如何将具有动态字段的JSON对象映射到Go结构体中

huangapple
go评论123阅读模式
英文:

How to map JSON objects with dynamic fields to Go structs

问题

我正在使用Go开发一个网站，并将其连接到Elastic Search。在Elastic Search中，我可以为索引类型设置动态字段。当我从Elastic Search中读取文档时，它将返回一个JSON对象作为结果，其中可能包含具有动态名称（或用户定义字段）的字段。

我可以获取JSON结果并将其解组为Go结构体，但我不知道将这些动态字段保留为Go结构体的最佳方法是什么。

这是我正在做的。例如，如果我从Elastic Search获取一个联系人文档，它可能如下所示：

  1. {  
  2.    "EmailAddress": "test@test.com",
  3.    "Name": "Test Contact",
  4.    "Phone": "17894785236",
  5.    "City": "San Francisco",
  6.    "State": "California"
  7. }

而Contact的Go结构体如下所示：

  1. type Contact struct {
  2. 	EmailAddress            string
  3. 	Name                    string
  4. 	Phone                   string
  5.     CustomFields            map[string]interface{}
  6. }

我实现了Marshaler和Unmarshaler来重写对象的编组和解组方式。

  1. func (c *Contact) MarshalJSON() ([]byte, error) {
  2. 	contactMap := make(map[string]interface{})
  3. 	contactMap["EmailAddress"] = c.EmailAddress
  4. 	contactMap["Name"] = c.Name
  5. 	contactMap["Phone"] = c.Phone
  7. 	for k, v := range c.CustomFields {
  8. 		contactMap[k] = v
  9. 	}
  11. 	return json.Marshal(contactMap)
  12. }
  14. func (c *Contact) UnmarshalJSON(data []byte) error {
  15. 	var contactMap map[string]interface{}
  17. 	if c == nil {
  18. 		return errors.New("RawString: UnmarshalJSON on nil pointer")
  19. 	}
  21. 	if err := json.Unmarshal(data, &contactMap); err != nil {
  22. 		return err
  23. 	}
  25. 	c.EmailAddress = contactMap["EmailAddress"].(string)
  26. 	c.Name = contactMap["Name"].(string)
  27. 	c.Phone = contactMap["Phone"].(string)
  29. 	for key, val := range contactMap {
  30. 		if key != "EmailAddress" && key != "Name" && key != "Phone" {
  31. 			c.CustomFields[key] = value
  32. 		}
  33. 	}
  35. 	return nil
  36. }

这是最佳的方法吗？你有什么建议？

英文:

I am developing a website using Go and connecting it to Elastic Search. In Elastic Search I can have dynamic fields for index types. When I read a document from Elastic Search it will return a JSON object as the result, which can include fields with dynamic names (or user defined fields).

I can get the JSON result and unmarshal it into a Go struct, but I do not know what is the best way to keep those dynamic fields as part of the Go struct.

This is what I am doing. For example, if I get a document for a Contact from Elastic Search it may look something like this:

  1. {  
  2.    "EmailAddress": "test@test.com",
  3.    "Name": "Test Contact",
  4.    "Phone": "17894785236",
  5.    "City": "San Francisco",
  6.    "State": "California"
  7. }

And the Go struct for Contact is:

  1. type Contact struct {
  2. 	EmailAddress            string
  3. 	Name                    string
  4. 	Phone                   string
  5.     CustomFields            map[string]interface{}
  6. }

And I implement Marshaler and Unmarshaler to override how the object is Marshaled and Unmarshalled.

  1. func (c *Contact) MarshalJSON() ([]byte, error) {
  2. 	contactMap := make(map[string]interface{})
  3. 	contactMap["EmailAddress"] = c.EmailAddress
  4. 	contactMap["Name"] = c.Name
  5. 	contactMap["Phone"] = c.Phone
  7. 	for k, v := range c.CustomFields {
  8. 		contactMap[k] = v
  9. 	}
  11. 	return json.Marshal(contactMap)
  12. }
  14. func (c *Contact) UnmarshalJSON(data []byte) error {
  15. 	var contactMap map[string]interface{}
  17. 	if c == nil {
  18. 		return errors.New("RawString: UnmarshalJSON on nil pointer")
  19. 	}
  21. 	if err := json.Unmarshal(data, &contactMap); err != nil {
  22. 		return err
  23. 	}
  25. 	c.EmailAddress = contactMap["EmailAddress"].(string)
  26. 	c.Name = contactMap["Name"].(string)
  27. 	c.Phone = contactMap["Phone"].(string)
  29. 	for key, val := range contactMap {
  30. 		if key != "EmailAddress" && key != "Name" && Key != "Phone" {
  31. 			c.CustomFields[key] = value
  32. 		}
  33. 	}
  35. 	return nil
  36. }

Is this the best way to do this? What would you recommend?

答案1

得分: 5

只需进行一些小的清理

  1. var contactMap map[string]interface{}
  3. if c == nil {
  4.     return errors.New("RawString: UnmarshalJSON on nil pointer")
  5. }
  7. if err := json.Unmarshal(data, &contactMap); err != nil {
  8.     return err
  9. }
  11. for key, val := range contactMap {
  12.     switch key {
  13.         case "EmailAddress":
  14.             c.EmailAddress = val.(string)
  15.         case "Name":
  16.             c.Name = val.(string)
  17.         case "Phone":
  18.             c.Phone = val.(string)
  19.         default:
  20.             c.CustomFields[key] = val
  21.     }
  22. }
  23. }
英文:

Just add little cleanup

  1. var contactMap map[string]interface{}
  3. if c == nil {
  4. 	return errors.New("RawString: UnmarshalJSON on nil pointer")
  5. }
  7. if err := json.Unmarshal(data, &contactMap); err != nil {
  8. 	return err
  9. }
  11. for key, val := range contactMap {
  12. 	switch key {
  13. 		case "EmailAddress":
  14. 			c.EmailAddress = val.(string)
  15. 		case "Name":
  16. 			c.Name = val.(string)
  17. 		case "Phone":
  18. 			c.Phone = val.(string)
  19. 		default:
  20. 			c.CustomFields[key] = val
  21. 		}
  22. 	}
  23. }

答案2

得分: 4

正如Simon在评论中指出的那样，如果JSON的结构是固定的，使用一个大的**map[string]interface{}**并不理想。最好的方法是使用结构体并使用http://golang.org/pkg/encoding/json/#Unmarshal进行解组（参考示例：http://play.golang.org/p/cDTe8x4xLk）。

但是对于结构未知的大型JSON数据块，您的实现完全有效。

编辑：添加了示例链接

英文:

As Simon has pointed out in comment, using one big map[string]interface{} isn't ideal if the structure of the json is fixed. Best way then is to use structure and unmarshall it using http://golang.org/pkg/encoding/json/#Unmarshal. (refer the example: http://play.golang.org/p/cDTe8x4xLk)

But for large json blob, for which structure is not known beforehand, your implementation works perfectly.

edit: added link to example

huangapple
  • 本文由 发表于 2014年12月16日 04:49:49
  • 转载请务必保留本文链接：https://go.coder-hub.com/27492888.html
  • elasticsearch
  • go
  • json
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