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I have a zip file that contains a CSV and I would like to unzip that into a specific folder and rename the file the same as the zip file name. For example.,

I have a zip file named: youtube_videos.csv.zip and I would like to unzip it to a certain folder called target and rename the file in that folder to youtube-videos.csv. I have it, but in two lines as below:

unzip datasets/youtube_videos.csv.zip -d target 
mv target/videos_data.csv target/youtube_videos.csv

Is there a one liner that could be used?

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    Why do you need a one liner? It can complicate things unnecessarilly. Nov 3, 2021 at 20:32
  • Is that good enough what I did?
    – joesan
    Nov 3, 2021 at 20:35
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    I don't see a problem with it. I posted an answer for a one liner, but as you can see, there are complications. Nov 3, 2021 at 20:52
  • why don't just use unzip datasets/youtube_videos.csv.zip -d target; mv target/videos_data.csv target/youtube_videos.csv?
    – phuclv
    Nov 4, 2021 at 3:36

2 Answers 2

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Here's a one liner option, but it doesn't rename the file, it outputs its contents to stdout. Also maybe can't be used with the -d option (I can't make it work):

$ unzip -p datasets/youtube_videos.csv.zip videos_data.csv > target/youtube_videos.csv

Or if there's just one file inside the zip, you can omit the name of the zipped files:

$ unzip -p datasets/youtube_videos.csv.zip > target/youtube_videos.csv

The option -p does, according to the manual:

-p extract files to pipe (stdout). Nothing but the file data is sent to stdout, and the files are always extracted in binary format, just as they are stored (no conversions).

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With libarchive bsdtar:

bsdtar -xvvf youtube_videos.csv.zip -C target -'s/videos_data/youtube_videos/g'

Would extract the archive into the target directory with videos_data substituted with youtube_videos in the name of archive members.

(note that the target directory must exist beforehand).

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