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

조회 수: 1(최근 30일)
Joanna Przeworska
Joanna Przeworska 2021년 5월 17일
댓글: Siddharth Bhutiya 2021년 5월 19일
Dear all,
How to convert this (below) into 'out_20210517'?
filename = sprintf('out_%s.xlsx', today('datetime'));
filename =
'out_17-May-2021.xlsx'

채택된 답변

Geoff Hayes
Geoff Hayes 2021년 5월 17일
Joanna - perhaps try using
datestr(now,'yyyymmdd')
instead/
  댓글 수: 2
Siddharth Bhutiya
Siddharth Bhutiya 2021년 5월 19일
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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VBBV
VBBV 2021년 5월 17일
%if true
k = datestr(filename(5:15),'yyyymmdd')
strcat(filename(1:4),k,filename(16:20))
You can use filename to reconstruct the date

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