Golang JSON Marshal/Unmarshal postgres now() - json

Question

I'm using postgres' now() as a default for my created timestamp, which is generates this:
id | user_id | title | slug | content | created
----+---------+-------+------+---------+----------------------------
1 | 1 | Foo | foo | bar | 2014-12-16 19:41:31.428883
2 | 1 | Bar | bar | whiz | 2014-12-17 02:03:31.566419
I tried to use json.Marshal and json.Unmarshal and ended up getting this error:
parsing time ""2014-12-16 19:41:31.428883"" as ""2006-01-02T15:04:05Z07:00"": cannot parse " 19:41:31.428883"" as "T"
So I decided to try and create a custom time, but can't seem to get anything working.
Post.go
package models
type Post struct {
Id int `json:"id"`
UserId int `json:"user_id"`
Title string `json:"title"`
Slug string `json:"slug"`
Content string `json:"content"`
Created Tick `json:"created"`
User User `json:"user"`
}
Tick.go
package models
import (
"fmt"
"time"
)
type Tick struct {
time.Time
}
var format = "2006-01-02T15:04:05.999999-07:00"
func (t *Tick) MarshalJSON() ([]byte, error) {
return []byte(t.Time.Format(format)), nil
}
func (t *Tick) UnmarshalJSON(b []byte) (err error) {
b = b[1 : len(b)-1]
t.Time, err = time.Parse(format, string(b))
return
}
Any help would be much appreciated, running what I've wrote here gives me this:
json: error calling MarshalJSON for type models.Tick: invalid character '0' after top-level value

Answer 1

JSON requires strings to be quoted (and in JSON a date is a string), however your MarshalJSON function returns an unquoted string.
I've slightly amended your code and it works fine now:
package models
import (
"fmt"
"time"
)
type Tick struct {
time.Time
}
var format = "2006-01-02T15:04:05.999999-07:00"
func (t *Tick) MarshalJSON() ([]byte, error) {
// using `append` to avoid string concatenation
b := make([]byte, 0, len(format)+2)
b = append(b, '"')
b = append(b, t.Time.Format(format)...)
b = append(b, '"')
return b, nil
}
func (t *Tick) UnmarshalJSON(b []byte) (err error) {
b = b[1 : len(b)-1]
t.Time, err = time.Parse(format, string(b))
return
}

Answer 2

it's seems like you're using the wrong format. Postgres uses RFC 3339, which is already defined in the time package.
This should work:
time.Parse(time.RFC3339, string(b))