Get date from file name

Question

I'm looking for a command or a simple procedure to get the date from the filename. My file name could have multiple date formats.

Eg:

Filename Format Output filename_2016-01-01 filename_%Y-%m-%d 2016-01-01 2016-01-01_filename %Y-%m-%d_filename 2016-01-01 20160101-filename %Y%m%d_filename 2016-01-01 1451606400-filename %s-filename 2016-01-01

I'm looking for some procedure say if I specify my "Format" it will return date from that format as output. I'm not worried about format of output. I'm not expecting a huge script. Please let me know if any such functionality already exists.

If there is not other procedure, I'm planning to build one.

So it is not the time a file was created, rather a subset of the actual filename you want to return? Is it to be stored in a file? Output on screen? Give some pseudo-wanted output. — mattias
– mattias, Commented Jun 30, 2016 at 20:12
Is the rest of the filename always known? For example, if it is always "Filename<datestuff>" or "<datestuff>_filename" or "<datestuff>-filename", can't you just strip away the known part? — mattias
– mattias, Commented Jun 30, 2016 at 20:14
@Prashanth I'm not aware of an existing command which could achieve that. — hek2mgl
– hek2mgl, Commented Jun 30, 2016 at 20:21

Jeff Y · Accepted Answer · 2016-06-30 20:51:19Z

Bash has variable dereferencing modification syntax that can handle some of what you want:

$ f=filename_2016-01-01 $ echo ${f#*_} 2016-01-01 $ f=2016-01-01_filename $ echo ${f%_*} 2016-01-01 $ f=20160101-filename $ echo ${f%-*} 20160101 $ f=1451606400-filename $ echo ${f%-*} 1451606400

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