The string '3/18/09 10:16 PM' is not a valid AllXsd value

Question

Obviously the reader doesn't like this format incoming from the response XML.

Wondering if I can reformat this. Trying to convert to DateTime using the following code with my XmlReader:

reader.ReadContentAsDateTime();

Solution 1

Xml readers generally expect dates/times in a very specific format; you can use this yourself using XmlConvert:

string s = XmlConvert.ToString(DateTime.Now);
DateTime when = XmlConvert.ToDateTime(s);

If you are using something else, you'll have to read it as a string and use DateTime.TryParseExact (or similar) to specify the actual format string:

string s = reader.ReadContentAsString();
DateTime when = DateTime.ParseExact(s, "M/d/yy hh:mm tt",
     CultureInfo.InvariantCulture);

If you are using XmlSerializer, you could use a shim property to do the conversion - let me know if this is what you are doing...

Solution 2

According to the XML schema spec, date time values should be in ISO8601 format, e.g., something like

2009-03-13T22:16:00

Comments

[Brian Leeming]

The data I was feeding in happened to be 10/29/2012 15:25 and that failed. When I changed it to 10/11/2012 15:25, it worked.

[burhan]

@MarcGravell, when i try to deserialize some rss feed using XmlSerializer, PubDate element causes error. how can i fix it?

[Marc Gravell]

@burhan by looking at what the value is coming in as, and handling it appropriately? Alternatively, the core framework includes RSS-targeted classes that may do a better job.

[burhan]

@MarcGravell, it gives the error System.InvalidOperationException. can you give the codes to fix this manually?

[Marc Gravell]

@burhan not without seeing the xml and your model, no

[burhan]

pls visit this link . my codes use the same algorithm.

[burhan]

@MarcGravell, did you visit that link? if so, what is the solution? (someone has the same problem in the comments at the end of the page)

[Marc Gravell]

@burhan the solution is, as always, to have xml and a model that match. Sometimes that means you can go direct - you might need to have the dates as a string type (rather than DateTime) and then do additional post-processing on it to parse the time. A random link, however, does not provide the exact xml you are struggling with.

[Zubin]

In ruby use iso8601 method, eg Time.now.iso8601.

[ajiang]

Out of curiosity, ISO8601 format should look something like 2009-03-18T22:16:00-05:00, but David's answer (omitting the timezone) works for me too (whereas the ISO8601 format doesn't). Is this because of a local serialization in the XmlConverter (which therefore doesn't need the timezone)?

[David Norman]

In ISO8601, the timezone is optional, so both my original string and your string are valid ISO8601 date time. According to w3.org/TR/xmlschema-2/#deviantformats, the timezone is also optional in XML schema. I'm not sure why the time zone version doesn't work for you.

[userlond]

For php Artisans:) there is toIso8601String method in nesbot/carbon php lib.

[relipse]

in php you can use date('c', $time);