"sed" command in bash

Could someone explain this command for me:

cat | sed -e 's,%,$,g' | sudo tee开发者_如何转开发 /etc/init.d/dropbox << EOF
   echo "Hello World"
EOF

What does the "sed" command do?

sed is the Stream EDitor. It can do a whole pile of really cool things, but the most common is text replacement.

The s,%,$,g part of the command line is the sed command to execute. The s stands for substitute, the , characters are delimiters (other characters can be used; /, : and @ are popular). The % is the pattern to match (here a literal percent sign) and the $ is the second pattern to match (here a literal dollar sign). The g at the end means to globally replace on each line (otherwise it would only update the first match).

Here sed is replacing all occurrences of % with $ in its standard input.

As an example

$ echo 'foo%bar%' | sed -e 's,%,$,g'

will produce "foo$bar$".

It reads Hello World (cat), replaces all (g) occurrences of % by $ and (over)writes it to /etc/init.d/dropbox as root.

sed is a stream editor. I would say try man sed.If you didn't find this man page in your system refer this URL:

http://unixhelp.ed.ac.uk/CGI/man-cgi?sed