json.Unmarshal uses RFC3339 to decode datetime but why mine is not working?

[Qiang Han]

package main

import "fmt"

import "time"

import "encoding/json"

type Foo struct {

Bar time.Time

}

func main() {

msg := "{\"bar\":\"2012-01-02T16:10:22Z07:00\"}"

var f Foo

err := json.Unmarshal([]byte(msg), &f)

if err != nil {

fmt.Printf("%+v", err)

} else {

fmt.Println(f)

}

}

output: parsing time ""2006-01-02T15:04:05Z07:00"" as ""2006-01-02T15:04:05Z07:00"": cannot parse "07:00"" as """

http://play.golang.org/p/mO38T1N5vn

Not sure what string are allowed after "Z"

[Konstantin Khomoutov]

JSON does not define any representation for encoding date/time
information.

It *appears* that people tend to use ISO-8601 for them [1].
May be that's what Go does.
Note that the documentation for the encoding/json package does not
mention dates and times at all.

1. http://en.wikipedia.org/wiki/ISO_8601

[Vasiliy]

Because your string is invalid. Timezone must be either in format
`±hh:mm` or be a single letter `Z` which means UTC. Everything works
fine if you remove `07:00` after `Z` or replace `Z` with `+` or `-`.

On Mon, 3 Feb 2014 00:48:18 -0800 (PST)
Qiang Han <qjavasw...@gmail.com> wrote:

[Qiang Han]

Make sense.

http://golang.org/pkg/time/#pkg-constants and the miss-leading error message: parsing time ""2006-01-02T15:04:05Z07:00"" as ""2006-01-02T15:04:05Z07:00"": cannot parse "07:00"" as """ make me think that it is parsed as RFC3339 format.

[Matt Harden]

See http://godoc.org/time#Time.MarshalJSON

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[tianzhaixi...@gmail.com]

thanks very much.

在 2014年2月3日星期一 UTC+8下午4:48:18，Qiang Han写道：