Question

我正在使用golang中的rest api，并且能够将数据封送至xml和json。我有多个结构的记录，但是xml编组并没有为这些记录添加根。因此，该XML无效。

我看到一些通过设置XMLName可以解决的问题，但我认为这只会将我的客户类型更改为其他类型。

我仍然可以尝试的是在xml编组的字节周围添加并合并。但是我不确定这是否是我必须要做的。

编辑：显然，由于我要编组内部有多个对象的结构，因此XML正确编组了。然后，封送处理为每个对象创建单独的XML文档。那么问题是，其余客户端可以使用哪些功能来消耗数据，或者在剩余响应中发送多个xml文档是不明智的做法？浏览器无法显示生成的xml。

要返回单个XML文档，我必须将列表放入结构中，然后可以将其编组。我不确定如何用sqlx初始化结构。

添加以下代码无效：

type custs struct {
    List []Customer `xml:"Customer"`
}

custs customers.[]Customer =  []Customer{}

代码：

package main

import (
    "database/sql"
    json "encoding/json"
    "encoding/xml"
    "fmt"
    "net/http"
)

type Customer struct {
    CustomerID int            `db:"CustomerId"`
    FirstName  string         `db:"FirstName"`
    LastName   string         `db:"LastName"`
    Company    sql.NullString `db:"Company"`
}

func tstSQLite(w http.ResponseWriter, r *http.Request) {

    contentType := r.Header.Get("Content-type")

    Customers := []Customer{}

    var err error

    fmt.Print("executing query")
    err = chinookdb.Select(&Customers, "SELECT CustomerId, FirstName,LastName, Company FROM customers")

    if err != nil {
        panic(err)
    }

    for _, g := range Customers {
        fmt.Println("FirstName:", g.FirstName, "LastName:", g.LastName)
    }

    var data []byte

    if contentType == "application/xml" {
        data, err = xml.Marshal(Customers)
    }

    if contentType == "application/json" || contentType != "application/xml" {
        data, err = json.Marshal(Customers)
    }


    w.Header().Set("Content-Type", "application/json")
    w.Write(data)
}

我得到的是以下xml：

   <Customer>
      <CustomerID>1</CustomerID>
      <FirstName>Luís</FirstName>
      <LastName>Gonçalves</LastName>
      <Company>
         <String>Embraer - Empresa Brasileira de Aeronáutica S.A.</String>
         <Valid>true</Valid>
      </Company>
   </Customer>
   <Customer>
      <CustomerID>2</CustomerID>
      <FirstName>Leonie</FirstName>
      <LastName>Köhler</LastName>
      <Company>
         <String />
         <Valid>false</Valid>
      </Company>
   </Customer>
   <Customer>
      <CustomerID>3</CustomerID>
      <FirstName>François</FirstName>
      <LastName>Tremblay</LastName>
      <Company>
         <String />
         <Valid>false</Valid>
      </Company>
   </Customer>

但是我认为应该是这样的：

<Customers>
   <Customer>
      <CustomerID>1</CustomerID>
      <FirstName>Luís</FirstName>
      <LastName>Gonçalves</LastName>
      <Company>
         <String>Embraer - Empresa Brasileira de Aeronáutica S.A.</String>
         <Valid>true</Valid>
      </Company>
   </Customer>
   <Customer>
      <CustomerID>2</CustomerID>
      <FirstName>Leonie</FirstName>
      <LastName>Köhler</LastName>
      <Company>
         <String />
         <Valid>false</Valid>
      </Company>
   </Customer>
   <Customer>
      <CustomerID>3</CustomerID>
      <FirstName>François</FirstName>
      <LastName>Tremblay</LastName>
      <Company>
         <String />
         <Valid>false</Valid>
      </Company>
   </Customer>
</Customers>

由于客户放在括号中，因此编组的json看起来正确：

[
   {
      "CustomerID":1,
      "FirstName":"Luís",
      "LastName":"Gonçalves",
      "Company":{
         "String":"Embraer - Empresa Brasileira de Aeronáutica S.A.",
         "Valid":true
      }
   },
   {
      "CustomerID":2,
      "FirstName":"Leonie",
      "LastName":"Köhler",
      "Company":{
         "String":"",
         "Valid":false
      }
   }
]

Answer 1

您必须将此客户集合包装在另一个结构中

package main

import (
    "encoding/xml"
    "fmt"
    "os"
)

func main() {

    type Customer struct {
        XMLName   xml.Name `xml:"Customer"`
        CustomerID int            `db:"CustomerId"`
        FirstName  string         `db:"FirstName"`
        LastName   string         `db:"LastName"`
        Company    string `db:"Company"`
    }

    type Customers struct {
        List []Customer
    }

    csts := []Customer{
        Customer{CustomerID: 1, FirstName: "John", LastName: "Doe", Company: "Demo Company"},
        Customer{CustomerID: 2, FirstName: "John2", LastName: "Doe2", Company: "Demo Company2"},
    }


    res := Customers {
        List: csts,
    }

    output, err := xml.MarshalIndent(res, "  ", "    ")
    if err != nil {
        fmt.Printf("error: %v\n", err)
    }

    os.Stdout.Write(output)
}

默认情况下，它可以使用构造的名称。您还可以使用xml.Name类型自定义名称。

XML封送处理会生成无根的XML

1 个答案: