PSQL_RETURN=$( "${cmd[#]}" 2>&1 )
Rather than evaluating the contents of a string, why not use a function?
call_psql() {
# optional, if variables are already defined in global scope
DB_NAME="$1"
SQL="$2"
EXPORT_FILE="$3"
psql "$DB_NAME" -F , -t --no-align -c "$SQL" -o "$EXPORT_FILE" 2>&1
}
then you can just call your function like:
PSQL_RETURN=$(call_psql "$DB_NAME" "$SQL" "$EXPORT_FILE")
It's entirely up to you how elaborate you make the function. You might like to check for the correct number of arguments (using something like (( $# == 3 ))) before calling the psql command.
Alternatively, perhaps you'd prefer just to make it as short as possible:
call_psql() { psql "$1" -F , -t --no-align -c "$2" -o "$3" 2>&1; }
In order to capture the command that is being executed for debugging purposes, you can use set -x in your script. This will the contents of the function including the expanded variables when the function (or any other command) is called. You can switch this behaviour off using set +x, or if you want it on for the whole duration of the script you can change the shebang to #!/bin/bash -x. This saves you explicitly echoing throughout your script to find out what commands are being run; you can just turn on set -x for a section.
A very simple example script using the shebang method:
#!/bin/bash -x
ec() {
echo "$1"
}
var=$(ec 2)
Running this script, either directly after making it executable or calling it with bash -x, gives:
++ ec 2
++ echo 2
+ var=2
Removing the -x from the shebang or the invocation results in the script running silently.
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
This bash script gives me Bad substitution error on ubuntu. Any help will be highly appreciated.
The default shell (/bin/sh) under Ubuntu points to dash, not bash.
me#pc:~$ readlink -f $(which sh)
/bin/dash
So if you chmod +x your_script_file.sh and then run it with ./your_script_file.sh, or if you run it with bash your_script_file.sh, it should work fine.
Running it with sh your_script_file.sh will not work because the hashbang line will be ignored and the script will be interpreted by dash, which does not support that string substitution syntax.
I had the same problem. Make sure your script didnt have
#!/bin/sh
at the top of your script. Instead, you should add
#!/bin/bash
For others that arrive here, this exact message will also appear when using the env variable syntax for commands, for example ${which sh} instead of the correct $(which sh)
Your script syntax is valid bash and good.
Possible causes for the failure:
Your bash is not really bash but ksh or some other shell which doesn't understand bash's parameter substitution. Because your script looks fine and works with bash.
Do ls -l /bin/bash and check it's really bash and not sym-linked to some other shell.
If you do have bash on your system, then you may be executing your script the wrong way like: ksh script.sh or sh script.sh (and your default shell is not bash). Since you have proper shebang, if you have bash ./script.sh or bash ./script.sh should be fine.
Try running the script explicitly using bash command rather than just executing it as executable.
Also, make sure you don't have an empty string for the first line of your script.
i.e. make sure #!/bin/bash is the very first line of your script.
Not relevant to your example, but you can also get the Bad substitution error in Bash for any substitution syntax that Bash does not recognize. This could be:
Stray whitespace. E.g. bash -c '${x }'
A typo. E.g. bash -c '${x;-}'
A feature that was added in a later Bash version. E.g. bash -c '${x#Q}' before Bash 4.4.
If you have multiple substitutions in the same expression, Bash may not be very helpful in pinpointing the problematic expression. E.g.:
$ bash -c '"${x } multiline string
$y"'
bash: line 1: ${x } multiline string
$y: bad substitution
Both - bash or dash - work, but the syntax needs to be:
FILENAME=/my/complex/path/name.ext
NEWNAME=${FILENAME%ext}new
I was adding a dollar sign twice in an expression with curly braces in bash:
cp -r $PROJECT_NAME ${$PROJECT_NAME}2
instead of
cp -r $PROJECT_NAME ${PROJECT_NAME}2
I have found that this issue is either caused by the marked answer or you have a line or space before the bash declaration
Looks like "+x" causes problems:
root#raspi1:~# cat > /tmp/btest
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
root#raspi1:~# chmod +x /tmp/btest
root#raspi1:~# /tmp/btest
root#raspi1:~# sh -x /tmp/btest
+ jobname=job_201312161447_0003
/tmp/btest: 4: /tmp/btest: Bad substitution
in my case (under ubuntu 18.04), I have mixed $( ${} ) that works fine:
BACKUPED_NB=$(ls ${HOST_BACKUP_DIR}*${CONTAINER_NAME}.backup.sql.gz | wc --lines)
full example here.
I used #!bin/bash as well tried all approaches like no line before or after #!bin/bash.
Then also tried using +x but still didn't work.
Finally i tried running the script ./script.sh it worked fine.
#!/bin/bash
jobname="job_201312161447_0003"
jobname_post=${jobname:17}
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# sh jaru.sh
jaru.sh: 3: jaru.sh: Bad substitution
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# ./jaru.sh
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts#
I'm trying to write a shell script that calls another script that then executes a rsync command.
The second script should run in its own terminal, so I use a gnome-terminal -e "..." command. One of the parameters of this script is a string containing the parameters that should be given to rsync. I put those into single quotes.
Up until here, everything worked fine until one of the rsync parameters was a directory path that contained a space. I tried numerous combinations of ',",\",\' but the script either doesn't run at all or only the first part of the path is taken.
Here's a slightly modified version of the code I'm using
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l '\''/media/MyAndroid/Internal storage'\''' "
Within Backup.sh this command is run
rsync $5 "$path"
where the destination $path is calculated from text in Stamp.
How can I achieve these three levels of nested quotations?
These are some question I looked at just now (I've tried other sources earlier as well)
https://unix.stackexchange.com/questions/23347/wrapping-a-command-that-includes-single-and-double-quotes-for-another-command
how to make nested double quotes survive the bash interpreter?
Using multiple layers of quotes in bash
Nested quotes bash
I was unsuccessful in applying the solutions to my problem.
Here is an example. caller.sh uses gnome-terminal to execute foo.sh, which in turn prints all the arguments and then calls rsync with the first argument.
caller.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh 'long path' arg2 arg3"
foo.sh:
#!/bin/bash
echo $# arguments
for i; do # same as: for i in "$#"; do
echo "$i"
done
rsync "$1" "some other path"
Edit: If $1 contains several parameters to rsync, some of which are long paths, the above won't work, since bash either passes "$1" as one parameter, or $1 as multiple parameters, splitting it without regard to contained quotes.
There is (at least) one workaround, you can trick bash as follows:
caller2.sh:
#!/bin/bash
gnome-terminal -t "TEST" -e "./foo.sh '--option1 --option2 \"long path\"' arg2 arg3"
foo2.sh:
#!/bin/bash
rsync_command="rsync $1"
eval "$rsync_command"
This will do the equivalent of typing rsync --option1 --option2 "long path" on the command line.
WARNING: This hack introduces a security vulnerability, $1 can be crafted to execute multiple commands if the user has any influence whatsoever over the string content (e.g. '--option1 --option2 \"long path\"; echo YOU HAVE BEEN OWNED' will run rsync and then execute the echo command).
Did you try escaping the space in the path with "\ " (no quotes)?
gnome-terminal -t 'Rsync scheduled backup' -e "nice -10 /Scripts/BackupScript/Backup.sh 0 0 '/Scripts/BackupScript/Stamp' '/Scripts/BackupScript/test' '--dry-run -g -o -p -t -R -u --inplace --delete -r -l ''/media/MyAndroid/Internal\ storage''' "
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
This bash script gives me Bad substitution error on ubuntu. Any help will be highly appreciated.
The default shell (/bin/sh) under Ubuntu points to dash, not bash.
me#pc:~$ readlink -f $(which sh)
/bin/dash
So if you chmod +x your_script_file.sh and then run it with ./your_script_file.sh, or if you run it with bash your_script_file.sh, it should work fine.
Running it with sh your_script_file.sh will not work because the hashbang line will be ignored and the script will be interpreted by dash, which does not support that string substitution syntax.
I had the same problem. Make sure your script didnt have
#!/bin/sh
at the top of your script. Instead, you should add
#!/bin/bash
For others that arrive here, this exact message will also appear when using the env variable syntax for commands, for example ${which sh} instead of the correct $(which sh)
Your script syntax is valid bash and good.
Possible causes for the failure:
Your bash is not really bash but ksh or some other shell which doesn't understand bash's parameter substitution. Because your script looks fine and works with bash.
Do ls -l /bin/bash and check it's really bash and not sym-linked to some other shell.
If you do have bash on your system, then you may be executing your script the wrong way like: ksh script.sh or sh script.sh (and your default shell is not bash). Since you have proper shebang, if you have bash ./script.sh or bash ./script.sh should be fine.
Try running the script explicitly using bash command rather than just executing it as executable.
Also, make sure you don't have an empty string for the first line of your script.
i.e. make sure #!/bin/bash is the very first line of your script.
Not relevant to your example, but you can also get the Bad substitution error in Bash for any substitution syntax that Bash does not recognize. This could be:
Stray whitespace. E.g. bash -c '${x }'
A typo. E.g. bash -c '${x;-}'
A feature that was added in a later Bash version. E.g. bash -c '${x#Q}' before Bash 4.4.
If you have multiple substitutions in the same expression, Bash may not be very helpful in pinpointing the problematic expression. E.g.:
$ bash -c '"${x } multiline string
$y"'
bash: line 1: ${x } multiline string
$y: bad substitution
Both - bash or dash - work, but the syntax needs to be:
FILENAME=/my/complex/path/name.ext
NEWNAME=${FILENAME%ext}new
I was adding a dollar sign twice in an expression with curly braces in bash:
cp -r $PROJECT_NAME ${$PROJECT_NAME}2
instead of
cp -r $PROJECT_NAME ${PROJECT_NAME}2
I have found that this issue is either caused by the marked answer or you have a line or space before the bash declaration
Looks like "+x" causes problems:
root#raspi1:~# cat > /tmp/btest
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${jobname:0:16}
jobname_post=${jobname:17}
root#raspi1:~# chmod +x /tmp/btest
root#raspi1:~# /tmp/btest
root#raspi1:~# sh -x /tmp/btest
+ jobname=job_201312161447_0003
/tmp/btest: 4: /tmp/btest: Bad substitution
in my case (under ubuntu 18.04), I have mixed $( ${} ) that works fine:
BACKUPED_NB=$(ls ${HOST_BACKUP_DIR}*${CONTAINER_NAME}.backup.sql.gz | wc --lines)
full example here.
I used #!bin/bash as well tried all approaches like no line before or after #!bin/bash.
Then also tried using +x but still didn't work.
Finally i tried running the script ./script.sh it worked fine.
#!/bin/bash
jobname="job_201312161447_0003"
jobname_post=${jobname:17}
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# sh jaru.sh
jaru.sh: 3: jaru.sh: Bad substitution
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# ./jaru.sh
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts#