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How to pass parameter (= path with * ) to bash script
Piping a script with “read” to bashbash: Source: command not foundBash script arguments not being passed to function?Bash Script if condition is always trueFolder exclusion formatting issueHow to escape a character in a heredoc bash scriptWhich shell interpreter runs a script with no hashbang… but run as sudo?To pass parameter to bash script fileHow do I get CRON to run a script that works in my regular shell, but not in CRON?Portable way to run command without PATH from bash script
can anyone here tell me why the * is not passed to the bash shell script?
My script named as top40 is
#!/bin/bash
sudo du -shx $1 | sort -rh | head -n 40
when I try to run it as top40 /var/*
the * is ignored. It is like top40 /var
but I want to see the top 40`s size of the directories.
when I do it whithout script and type it on the prompt, it works fine.
I really can not find the reason for this.
Thanks for opening my eyes. I use Ubuntu 18.04 LTS
bash ubuntu
New contributor
add a comment |
can anyone here tell me why the * is not passed to the bash shell script?
My script named as top40 is
#!/bin/bash
sudo du -shx $1 | sort -rh | head -n 40
when I try to run it as top40 /var/*
the * is ignored. It is like top40 /var
but I want to see the top 40`s size of the directories.
when I do it whithout script and type it on the prompt, it works fine.
I really can not find the reason for this.
Thanks for opening my eyes. I use Ubuntu 18.04 LTS
bash ubuntu
New contributor
Wildcards are expanded before your script is called. (You'll find that$2
,$3
etc are defined.)
– Ulrich Schwarz
yesterday
1
Change$1
to"$@"
.
– jordanm
yesterday
@jordanm thanks a lot!
– Walter Schrabmair
yesterday
add a comment |
can anyone here tell me why the * is not passed to the bash shell script?
My script named as top40 is
#!/bin/bash
sudo du -shx $1 | sort -rh | head -n 40
when I try to run it as top40 /var/*
the * is ignored. It is like top40 /var
but I want to see the top 40`s size of the directories.
when I do it whithout script and type it on the prompt, it works fine.
I really can not find the reason for this.
Thanks for opening my eyes. I use Ubuntu 18.04 LTS
bash ubuntu
New contributor
can anyone here tell me why the * is not passed to the bash shell script?
My script named as top40 is
#!/bin/bash
sudo du -shx $1 | sort -rh | head -n 40
when I try to run it as top40 /var/*
the * is ignored. It is like top40 /var
but I want to see the top 40`s size of the directories.
when I do it whithout script and type it on the prompt, it works fine.
I really can not find the reason for this.
Thanks for opening my eyes. I use Ubuntu 18.04 LTS
bash ubuntu
bash ubuntu
New contributor
New contributor
edited yesterday
Walter Schrabmair
New contributor
asked yesterday
Walter SchrabmairWalter Schrabmair
1033
1033
New contributor
New contributor
Wildcards are expanded before your script is called. (You'll find that$2
,$3
etc are defined.)
– Ulrich Schwarz
yesterday
1
Change$1
to"$@"
.
– jordanm
yesterday
@jordanm thanks a lot!
– Walter Schrabmair
yesterday
add a comment |
Wildcards are expanded before your script is called. (You'll find that$2
,$3
etc are defined.)
– Ulrich Schwarz
yesterday
1
Change$1
to"$@"
.
– jordanm
yesterday
@jordanm thanks a lot!
– Walter Schrabmair
yesterday
Wildcards are expanded before your script is called. (You'll find that
$2
, $3
etc are defined.)– Ulrich Schwarz
yesterday
Wildcards are expanded before your script is called. (You'll find that
$2
, $3
etc are defined.)– Ulrich Schwarz
yesterday
1
1
Change
$1
to "$@"
.– jordanm
yesterday
Change
$1
to "$@"
.– jordanm
yesterday
@jordanm thanks a lot!
– Walter Schrabmair
yesterday
@jordanm thanks a lot!
– Walter Schrabmair
yesterday
add a comment |
1 Answer
1
active
oldest
votes
This is basic shell expansion. Because the calling shell recognizes the *
as a wildcard character, it's actually interpreting that for you, and passing a list of all the items in /var/ to your script.
If you escape the wildcard, or quote the wildcard to prevent it from being expanded in the calling shell, it will pass to your script the way you expect it to.
Here are two examples that will probably do what you want if my understanding of your intent is correct.
/path/to/script.sh '/var/run/*'
- this example prevents shell interpretation by putting the asterisk in a single quoted string, which bypasses expansion and gets passed as-is to the script.
/path/to/script.sh /var/run/*
- this example just escapes the single asterisk character from shell expansion.
Both of these examples result in the string /var/run/*
being passed AS-IS, to your script where they become $1
.
1
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
This is basic shell expansion. Because the calling shell recognizes the *
as a wildcard character, it's actually interpreting that for you, and passing a list of all the items in /var/ to your script.
If you escape the wildcard, or quote the wildcard to prevent it from being expanded in the calling shell, it will pass to your script the way you expect it to.
Here are two examples that will probably do what you want if my understanding of your intent is correct.
/path/to/script.sh '/var/run/*'
- this example prevents shell interpretation by putting the asterisk in a single quoted string, which bypasses expansion and gets passed as-is to the script.
/path/to/script.sh /var/run/*
- this example just escapes the single asterisk character from shell expansion.
Both of these examples result in the string /var/run/*
being passed AS-IS, to your script where they become $1
.
1
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
add a comment |
This is basic shell expansion. Because the calling shell recognizes the *
as a wildcard character, it's actually interpreting that for you, and passing a list of all the items in /var/ to your script.
If you escape the wildcard, or quote the wildcard to prevent it from being expanded in the calling shell, it will pass to your script the way you expect it to.
Here are two examples that will probably do what you want if my understanding of your intent is correct.
/path/to/script.sh '/var/run/*'
- this example prevents shell interpretation by putting the asterisk in a single quoted string, which bypasses expansion and gets passed as-is to the script.
/path/to/script.sh /var/run/*
- this example just escapes the single asterisk character from shell expansion.
Both of these examples result in the string /var/run/*
being passed AS-IS, to your script where they become $1
.
1
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
add a comment |
This is basic shell expansion. Because the calling shell recognizes the *
as a wildcard character, it's actually interpreting that for you, and passing a list of all the items in /var/ to your script.
If you escape the wildcard, or quote the wildcard to prevent it from being expanded in the calling shell, it will pass to your script the way you expect it to.
Here are two examples that will probably do what you want if my understanding of your intent is correct.
/path/to/script.sh '/var/run/*'
- this example prevents shell interpretation by putting the asterisk in a single quoted string, which bypasses expansion and gets passed as-is to the script.
/path/to/script.sh /var/run/*
- this example just escapes the single asterisk character from shell expansion.
Both of these examples result in the string /var/run/*
being passed AS-IS, to your script where they become $1
.
This is basic shell expansion. Because the calling shell recognizes the *
as a wildcard character, it's actually interpreting that for you, and passing a list of all the items in /var/ to your script.
If you escape the wildcard, or quote the wildcard to prevent it from being expanded in the calling shell, it will pass to your script the way you expect it to.
Here are two examples that will probably do what you want if my understanding of your intent is correct.
/path/to/script.sh '/var/run/*'
- this example prevents shell interpretation by putting the asterisk in a single quoted string, which bypasses expansion and gets passed as-is to the script.
/path/to/script.sh /var/run/*
- this example just escapes the single asterisk character from shell expansion.
Both of these examples result in the string /var/run/*
being passed AS-IS, to your script where they become $1
.
answered yesterday
Tim KennedyTim Kennedy
14.7k23152
14.7k23152
1
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
add a comment |
1
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
1
1
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
Thank you - I learnt a lot with your answer.
– Walter Schrabmair
yesterday
add a comment |
Walter Schrabmair is a new contributor. Be nice, and check out our Code of Conduct.
Walter Schrabmair is a new contributor. Be nice, and check out our Code of Conduct.
Walter Schrabmair is a new contributor. Be nice, and check out our Code of Conduct.
Walter Schrabmair is a new contributor. Be nice, and check out our Code of Conduct.
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-bash, ubuntu
Wildcards are expanded before your script is called. (You'll find that
$2
,$3
etc are defined.)– Ulrich Schwarz
yesterday
1
Change
$1
to"$@"
.– jordanm
yesterday
@jordanm thanks a lot!
– Walter Schrabmair
yesterday