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In addition to my own computer, I sometimes use an Ubuntu cluster at my school. Rather than manually keep my .bashrc's in sync, I would like to make the school cluster's .bashrc source my personal .bashrc from DropBox via a URL. However, when I naively try source http://myurl, I just get an error: http://myurl: No such file or directory. How can I can get bash to source from a script located online?

Worst case, I could curl to a named pipe and source that. Is there anything more elegant?

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2 Answers 2

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You can use process substitution with source:

source <(curl -s http://example.com/foo)

Note: I consider directly running code retrieved over the internet to be a serious security risk. It's probably less risky if this is done over an internal network (depending on its overall security).

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  • Clever! Good point though that an attacker could easily feed me any code they want.
    – AlcubierreDrive
    Commented Mar 9, 2011 at 14:49
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    Works on Linux, fails on OSX. Writing to a tempfile for OSX instead.
    – Edward Anderson
    Commented Sep 8, 2014 at 20:29
  • @nilbus: According to this, you might try source <(curl -s http://example.com/foo | iconv -f windows-1251)
    – Dennis Williamson
    Commented Sep 8, 2014 at 20:57
  • That got rid of the error, but the script didn't run still on OSX.
    – Edward Anderson
    Commented Sep 9, 2014 at 0:31
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    @Dennis Williamson: fixed by using source /dev/stdin <<< "$(curl -s http://xxx.sh)", it works fine on both!
    – Miao1007
    Commented Dec 14, 2016 at 10:53
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Can't you just use 

curl http://example.com/whatever.sh | bash
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  • That works and it's what I'm currently doing, but I was wondering if there's a more elegant way. But maybe this is the best, so if nothing turns up soon I'll accept your answer.
    – AlcubierreDrive
    Commented Mar 9, 2011 at 14:12
  • Also, note that this is going in my .bashrc, so technically your solution is infinite recursion. ;) But I know what you meant.
    – AlcubierreDrive
    Commented Mar 9, 2011 at 14:45
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    Piping commands to bash will execute them in a separate process, which makes this method unusable for executing bashrc.
    – grawity_u1686
    Commented Mar 9, 2011 at 15:15
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    Depends on what the commands are - but yes, sorry I agree
    – John Burton
    Commented Mar 9, 2011 at 16:24

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