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Can I set my bashrc to echo $? after a particular command?

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1

I'm familiar with setting simple aliases in my .bashrc, and think it might be capable of doing what I want, but not quite sure how to do it. The desire: always echo $? after a particular program finishes executing. The particular program I want to run this after prints out fairly awkward STDERR messages that look at first glance like an error even upon success. I always echo $? after as a first check before I actually read the stuff, but at least a quarter of the time mistype and lose the echo opportunity. What I'd like to do is set an alias so that when I type

$ foo --option [argument]

It actually does

$ foo --option [argument]

working blah blah blah

$ echo $?

Is this possible?

bash alias echo

share|improve this question

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

add a comment | 

1

I'm familiar with setting simple aliases in my .bashrc, and think it might be capable of doing what I want, but not quite sure how to do it. The desire: always echo $? after a particular program finishes executing. The particular program I want to run this after prints out fairly awkward STDERR messages that look at first glance like an error even upon success. I always echo $? after as a first check before I actually read the stuff, but at least a quarter of the time mistype and lose the echo opportunity. What I'd like to do is set an alias so that when I type

$ foo --option [argument]

It actually does

$ foo --option [argument]

working blah blah blah

$ echo $?

Is this possible?

bash alias echo

share|improve this question

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

add a comment | 

I'm familiar with setting simple aliases in my .bashrc, and think it might be capable of doing what I want, but not quite sure how to do it. The desire: always echo $? after a particular program finishes executing. The particular program I want to run this after prints out fairly awkward STDERR messages that look at first glance like an error even upon success. I always echo $? after as a first check before I actually read the stuff, but at least a quarter of the time mistype and lose the echo opportunity. What I'd like to do is set an alias so that when I type

$ foo --option [argument]

It actually does

$ foo --option [argument]

working blah blah blah

$ echo $?

Is this possible?

bash alias echo

share|improve this question

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

$ foo --option [argument]

$ foo --option [argument]

working blah blah blah

$ echo $?

share|improve this question

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

share|improve this question

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

edited 6 hours ago

Jesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

asked 8 hours ago

Willoughby WillWilloughby Will

833 bronze badges

1 Answer
1

active

oldest

votes

5

Using a function:

foo () 
 command foo "$@"
 echo "$?"

This will just execute the foo command with any arguments provided and then echo the exit status afterward.

You may also want to have the foo function return with the same exit status as the foo command with:

foo() 
 local ret
 command foo "$@"
 ret="$?"
 echo "$ret"
 return "$ret"

share|improve this answer

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Works perfectly, thanks!
  
  – Willoughby Will
  8 hours ago
  

  
  
  
  

add a comment | 

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5

Using a function:

foo () 
 command foo "$@"
 echo "$?"

This will just execute the foo command with any arguments provided and then echo the exit status afterward.

You may also want to have the foo function return with the same exit status as the foo command with:

foo() 
 local ret
 command foo "$@"
 ret="$?"
 echo "$ret"
 return "$ret"

share|improve this answer

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Works perfectly, thanks!
  
  – Willoughby Will
  8 hours ago
  

  
  
  
  

add a comment | 

5

Using a function:

foo () 
 command foo "$@"
 echo "$?"

This will just execute the foo command with any arguments provided and then echo the exit status afterward.

You may also want to have the foo function return with the same exit status as the foo command with:

foo() 
 local ret
 command foo "$@"
 ret="$?"
 echo "$ret"
 return "$ret"

share|improve this answer

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Works perfectly, thanks!
  
  – Willoughby Will
  8 hours ago
  

  
  
  
  

add a comment | 

Using a function:

foo () 
 command foo "$@"
 echo "$?"

This will just execute the foo command with any arguments provided and then echo the exit status afterward.

You may also want to have the foo function return with the same exit status as the foo command with:

foo() 
 local ret
 command foo "$@"
 ret="$?"
 echo "$ret"
 return "$ret"

share|improve this answer

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

foo () 
 command foo "$@"
 echo "$?"

foo() 
 local ret
 command foo "$@"
 ret="$?"
 echo "$ret"
 return "$ret"

share|improve this answer

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

edited 7 hours ago

Stéphane Chazelas

327k5757 gold badges635635 silver badges10021002 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges

answered 8 hours ago

Jesse_bJesse_b

18.2k33 gold badges4444 silver badges8585 bronze badges