CODEWITHSUNDEEP
c#jsondatetimejson.net
g.pickardou
g.pickardou

Reputation: 35873

Deserialize a single DateTime object with JsonConvert

Context

The line JsonConvert.SerializeObject(DateTime.Now) gives the following result:

"2018-05-25T07:59:27.2175427+02:00"

However when I try to deserialize this JSON string to a DateTime with the line: JsonConvert.DeserializeObject<DateTime>("2018-05-25T07:59:27.2175427+02:00")

it gives an Newtonsoft.Json.JsonReaderException with the following message:

Unexpected character encountered while parsing value: 2. Path '', line 1, position 1.

What else I've tried so far

"2018-05-25T07:59:27"

causes the very same exception

Question

Having the datetime string in JSON serialized format, I would like to have a DateTime variable and the correct value in it. How can I accomplish this task?

Upvotes: 1

Views: 1335

Answers (2)

dbc
dbc

Reputation: 116785

As shown in the JSON standard, a JSON string literal must be quoted:

A string is a sequence of zero or more Unicode characters, wrapped in double quotes, using backslash escapes. A character is represented as a single character string. A string is very much like a C or Java string.

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Thus, to be valid JSON, your c# string literal must include the surrounding double quotes, like so:

var dateTime = JsonConvert.DeserializeObject<DateTime>("\"2018-05-25T07:59:27.2175427+02:00\"");

It's easy to confuse the outermost quotes, which are part of the c# language and delimit the string in your c# code but are not included in the string itself, with the inner quotes, which are part of the string literal itself.

Sample fiddle here.

Upvotes: 3

Jon Skeet
Jon Skeet

Reputation: 1500665

The problem is that JsonConvert.DeserializeObject looks like it wants a JSON object rather than just any JSON value. (It's a shame that SerializeObject doesn't always produce an object, but...)

You can parse it like this:

DateTime dt = new JValue("2018-05-25T07:59:27.2175427+02:00").ToObject<DateTime>();

Or (equivalently? I'm not entirely sure):

DateTime dt = (DateTime) new JValue("2018-05-25T07:59:27.2175427+02:00");

There may be a better way of doing so, but that at least works.

Upvotes: 3

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