I was wondering if there were specific permissions that were associated with a shell script or if some variable references are taken as being syntactically different.
I tried my short renaming script below:
#!/bin/bash
echo "Starting Renaming Script"
for file in ./*
do
rename=$(echo $file | sed 's/\(img_\)\([0-9]*-[0-9]*\)-\([0-9]*\)_\([0-9]*\).jpg/newyears_20\3-\2_0\4.jpg/')
mv $file $rename
done
All it does is rename a few files, but I noticed that it would work on the command line, but not in the shell script when I ran sh rename.sh
I got the error
rename.sh: syntax error at line 7: `rename=$' unexpected
Is variable assignment handled differently in the shell than on the command line?
Different shells handle commands differently. Your script is a bash script (as identified on the first line #!/bin/bash), therefore it needs to be run by bash, not sh.
bash rename.sh
Related
I want to find bash script files under folders in Array.
But bash script files do not have a specified extension.
I wrote something like this:
for i in "${array[#]}"
do
# Here I will write the condition that the file is found in the folder $k
done
If your scripts have #!/bin/bash or #!/bin/sh in their first line (as they should), then you can use the file command to check if a file is a script or not.
For example, take this script:
#!/bin/bash
echo "I am a script!"
Output of file filename.sh will be filename.sh: Bourne-Again shell script, ASCII text executable, which is indicating it is a shell script. Note that the file command does not use the extension of the file to indicate its format, but uses the content of it.
If you don't have those lines at the beginning of your file, You can try to run every file (command: bash filename.ext) and the check if it was run successfully or not by checking the value of the variable ${?}. This is not a clean method but it sure can help if you have no other choices.
The file command determines a file type.
e.g
#!/bin/bash
arr=(~/*)
for i in "${arr[#]}"
do
type=`file -b $i | awk '{print $2}'`
if [[ $type = shell ]];then
echo $i is a shell script
fi
done
#!/bin/bash
jobname="job_201312161447_0003"
jobname_pre=${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:
Run bash commands from txt file
(4 answers)
Closed 4 years ago.
I tried to execute commands read it from txt file. But only 1st command is executing, after that script is terminated. My script file name is shellEx.sh is follows:
echo "pwd" > temp.txt
echo "ls" >> temp.txt
exec < temp.txt
while read line
do
exec $line
done
echo "printed"
if I keep echo in the place of exec, just it prints both pwd and ls. But i want to execute pwd and ls one by one.
o/p am getting is:
$ bash shellEx.sh
/c/Users/Aditya Gudipati/Desktop
But after pwd, ls also need to execute for me.
Anyone can please give better solution for this?
exec in bash is meant in the Unix sense where it means "stop running this program and start running another instead". This is why your script exits.
If you want to execute line as a shell command, you can use:
line="find . | wc -l"
eval "$line"
($line by itself will not allow using pipes, quotes, expansions or other shell syntax)
To execute the entire file including multiline commands, use one of:
source ./myfile # keep variables, allow exiting script
bash myfile # discard variables, limit exit to myfile
A file with one valid command per line is itself a shell script. Just use the . command to execute it in the current shell.
$ echo "pwd" > temp.txt
$ echo "ls" >> temp.txt
$ . temp.txt
#!/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=${jobname:17}
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# sh jaru.sh
jaru.sh: 3: jaru.sh: Bad substitution
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts# ./jaru.sh
root#ip-10-2-250-36:/home/bitnami/python-module/workflow_scripts#
I'm trying some staff that is working perfectly when I write it in the regular shell, but when I include it in a bash script file, it doesn't.
First example:
m=`date +%m`
m_1=$((m-1))
echo $m_1
This gives me the value of the last month (actual minus one), but doesn't work if its executed from a script.
Second example:
m=6
m=$m"t"
echo m
This returns "6t" in the shell (concatenates $m with "t"), but just gives me "t" when executing from a script.
I assume all these may be answered easily by an experienced Linux user, but I'm just learning as I go.
Thanks in advance.
Re-check your syntax.
Your first code snippet works either from command line, from bash and from sh since your syntax is valid sh. In my opinion you probably have typos in your script file:
~$ m=`date +%m`; m_1=$((m-1)); echo $m_1
4
~$ cat > foo.sh
m=`date +%m`; m_1=$((m-1)); echo $m_1
^C
~$ bash foo.sh
4
~$ sh foo.sh
4
The same can apply to the other snippet with corrections:
~$ m=6; m=$m"t"; echo $m
6t
~$ cat > foo.sh
m=6; m=$m"t"; echo $m
^C
~$ bash foo.sh
6t
~$ sh foo.sh
6t
Make sure the first line of your script is
#!/bin/bash
rather than
#!/bin/sh
Bash will only enable its extended features if explicitly run as bash. If run as sh, it will operate in POSIX compatibility mode.
First of all, it works fine for me in a script, and on the terminal.
Second of all, your last line, echo m will just output "m". I think you meant "$m"..