Difference between ` and $ in UNIX [duplicate] - bash

This question already has answers here:
What is the difference between $(command) and `command` in shell programming?
(6 answers)
What is the benefit of using $() instead of backticks in shell scripts? [duplicate]
(9 answers)
Closed 3 years ago.
In Unix we have 2 ways to execute a command and capture its output in a variable:
1.)
x=`wc-l`
2.)
x=$(wc -l)
Can anyone help me understand the basic difference between the two, and when to use which syntax.

Related

Use command line argument in bash script substituted with value from script [duplicate]

This question already has answers here:
Dynamic variable names in Bash
(19 answers)
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
Bash - variable variables [duplicate]
(4 answers)
Closed 2 years ago.
I recently started bash scripting and got stuck with a very basic usecase, searched stackoverflow/google but couldn't find a way to achieve what I am trying to do.
I have a script color.sh
#!/bin/bash
Apple="Red"
Orange="Orange"
Banana="Yello"
echo $$1
What I am trying to achieve is print the color of fruit and accept fruit from command line. The output I want is
./color.sh Apple -> Red, but what I get is some random number which I think is process Id.

Find out what the (shell) script was invoked with [duplicate]

This question already has answers here:
How do I parse command line arguments in Bash?
(40 answers)
How to get exact command line string from shell?
(2 answers)
Closed 3 years ago.
Suppose my script.sh could take a number of options and arguments. What is the best way to find out what the script was invoked with (form inside the script)?
For eg., someone called it with script.sh --foo_option bar_arg
Is there a way to echo that exact command they typed from inside the script?
I've tried echo !! which does not work inside a script.

Bash expand variable in an another variable [duplicate]

This question already has answers here:
How to use a variable's value as another variable's name in bash [duplicate]
(6 answers)
What does "${!var}" mean in shell script? [duplicate]
(1 answer)
Closed 4 years ago.
I have the following variables
my_country_code="green"
x="country"
echo ${my_$x_code}
bash: ${my_$x_code}: bad substitution
echo should print green as output, but unable to find any technique which will give the correct output
my_x_code="my_${x}_code"
echo ${!my_x_code}

Why we use #!/bin/bash in Bash script? [duplicate]

This question already has answers here:
Why do you need to put #!/bin/bash at the beginning of a script file?
(10 answers)
What is the preferred Bash shebang ("#!")?
(6 answers)
Closed 4 years ago.
What is the significant of using #!/bin/bash in the starting of bash script? Can we write a bash script without #!/bin/bash ?
This line is called shebang. It’s a ‚magic‘ line telling the program loader (kernel) how to execute a script on unixoid systems.
Cf. https://en.m.wikipedia.org/wiki/Shebang_(Unix)

Esoteric shell script syntax [duplicate]

This question already has answers here:
Trouble understanding parameter substitution in a script
(1 answer)
Usage of :- (colon dash) in bash
(2 answers)
Closed 9 years ago.
I've found this in a shell script that I use and I'm having trouble finding a formal description/definition of this syntax:
ACTION=${1:-update}
I'm assuming that if $1 variable does not exist (no command line arguments) then "-update" is used.
It's not esoteric. It's POSIX, and even Bourne. In every shell manpage ever. man bash or man ksh. The assumption is mostly right, if the parameter 1 is unset or empty string, then expand the alternate.
http://pubs.opengroup.org/onlinepubs/9699919799/utilities/V3_chap02.html#tag_18_06_02

Resources