grep for portion of a line with specific syntax

I am having trouble trying to grep (or whatever is best) the date out of a specific output


499942634/1322025954/FFF035/11/23/2011 12:25:53 AM/full
3423594078/1326441742/FFF034/01/13/2012 02:37:07 AM/full


I was thinking use cut from position but the date is at two different positions in the two lines above.

Please let me know how to get that date!
Justin_EdmandsAsked:
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farzanjCommented:
Try this:
cut -d'/' -f4-6 filename

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Justin_EdmandsAuthor Commented:
almost, strip off the end?
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farzanjCommented:
With grep you could do something like

grep -Po '\d\d\/\d\d\/\d{4}' filename

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Justin_EdmandsAuthor Commented:
perfect.
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farzanjCommented:
Ok, Try this:
cut -d'/' -f4-6 filename | cut -d' ' -f1

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Justin_EdmandsAuthor Commented:
Thank you very much.
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farzanjCommented:
Glad to help :)
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