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

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Dear all,
How to convert this (below) into 'out_20210517'?
filename = sprintf('out_%s.xlsx', today('datetime'));
filename =
'out_17-May-2021.xlsx'

Respuesta aceptada

Geoff Hayes
Geoff Hayes el 17 de Mayo de 2021
Joanna - perhaps try using
datestr(now,'yyyymmdd')
instead/
  2 comentarios
Joanna Przeworska
Joanna Przeworska el 17 de Mayo de 2021
Of course! It works. Thanks a lot Geoff.
Best regards,
JP
Siddharth Bhutiya
Siddharth Bhutiya el 19 de Mayo de 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.

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