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How to pass parameter (= path with * ) to bash script

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0

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

share|improve this question

edited yesterday

Walter Schrabmair

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

New contributor

Walter Schrabmair is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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 | 

0

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

share|improve this question

edited yesterday

Walter Schrabmair

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

New contributor

Walter Schrabmair is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  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

share|improve this question

edited yesterday

Walter Schrabmair

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

New contributor

Walter Schrabmair is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

 #!/bin/bash
 sudo du -shx $1 | sort -rh | head -n 40

share|improve this question

edited yesterday

Walter Schrabmair

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

New contributor

Walter Schrabmair is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

share|improve this question

edited yesterday

Walter Schrabmair

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

New contributor

Walter Schrabmair is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

New contributor

Walter Schrabmair is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked yesterday

Walter SchrabmairWalter Schrabmair

1033

1 Answer
1

active

oldest

votes

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.

share|improve this answer

answered yesterday

Tim KennedyTim Kennedy

14.7k23152

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thank you - I learnt a lot with your answer.
  
  – Walter Schrabmair
  yesterday
  

  
  
  
  

add a comment | 

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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.

share|improve this answer

answered yesterday

Tim KennedyTim Kennedy

14.7k23152

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Thank you - I learnt a lot with your answer.
  
  – Walter Schrabmair
  yesterday
  

  
  
  
  

add a comment | 

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.

share|improve this answer

answered yesterday

Tim KennedyTim Kennedy

14.7k23152

• 
  
  
  

  
  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.

share|improve this answer

answered yesterday

Tim KennedyTim Kennedy

14.7k23152

share|improve this answer

answered yesterday

Tim KennedyTim Kennedy

14.7k23152

answered yesterday

Tim KennedyTim Kennedy

14.7k23152

answered yesterday

Tim KennedyTim Kennedy

14.7k23152