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1

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 quotation marks.

Put a variable between $(), I get it, but why the quotation marks are needed, even in the if statement ?

To make a nested command ??

Thanks

bash scripts

share|improve this question

edited 4 mins ago

Shankhara

asked 11 mins ago

ShankharaShankhara

61

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 | 

1

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 quotation marks.

Put a variable between $(), I get it, but why the quotation marks are needed, even in the if statement ?

To make a nested command ??

Thanks

bash scripts

share|improve this question

edited 4 mins ago

Shankhara

asked 11 mins ago

ShankharaShankhara

61

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 quotation marks.

Put a variable between $(), I get it, but why the quotation marks are needed, even in the if statement ?

To make a nested command ??

Thanks

bash scripts

share|improve this question

edited 4 mins ago

Shankhara

asked 11 mins ago

ShankharaShankhara

61

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.

#!/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

share|improve this question

edited 4 mins ago

Shankhara

asked 11 mins ago

ShankharaShankhara

61

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.

share|improve this question

edited 4 mins ago

Shankhara

asked 11 mins ago

ShankharaShankhara

61

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 11 mins ago

ShankharaShankhara

61

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 11 mins ago

ShankharaShankhara

61

2 Answers
2

active

oldest

votes

1

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

To answer the other part of your question, if [ ] is just how syntax for if is defined in bash. See more in https://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html

share

answered 5 mins ago

marosgmarosg

45437

add a comment | 

1

The following runs the external command command and returns its output.

"$(command)"

Without the brackets/parentheses, 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 inserted into a single string even if the contents include spaces. When using $variable the contents of the variable may be expanded into an argument list of multiple arguments.

share

edited 5 mins ago

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  HI @thomasrutter, I'm sorry, i meant quotation mark ... I edit my comment now !
  
  – Shankhara
  5 mins ago
  

  
  
  
  

add a comment | 

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1

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

To answer the other part of your question, if [ ] is just how syntax for if is defined in bash. See more in https://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html

share

answered 5 mins ago

marosgmarosg

45437

add a comment | 

1

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

To answer the other part of your question, if [ ] is just how syntax for if is defined in bash. See more in https://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html

share

answered 5 mins ago

marosgmarosg

45437

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

To answer the other part of your question, if [ ] is just how syntax for if is defined in bash. See more in https://www.tldp.org/LDP/Bash-Beginners-Guide/html/sect_07_01.html

share

answered 5 mins ago

marosgmarosg

45437

currentoutput=`lsusb`

share

answered 5 mins ago

marosgmarosg

45437

answered 5 mins ago

marosgmarosg

45437

answered 5 mins ago

marosgmarosg

45437

1

The following runs the external command command and returns its output.

"$(command)"

Without the brackets/parentheses, 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 inserted into a single string even if the contents include spaces. When using $variable the contents of the variable may be expanded into an argument list of multiple arguments.

share

edited 5 mins ago

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  HI @thomasrutter, I'm sorry, i meant quotation mark ... I edit my comment now !
  
  – Shankhara
  5 mins ago
  

  
  
  
  

add a comment | 

1

The following runs the external command command and returns its output.

"$(command)"

Without the brackets/parentheses, 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 inserted into a single string even if the contents include spaces. When using $variable the contents of the variable may be expanded into an argument list of multiple arguments.

share

edited 5 mins ago

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  HI @thomasrutter, I'm sorry, i meant quotation mark ... I edit my comment now !
  
  – Shankhara
  5 mins ago
  

  
  
  
  

add a comment | 

The following runs the external command command and returns its output.

"$(command)"

Without the brackets/parentheses, 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 inserted into a single string even if the contents include spaces. When using $variable the contents of the variable may be expanded into an argument list of multiple arguments.

share

edited 5 mins ago

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089

"$(command)"

"$variable"

share

edited 5 mins ago

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089

answered 8 mins ago

thomasrutterthomasrutter

27.3k47089