Parse date with format yyyyMMdd

[Guest]

Dear all,

i have a date in the format yyyyMMdd. When I try to parse this I get the
following error: string was not recognized as a valid datetime.

I use the following code:

CultureInfo culture = new CultureInfo("en-US");
culture.DateTimeFormat.DateSeparator = string.Empty;
culture.DateTimeFormat.ShortDatePattern = "yyyyMMdd";

// Here the error occurs:
DateTime date = DateTime.Parse("20041123", culture.DateTimeFormat);

Thanx,
René

 

[Guest]

Hey René,

Try this instead:

CultureInfo nfo = new CultureInfo("en-US");
DateTime d = DateTime.ParseExact("20041123", "yyyyMMdd", nfo);

HTH, Jakob.

 

[Guest]

Thanx,

this works.

To be honest this was not really my problem but I did try to understand date
formatting and then I ran into this problem.

Can you please look at my original problem. I have a xml with a date. When I
read it into a dataset with ReadXML I get a datetime conversion error (the
column is defined as xs:date in the dataset schema). The format is
"yyyyMMdd". I did try to solve the error by setting the cultureinfo of the
current threat, my code:

CultureInfo culture = new CultureInfo("en-US", false);
culture.DateTimeFormat.DateSeparator = string.Empty;
culture.DateTimeFormat.ShortDatePattern = "yyyyMMdd";
culture.DateTimeFormat.LongDatePattern = "yyyyMMdd";

System.Threading.Thread.CurrentThread.CurrentCulture = culture;
System.Threading.Thread.CurrentThread.CurrentUICulture = culture;

taskdata.ReadXml(reader, XmlReadMode.Auto);

Thanks,
René

 

[Guest]

As far as I know, the format of the dates in the XML must be 'yyyy-MM-dd'.
Otherwise you can not use ReadXml. Are you generating the XML yourself?

Regards, Jakob.

 

[Guest]

No, the xml comes from the back-office.

I'm not sure if a date always have to be something with a '-'.

I try to translate it to another problem:

If I want to do:
DateTime.Parse("20041123")

Do you think that it should work by changing the current culture:

CultureInfo culture = new CultureInfo("en-US", false);
culture.DateTimeFormat.DateSeparator = string.Empty;
culture.DateTimeFormat.ShortDatePattern = "yyyyMMdd";
culture.DateTimeFormat.LongDatePattern = "yyyyMMdd";

System.Threading.Thread.CurrentThread.CurrentCulture = culture;
System.Threading.Thread.CurrentThread.CurrentUICulture = culture;

DateTime.Parse("20041123")

This should work, don't you think (for me it doesn't but why)? Next, if this
works and readXML doesn't then I'm sure readXML needs a seperator.

Hope you find some time to share your thoughts,
René

 

[Guest]

No, I am afraid it can not be done that way.
DateTimeFormatInfo.ShortDatePattern is used only for outputting and not for
parsing.

But it is true that DateTime.Parse will use the current culture if nothing
else has been specified. The method
DateTimeFormatInfo.GetAllDateTimePatterns gives you the supported patterns
for a culture. I created a small piece of code that ran through all cultures
and all patterns, and none of them matched 'yyyyMMdd'.

Do you know for a fact that ReadXml uses DateTime.Parse?

Regards, Jakob.

 

[Guest]

Jakob,

thanks again for your response. I think you're right, all the patterns you
can set influence the display not the parsing. Strange, I think, that you
can't influence the parsing.

ReadXML isn't actually using datetime.parse but XMLConvert.ToDateTime.
This one does work when I use the format attribute:
DateTime dt = XmlConvert.ToDateTime("20041119", "yyyyMMdd");

But when I use dataset.readxml it doesn't work because it uses
XmlConvert.ToDateTime("20041119");
There is no way I can add a parameter.

So now I'm back where I started: can I add a custom date pattern to
DateTimeFormat (and then use this the current culture).

Regards,
René

 

[Guest]

René,

According to the XML Schema, dates in XML must be specified in the format
'yyyy-MM-yy' (http://www.w3.org/TR/xmlschema-2/#date).

I recommend not using datasets for this if you can not control the XML
(actually, I hardly ever recommend datasets but that's a different story).
If you parse the XML manually, you gain greater control.

Regards, Jakob.