I have XML data I am retrieving via a REST API that I am unmarshal-ing into a GO struct. One of the fields is a date field, however the date format returned by the API does not match the default time.Time parse format and thus the unmarshal fails.
Is there any way to specify to the unmarshal function which date format to use in the time.Time parsing? I'd like to use properly defined types and using a string to hold a datetime field feels wrong.
Sample struct:
type Transaction struct {
Id int64 `xml:"sequencenumber"`
ReferenceNumber string `xml:"ourref"`
Description string `xml:"description"`
Type string `xml:"type"`
CustomerID string `xml:"namecode"`
DateEntered time.Time `xml:"enterdate"` //this is the field in question
Gross float64 `xml:"gross"`
Container TransactionDetailContainer `xml:"subfile"`
}
The date format returned is "yyyymmdd".
I had the same problem.
time.Time
doesn't satisfy the xml.Unmarshaler
interface. And you can not specify a date fomat.
If you don't want to handle the parsing afterward and you prefer to let the xml.encoding
do it, one solution is to create a struct with an anonymous time.Time
field and implement your own UnmarshalXML
with your custom date format.
type Transaction struct {
//...
DateEntered customTime `xml:"enterdate"` // use your own type that satisfies UnmarshalXML
//...
}
type customTime struct {
time.Time
}
func (c *customTime) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
const shortForm = "20060102" // yyyymmdd date format
var v string
d.DecodeElement(&v, &start)
parse, err := time.Parse(shortForm, v)
if err != nil {
return err
}
*c = customTime{parse}
return nil
}
If your XML element uses an attribut as a date, you have to implement UnmarshalXMLAttr the same way.
From what I have read the encoding/xml has some known issues that have been put off until a later date...
To get around this issue, instead of using the type time.Time
use string
and handle the parsing afterwards.
I had quite a bit of trouble getting time.Parse to work with dates in the following format: "Fri, 09 Aug 2013 19:39:39 GMT"
Oddly enough I found that "net/http" has a ParseTime function that takes a string that worked perfectly... http://golang.org/pkg/net/http/#ParseTime
I've implemented a xml dateTime format conforming a spec, you can find it on GitHub: https://github.com/datainq/xml-date-time
You can find XML dateTime in W3C spec
const shortForm = "20060102" // yyyymmdd date format
It is unreadable. But it is right in Go. You can read the source in http://golang.org/src/time/format.go
来源:https://stackoverflow.com/questions/17301149/golang-xml-unmarshal-and-time-time-fields