How to get the latest file from a directory using groovy?

Question

I have a directory that contains a list of files. I wanted to get the latest file out of all the contents of the said directory. How will I do that?

I am using this code, but I am not getting the latest file out of it. Please help.

def fileDir = new File("A/B").listFiles().first()

Thanks.

there is a similar question on how to do that in java. you can use that in groovy too and propably simplify it a bit... http://stackoverflow.com/questions/203030/best-way-to-list-files-in-java-sorted-by-date-modified — moeTi, Nov 08 '12 at 08:53

score 13 · Accepted Answer · answered Nov 08 '12 at 09:20

13

As simple as:

new File( 'A/B' ).listFiles()?.sort { -it.lastModified() }?.head()

(taking the negative lastModified, as we want the newest file first)

answered Nov 08 '12 at 09:20

tim_yates

While using this my fileDir is = null. Any suggestions? – Rod Kimble Sep 14 '17 at 06:12
@RodKimble not sure what you mean... maybe a new question? – tim_yates Sep 14 '17 at 08:18

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