CODEWITHSUNDEEP
linuxbashshell
Rambo
Rambo

Reputation: 11

Can anyone explain me about below exression

I am confused about the below regular expression I found in a shell script 

#!/bin/sh
prog=${0%%.sh}-via.sh

Anyone give me a valuable information

Upvotes: 0

Views: 65

Answers (3)

vdavid
vdavid

Reputation: 2544

When you run a script, the variable $0 is automatically set with the name of the script (which may or may not contain the folder, but I think this is not relevant to your question).

In bash the ${0} syntax can be used to fetch the same variable, but in the braces you can add extra processing to the variable. In your example:

${0%%.sh}

will remove the ".sh" part at the end of the string, using parameter substitution.

prog=${0%%.sh}-via.sh

will add -via.sh at the end, and assign it to prog.

All in all, what it does is replacing the .sh by -via.sh. Let's say your script is called "myscript.sh", then prog will be assigned the value "myscript-via.sh".

As a side note, I would recommend starting your script with #!/bin/bash instead of #!/bin/sh because parameter substitution is a bash feature. More info about why it matters here.

Upvotes: 2

NaN
NaN

Reputation: 8581

${0%%.sh} removes the suffix .sh from the variable 0 (it contains the name of the script) and appends -via.sh. The result is stored in prog which is <script_name_without_extension>-via.sh

For further reference see this related question: Remove a fixed prefix/suffix from a string in Bash

Upvotes: 1

Poshi
Poshi

Reputation: 5762

There are no regular expressions involved in that script.

${0} is the name of the called script. The %% operator removes a matching suffix pattern., so ${0%%.sh} is the name of the called script without the final ".sh". Finally, ${0%%.sh}-via.sh is the name of the called script without the final ".sh" and appending "-via.sh".

That value is assigned to variable prog.

Upvotes: 4

Related Questions