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Variable with brackets “$()”
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Previously a member @Evan Chen, wrote this script :
#!/bin/bash
while [ true ]
do
currentoutput="$(lsusb)"
if [ "$currentoutput" != "$lastoutput" ]
then
echo "" date and Time >> test.log
date +%x_r >> test.log
lastoutput="$(lsusb)"
lsusb >> test.log
fi
sleep 5
done
I'm a newbie, trying to learn fast and I got a question about the variable's brackets.
Put a variable between $(), I get it, but why the brackets are needed, even in the if statement ?
To make a nested command ??
Thanks
bash scripts
asked 5 mins ago
ShankharaShankhara
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Previously a member @Evan Chen, wrote this script :
#!/bin/bash
while [ true ]
do
currentoutput="$(lsusb)"
if [ "$currentoutput" != "$lastoutput" ]
then
echo "" date and Time >> test.log
date +%x_r >> test.log
lastoutput="$(lsusb)"
lsusb >> test.log
fi
sleep 5
done
I'm a newbie, trying to learn fast and I got a question about the variable's brackets.
Put a variable between $(), I get it, but why the brackets are needed, even in the if statement ?
To make a nested command ??
Thanks
bash scripts
asked 5 mins ago
ShankharaShankhara
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
Previously a member @Evan Chen, wrote this script :
#!/bin/bash
while [ true ]
do
currentoutput="$(lsusb)"
if [ "$currentoutput" != "$lastoutput" ]
then
echo "" date and Time >> test.log
date +%x_r >> test.log
lastoutput="$(lsusb)"
lsusb >> test.log
fi
sleep 5
done
I'm a newbie, trying to learn fast and I got a question about the variable's brackets.
Put a variable between $(), I get it, but why the brackets are needed, even in the if statement ?
To make a nested command ??
Thanks
bash scripts
asked 5 mins ago
ShankharaShankhara
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Previously a member @Evan Chen, wrote this script :
#!/bin/bash
while [ true ]
do
currentoutput="$(lsusb)"
if [ "$currentoutput" != "$lastoutput" ]
then
echo "" date and Time >> test.log
date +%x_r >> test.log
lastoutput="$(lsusb)"
lsusb >> test.log
fi
sleep 5
done
I'm a newbie, trying to learn fast and I got a question about the variable's brackets.
Put a variable between $(), I get it, but why the brackets are needed, even in the if statement ?
To make a nested command ??
Thanks
bash scripts
bash scripts
asked 5 mins ago
ShankharaShankhara
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
ShankharaShankhara
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
ShankharaShankhara
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 5 mins ago
ShankharaShankhara
1
asked 5 mins ago
ShankharaShankhara
1
1
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Shankhara is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The following runs the external command command and returns its output.
"$(command)"
Without the brackets, this would look for a variable instead of running a command:
"$variable"
As for the difference between $variable and "$variable", this becomes relevant when $variable contains spaces. When using "$variable", the entire variable contents will be interpreted as a single string. When using $variable they may expand into multiple arguments.
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
add a comment |
In currentoutput="$(lsusb)" lsusb is not a variable, it is a command. What this statement does, it executes lsusb command and assigns its output to currentoutput variable.
Older syntax for this was
currentoutput=`lsusb`
you can find it in many examples and scripts
answered 19 secs ago
marosgmarosg
44437
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The following runs the external command command and returns its output.
"$(command)"
Without the brackets, this would look for a variable instead of running a command:
"$variable"
As for the difference between $variable and "$variable", this becomes relevant when $variable contains spaces. When using "$variable", the entire variable contents will be interpreted as a single string. When using $variable they may expand into multiple arguments.
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
add a comment |
The following runs the external command command and returns its output.
"$(command)"
Without the brackets, this would look for a variable instead of running a command:
"$variable"
As for the difference between $variable and "$variable", this becomes relevant when $variable contains spaces. When using "$variable", the entire variable contents will be interpreted as a single string. When using $variable they may expand into multiple arguments.
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
add a comment |
The following runs the external command command and returns its output.
"$(command)"
Without the brackets, this would look for a variable instead of running a command:
"$variable"
As for the difference between $variable and "$variable", this becomes relevant when $variable contains spaces. When using "$variable", the entire variable contents will be interpreted as a single string. When using $variable they may expand into multiple arguments.
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
The following runs the external command command and returns its output.
"$(command)"
Without the brackets, this would look for a variable instead of running a command:
"$variable"
As for the difference between $variable and "$variable", this becomes relevant when $variable contains spaces. When using "$variable", the entire variable contents will be interpreted as a single string. When using $variable they may expand into multiple arguments.
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
answered 2 mins ago
thomasrutterthomasrutter
27.3k47089
27.3k47089
add a comment |
add a comment |
In currentoutput="$(lsusb)" lsusb is not a variable, it is a command. What this statement does, it executes lsusb command and assigns its output to currentoutput variable.
Older syntax for this was
currentoutput=`lsusb`
you can find it in many examples and scripts
answered 19 secs ago
marosgmarosg
44437
add a comment |
In currentoutput="$(lsusb)" lsusb is not a variable, it is a command. What this statement does, it executes lsusb command and assigns its output to currentoutput variable.
Older syntax for this was
currentoutput=`lsusb`
you can find it in many examples and scripts
answered 19 secs ago
marosgmarosg
44437
add a comment |
In currentoutput="$(lsusb)" lsusb is not a variable, it is a command. What this statement does, it executes lsusb command and assigns its output to currentoutput variable.
Older syntax for this was
currentoutput=`lsusb`
you can find it in many examples and scripts
answered 19 secs ago
marosgmarosg
44437
In currentoutput="$(lsusb)" lsusb is not a variable, it is a command. What this statement does, it executes lsusb command and assigns its output to currentoutput variable.
Older syntax for this was
currentoutput=`lsusb`
you can find it in many examples and scripts
answered 19 secs ago
marosgmarosg
44437
answered 19 secs ago
marosgmarosg
44437
answered 19 secs ago
marosgmarosg
44437
answered 19 secs ago
marosgmarosg
44437
44437
add a comment |
add a comment |
Shankhara is a new contributor. Be nice, and check out our Code of Conduct.
Shankhara is a new contributor. Be nice, and check out our Code of Conduct.
Shankhara is a new contributor. Be nice, and check out our Code of Conduct.
Shankhara is a new contributor. Be nice, and check out our Code of Conduct.
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