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Deciphering lines of code

Coyote1967
Coyote1967 Posts: 1
in Software Development

Hey everyone I don't know if this is the right place to ask this question but I am trying to learn what different lines of code do and I need some help. I am looking at the Save command specifically. It was written to save files and back then up to a directory for later backup usage. The problem that I am having is in the first part of the code. For instance on this line

if [! -d $1];then echo $1 is not a directory; exit 1; fi

I know the fi means that the line is finished and the echo is a command to run the $1 file but what is the first part? Any help will be appreciated. Thanks

Comments

  • darkixion
    darkixion Posts: 41
    Well fi actually marks the end of the if statement (it's just "if" backwards). The [! -d $1] is performing a test on a passed value, which is being referred to by $1 (the 1st parameter). -d tests to see whether it is a directory. The ! negates the statement to make it check whether it is not a directory.

    In English, it's saying "If the parameter passed is not a directory, then say '[value of parameter passed] is not a directory."

    Here are more options you can use (although probably not a complete list):
    -e file
-s non-empty file
-b block device 
-c character device
-f a non-directory and non-device file
-p pipe
-h symbolic link
-r file has read permission
-w file has write permission
-x file has execute permission
-k has sticky bit set
-O you are the file owner
-G same group as you
-N modified since last read
  • darkixion
    darkixion Posts: 41
    Oh, I should have clarified the square brackets. Those just indicate that a test is being performed inside them, and it will return 0 (true) or 1 (false) to the if clause.

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