Dissect a bad oneliner¶
$ ls *.zip | while read i; do j=`echo $i | sed 's/.zip//g'`; mkdir $j; cd $j; unzip ../$i; cd ..; done
This is an actual one-liner someone asked about in #bash. There are several things wrong with it. Let's break it down!
(Please read http://mywiki.wooledge.org/ParsingLs.) This command executes ls on the expansion of *.zip. Assuming there are filenames in the current directory that end in '.zip', ls will give a human-readable list of those names. The output of ls is not for parsing. But in sh and bash alike, we can loop safely over the glob itself:
Let's break it down some more!
The goal here seems to be get the filename without its .zip extension. In fact, there is a POSIX(r)-compliant command to do this: basename The implementation here is suboptimal in several ways, but the only thing that's genuinely error-prone with this is "echo $i". Echoing an unquoted variable means wordsplitting will take place, so any whitespace in $i will essentially be normalized. In sh it is necessary to use an external command and a subshell to achieve the goal, but we can eliminate the pipe (subshells, external commands, and pipes carry extra overhead when they launch, so they can really hurt performance in a loop). Just for good measure, let's use the more readable, modern $() construct instead of the old style backticks:
In Bash we don't need the subshell or the external basename command. See Substring removal with parameter expansion:
Let's keep going:
As a programmer, you never know the situation under which your program will run. Even if you do, the following best practice will never hurt: When a following command depends on the success of a previous command(s), check for success! You can do this with the "&&" conjunction, that way, if the previous command fails, bash will not try to execute the following command(s). It's fully POSIX(r). Oh, and remember what I said about wordsplitting in the previous step? Well, if you don't quote $j, wordsplitting can happen again.
That's almost right, but there's one problem -- what happens if $j contains a slash? Then cd .. will not return to the original directory. That's wrong! cd - causes cd to return to the previous working directory, so it's a much better choice:
(If it occurred to you that I forgot to check for success after cd -, good job! You could do this with { cd - || break; }, but I'm going to leave that out because it's verbose and I think it's likely that we will be able to get back to our original working directory without a problem.)
So now we have:
sh $ for i in *.zip; do j=$(basename "$i" ".zip"); mkdir "$j" && cd "$j" && unzip ../$i && cd -; done
Let's throw the unzip command back in the mix:
Well, besides word splitting, there's nothing terribly wrong with this. Still, did it occur to you that unzip might already be able to target a directory? There isn't a standard for the unzip command, but all the implementations I've seen can do it with the -d flag. So we can drop the cd commands entirely:
There! That's as good as it gets.