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How to convert this 'out_17-May-2021.xlsx' into 'out_20210517.xlsx'?

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Joanna Przeworska
Joanna Przeworska il 17 Mag 2021
Commentato: Siddharth Bhutiya il 19 Mag 2021
Dear all,
How to convert this (below) into 'out_20210517'?
filename = sprintf('out_%s.xlsx', today('datetime'));
filename =
'out_17-May-2021.xlsx'

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Geoff Hayes
Geoff Hayes il 17 Mag 2021
Joanna - perhaps try using
datestr(now,'yyyymmdd')
instead/
  2 Commenti
Joanna Przeworska
Joanna Przeworska il 17 Mag 2021
Of course! It works. Thanks a lot Geoff.
Best regards,
JP
Siddharth Bhutiya
Siddharth Bhutiya il 19 Mag 2021
You could also do it using datetime by specifying the display format using the Format name-value pair.
>> filename = sprintf('out_%s.xlsx', datetime('now','Format','yyyyMMdd'))
filename =
'out_20210519.xlsx'
This seems like a simple workflow so it might not matter, but if you are working with dates and times, datetime would be recommended over using something like datestr or datenum.

Accedi per commentare.

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