How do I convert a string to a date?
显示 更早的评论
Currently I have the string:
2017-03-18 03:26:42.000000
I want to convert this to a date, and I use the following function:
datetime(my_string,'InputFormat','yyyy-mm-dd HH:MM:SS')
However, when I do this I get the following error:
Unable to parse '2017-03-18 03:26:42.000000' as a date/time using the format 'yyyy-mm-dd HH:MM:SS'.
I have also tried
datetime(my_string,'InputFormat','yyyy-mm-dd HH:MM:SSSSSSSSS')
and
datetime(my_string,'InputFormat','yyyy-mm-dd HH:MM:SS.SSSSSSS')
But none of these seem to work. How would I go about doing this?
采纳的回答
Carefully read the documentation of the Format property. The month identifier is uppercase M and the minute identifier is lowercase m. The seconds field is lowercase s with the fractional part as uppercase S. So:
datetime(my_string,'InputFormat','yyyy-MM-dd HH:mm:ss.S')
if you want to display the fractional seconds by default:
datetime(my_string,'InputFormat','yyyy-MM-dd HH:mm:ss.S', 'Format', 'yyyy-MM-dd HH:mm:ss.SSS')
2 个评论
Douglas Leaffer
2022-1-22
The date and time vectors are the same length. I think perhaps the problem lies with the way the .txt file is imported. I have attached an excerpt of the file if you have bandwidth to suggest further help. The coumns of interest are date = column 7; time = column 8
Michael Connolly
2023-3-28
Thank you for posting this! It fixed my problem.
更多回答（1 个）
Rik
2017-7-17
It is a bit confusing, but the capital S only denotes the fraction of a second, so not the second itself. The code below works.
my_string='2017-03-18 03:26:42.000000';
datetime(my_string,'InputFormat','yyyy-MM-dd HH:mm:ss.SSSSSS')
5 个评论
Guillaume
2017-7-17
Note that for decoding you only need one S regardless of the number of fractional digits. The number of S only affects the datetime display when it is part of the Format property
Rik
2017-7-17
Good to know, thanks Guillaume.
Amini
2022-2-3
thanks for your answers, I have to use the function datetime, actually I want to find the differences between timestamps. (for example: row 2 minuse row 1). the problem is two characters before the DAY 24 '[ and also the format of MONTH Sep. how should I use this function to approach this problem? thanks in advance. @Rik @Guillaume @Douglas Leaffer
Amini
2022-2-3
in this way it works:
DateStrings = {'24/09/2021:10:56:23'};
t = datetime(DateStrings,'InputFormat','dd/MM/yyyy:HH:mm:ss')
But my date is 24/Sep/2021:10:56:23
and in this way it doesn't work
DateStrings = {'24/Sep/2021:10:56:23'};
t = datetime(DateStrings,'InputFormat','dd/MM/yyyy:HH:mm:ss')
%again, beside the problem of two extra characters
Amini
2022-2-3
I found it how to eliminate the extra characters, using erase but still stacked with MONTH part
类别
在 帮助中心 和 File Exchange 中查找有关 Dates and Time 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!
Translated by 
