How do I set a bash variable that contains another variable?
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
asked 36 mins ago
fuzzifuzzi
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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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
asked 36 mins ago
fuzzifuzzi
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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
asked 36 mins ago
fuzzifuzzi
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 $var1 | cut -d '/' -f 1 is "first" which is correct. However, I haven't been able to set the result of this to another variable. In the case above, var2 is empty.
What would the correct syntax for line 3 be so that the part of the string before the is returned as the value of var2?
bash shell-script
bash shell-script
asked 36 mins ago
fuzzifuzzi
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 36 mins ago
fuzzifuzzi
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 36 mins ago
fuzzifuzzi
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 36 mins ago
fuzzifuzzi
13
asked 36 mins ago
fuzzifuzzi
13
13
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
fuzzi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
1
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oldest
votes
You could use command substitution
var2=$(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 "$var3"
second
(removes the shortest leading substring matching */) should you need it as well.
answered 23 mins ago
steeldriversteeldriver
36.8k45287
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
You could use command substitution
var2=$(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 "$var3"
second
(removes the shortest leading substring matching */) should you need it as well.
answered 23 mins ago
steeldriversteeldriver
36.8k45287
add a comment |
You could use command substitution
var2=$(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 "$var3"
second
(removes the shortest leading substring matching */) should you need it as well.
answered 23 mins ago
steeldriversteeldriver
36.8k45287
add a comment |
You could use command substitution
var2=$(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 "$var3"
second
(removes the shortest leading substring matching */) should you need it as well.
answered 23 mins ago
steeldriversteeldriver
36.8k45287
You could use command substitution
var2=$(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.
answered 23 mins ago
steeldriversteeldriver
36.8k45287
answered 23 mins ago
steeldriversteeldriver
36.8k45287
answered 23 mins ago
steeldriversteeldriver
36.8k45287
answered 23 mins ago
steeldriversteeldriver
36.8k45287
36.8k45287
add a comment |
add a comment |
fuzzi is a new contributor. Be nice, and check out our Code of Conduct.
fuzzi is a new contributor. Be nice, and check out our Code of Conduct.
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