unix - What is the modification date ls reports for directories in Bash shell?

09
2014-02

Nagel

When I run ls -l in Bash shell, it lists a modification time for both files and directories. As noted in this thread, the modification time for directories does not reflect all of its contents. It also does not seem to be affected by changing the name of the directory. What exactly does the modification time for a directory reflect?

Is it the latest modification time of a file (or directory) in the top level of the directory?

Answers

Dolda2000

Note that, in Unix, a directory does not "contain" the files in it. Rather, it contains links to them. See the link(2) system call for further information.

This means that the "direct" contents of a directory is a list of filenames and corresponding i-numbers. The modification date of the directory, therefore, indicates whenever this list changed. Operations that would cause such changes would include, but not necessarily be limited to, the following:

Creating a new file in the directory
Removing (or, rather, unlink(2)ing) a file from the directory
Renaming a file in the directory
Hard-linking a file elsewhere into the directory

user32019

Think of directory as of simple file containing list of other files. Whenever you change its contents (rename files, add or remove contained files) the modification time of a directory changes.

Related Answers

glenn jackman

A compact version:

options=(
    "Display a long listing of the current directory"
    "Display who is logged on the system"
    "Display the current date and time"
    "Quit" 
)

select option in "${options[@]}"; do
    case "$REPLY" in 
        1) ls -l ;;
        2) who ;;
        3) date ;;
        4) break ;;
    esac
done

bmk

I guess it's due to the while loop ;)

The value of $input doesn't change inside of the loop. Either you insert something like a read input into the loop or you change the while ... do into a if ... then. If I understand the intention of your script correctly you don't need those while loops where they are now. You need one big while loop covering everything from echo "Menu for $name till the end.

My version of the script would be:

#!/bin/bash
echo -n "Name please? "
read name
input=""
while [ "$input" != "4" ]; do
  echo "Menu for $name
    1. Display a long listing of the current directory
    2. Display who is logged on the system
    3. Display the current date and time
    4. Quit "
  read input
  input1="1"
  input2="2"
  input3=$(date)
  input4=$(exit)
  if [ "$input" = "3" ]; then
    echo "Current date and time: $input3"
  fi

  if [ "$input" = "4" ]; then
    echo "Goodbye $input4"
  fi
done

Ranjithkumar T

case is here to help you, try

 # cat list.sh
 #!/bin/bash
 echo -n "Name please? "
 read name
 echo "Menu for $name"
 echo -e "Enter ( 1 ) to Display a long listing of the current directory \nEnter ( 2 )             to Display who is logged on the system\nEnter ( 3 ) to Display the current date and     time\nEnter ( 4 ) to Quit"
 read en
 case "$en" in
 1)ls -lR;;
 2)who;;
 3)date;;
 4)echo "Good Bye..."
 sleep 0;;
 *)echo "Wrong Option selected" ;;
 esac

Output:

 # ./list.sh
 Name please? Ranjith
 Menu for Ranjith
 Enter ( 1 ) to Display a long listing of the current directory
 Enter ( 2 ) to Display who is logged on the system
 Enter ( 3 ) to Display the current date and time
 Enter ( 4 ) to Quit

If you enter other options then the menu shows, then

 # ./list.sh
 Name please? Ranjith
 Menu for Ranjith
 Enter ( 1 ) to Display a long listing of the current directory
 Enter ( 2 ) to Display who is logged on the system
 Enter ( 3 ) to Display the current date and time
 Enter ( 4 ) to Quit
 5
 Wrong Option selected

...

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unix - What is the modification date ls reports for directories in Bash shell?