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How do I set a bash variable that contains another variable? [duplicate]

1

This question already has an answer here:

• 
  How can I assign the output of a command to a shell variable?
  
  2 answers
  
  

I am trying to set the variable service to be the value of

var1=first/second
echo $var1 | cut -d '/' -f 1
var2=$var1 | cut -d '/' -f 1"
echo $var2

The result of echo $var1 | cut -d '/' -f 1 is "first" which is correct. However, I haven't been able to set the result of this to another variable. In the case above, var2 is empty.

What would the correct syntax for line 3 be so that the part of the string before the is returned as the value of var2?

bash shell-script

share|improve this question

asked Mar 12 at 22:36

fuzzifuzzi

85

marked as duplicate by G-Man, jimmij, ilkkachu bash
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Mar 13 at 12:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  See also Storing output of command in shell variable, Calculate variable and output it to another variable, Bash: Assign output of pipe to a variable, setting output of a command to a variable, and more.
  
  – G-Man
  Mar 12 at 23:32
  

  
  
  
  

add a comment | 

1

This question already has an answer here:

• 
  How can I assign the output of a command to a shell variable?
  
  2 answers
  
  

I am trying to set the variable service to be the value of

var1=first/second
echo $var1 | cut -d '/' -f 1
var2=$var1 | cut -d '/' -f 1"
echo $var2

The result of echo $var1 | cut -d '/' -f 1 is "first" which is correct. However, I haven't been able to set the result of this to another variable. In the case above, var2 is empty.

What would the correct syntax for line 3 be so that the part of the string before the is returned as the value of var2?

bash shell-script

share|improve this question

asked Mar 12 at 22:36

fuzzifuzzi

85

marked as duplicate by G-Man, jimmij, ilkkachu bash
Users with the  bash badge can single-handedly close bash questions as duplicates and reopen them as needed.

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Mar 13 at 12:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  See also Storing output of command in shell variable, Calculate variable and output it to another variable, Bash: Assign output of pipe to a variable, setting output of a command to a variable, and more.
  
  – G-Man
  Mar 12 at 23:32
  

  
  
  
  

add a comment | 

This question already has an answer here:

• 
  How can I assign the output of a command to a shell variable?
  
  2 answers
  
  

I am trying to set the variable service to be the value of

var1=first/second
echo $var1 | cut -d '/' -f 1
var2=$var1 | cut -d '/' -f 1"
echo $var2

The result of echo $var1 | cut -d '/' -f 1 is "first" which is correct. However, I haven't been able to set the result of this to another variable. In the case above, var2 is empty.

What would the correct syntax for line 3 be so that the part of the string before the is returned as the value of var2?

bash shell-script

share|improve this question

asked Mar 12 at 22:36

fuzzifuzzi

85

var1=first/second
echo $var1 | cut -d '/' -f 1
var2=$var1 | cut -d '/' -f 1"
echo $var2

share|improve this question

asked Mar 12 at 22:36

fuzzifuzzi

85

share|improve this question

asked Mar 12 at 22:36

fuzzifuzzi

85

asked Mar 12 at 22:36

fuzzifuzzi

85

asked Mar 12 at 22:36

fuzzifuzzi

85

1 Answer
1

active

oldest

votes

2

You could use command substitution

var2=$(echo "$var1" | cut -d '/' -f 1)

However in this case it would be better to use the shell's parameter substitution directly:

$ var2=$var1%/*
$ echo "$var2"
first

(removes the shortest trailing substring matching /*) and

$ var3=$var1#*/
$ echo "$var3"
second

(removes the shortest leading substring matching */) should you need it as well.

share|improve this answer

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287

add a comment | 

2

You could use command substitution

var2=$(echo "$var1" | cut -d '/' -f 1)

However in this case it would be better to use the shell's parameter substitution directly:

$ var2=$var1%/*
$ echo "$var2"
first

(removes the shortest trailing substring matching /*) and

$ var3=$var1#*/
$ echo "$var3"
second

(removes the shortest leading substring matching */) should you need it as well.

share|improve this answer

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287

add a comment | 

2

You could use command substitution

var2=$(echo "$var1" | cut -d '/' -f 1)

However in this case it would be better to use the shell's parameter substitution directly:

$ var2=$var1%/*
$ echo "$var2"
first

(removes the shortest trailing substring matching /*) and

$ var3=$var1#*/
$ echo "$var3"
second

(removes the shortest leading substring matching */) should you need it as well.

share|improve this answer

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287

add a comment | 

You could use command substitution

var2=$(echo "$var1" | cut -d '/' -f 1)

However in this case it would be better to use the shell's parameter substitution directly:

$ var2=$var1%/*
$ echo "$var2"
first

(removes the shortest trailing substring matching /*) and

$ var3=$var1#*/
$ echo "$var3"
second

(removes the shortest leading substring matching */) should you need it as well.

share|improve this answer

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287

var2=$(echo "$var1" | cut -d '/' -f 1)

$ var2=$var1%/*
$ echo "$var2"
first

$ var3=$var1#*/
$ echo "$var3"
second

share|improve this answer

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287

answered Mar 12 at 22:49

steeldriversteeldriver

37.2k45287