Question

I am new in bash scripting and I have to write a script that finds the newest .mpg file in directory and then sends it to a ftp server. So far I found command that find my file.

#! /bin/bash

find /home/user/directory -type f -mmin -50 -name *.mpg \( ! -regex ".*/\..*" \)

This works but I can’t figure out how to send the output of that command to ftp server. I’ve read that I have to use variables, but I can’t understand them so far.

Answer 1

Use can pipe the output from one command as the input to another with |

find /home/user/directory -type f -mmin -50 -name *.mpg \( ! -regex ".*/\..*" \) | ftp ...

A simpler way to find the latest mpg file:

ls -rt /home/user/directory/*.mpg | head -1 | ftp ...

If you want to send multiple files xargs will be needed i.e for the 5 newest files

ls -rt /home/user/directory/*.mpg | head -5 | xargs ftp ...

ls lists the files in the /home/user/directory/*.mpg where * is expanded to any filename where with the .mpg extention. -t tells ls to list in time order and -r is reverse sort as we want the newest first not the oldest. The head command is used to only show the number of results we want like one -1 or five -5. head is usually used for viewing only a certain number of lines in a file, to view the first line in a file we would run head -1 file.txt.

Pipe example:

grep is used to search for text in files/stdout

So if you wanted to know if firefox is contained in the first line of file.txt we would pipe the output of head -1 file.txt to grep firefox

head -1 file.txt | grep firefox

You can pipe multiple commands together to achieve the result you want.