duanbei2914 2019-03-13 21:11
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Golang中数组的XML解组属性不同

I have a web service where I expect to receive two different requests; one is called Request containing just one Request and the other is called RequestBulk which contains an array of Requests. Both mapped to Golang structs as follows:

type RequestBulk struct {
    XMLName  xml.Name  `xml:"https://item.com RequestBulk"`
    Message  string    `xml:"Message"`
    Request  []Request `xml:"Request,omitempty"`
}

type Request struct {
    XMLName xml.Name `xml:"https://item.com Request"`
    Text    string   `xml:"text"`
}

Unmarshalling the following XMLs works as expected:

<Request xmlns="https://item.com">
  <text>Some request text</text>
</Request>
<RequestBulk xmlns="https://item.com">
  <Message>Some Text</Message>
  <Request xmlns="https://item.com">
    <text>Some request text</text>
  </Request>
  <Request xmlns="https://item.com">
    <text>Some other request text</text>
  </Request>
</RequestBulk>

The Problem

In RequestBulk, if I change

Request []Request `xml:"Request,omitempty"`

to

RequestMessage []Request `xml:"RequestMessage,omitempty"`

and change the XML to:

<RequestBulk xmlns="https://item.com">
  <Message>Some Text</Message>
  <RequestMessage xmlns="https://item.com">
    <text>Some request text</text>
  </RequestMessage>
  <RequestMessage xmlns="https://item.com">
    <text>Some other request text</text>
  </RequestMessage>
</RequestBulk>

I get the following error:

expected element type <Request> but have <RequestMessage>

Obviously because of the XMLName xml.Name `xml:"https://item.com Request"`

The Question

How do I keep the Request struct unchanged and still accept messages of type RequestBulk with a different name for the struct Request, namely, RequestMessage?

In other words; How do I use the same struct with different namespaces?


Run it on Go Playground.

  • 写回答

1条回答 默认 最新

  • dpklt4291 2019-03-13 21:26
    关注

    You can implement the Unmarshaler interface to overwrite the element's local name before passing the element on to the decoder for the actual unmarshaling.

    func (r *Request) UnmarshalXML(d *xml.Decoder, start xml.StartElement) error {
        if start.Name.Local == "RequestMessage" {
            start.Name.Local = "Request" // overwrite
        }
        type tmp Request // avoid infinite recursive calls to Request.UnmarshalXML
        return d.DecodeElement((*tmp)(r), &start) // unmarshal
    }
    

    https://play.golang.org/p/0a_gpgkywwf

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
    评论

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