英文:
MongoDB Return entire JSON from query using bson.Raw
问题
我正在尝试搜索这个文档:
"meta": {
"pageId": "...",
"userId": "...",
"ver": "0",
},
"dialog": {
...
}
并将整个"dialog"
作为JSON获取,所以我创建了这个结构体:
type Dialog struct {
Dialog bson.Raw `json:"dialog" bson:"dialog"`
}
然后我这样查询文档:
dialog := Dialog{}
query := c.Find(locate).One(&dialog)
当我打印dialog
时,我得到一堆数字,我相信这些是查询结果的原始字节。
问题是:如何将其解组为JSON对象?
我找到的唯一相关信息是 https://stackoverflow.com/questions/25276884/marshal-into-a-bson-raw(它没有解释如何解组为JSON)。
更新
fmt.Println(bson.UnmarshalJSON(dialog.Dialog.Data, &a))
结果是:
json: unknown constant "l"
正如你所看到的,我不得不从Raw类型中提取Data
,我不认为这是最好的方法,因为Raw
类型中还有未使用的Kind
字段。另外,这个l
是什么意思?
更新2
我原以为必须将其解组为JSON类型才能使用,但我发现直接解组为map更好,所以这是代码:
var a bson.M
fmt.Println(bson.Unmarshal(dialog.Dialog.Data, &a))
fmt.Println(a)
这对我起作用了
然而,我仍然忽略了Raw
类型的Kind
字段。有没有更好的方法?
英文:
I'm trying to search for this document:
"meta": {
"pageId": "...",
"userId": "...",
"ver": "0",
},
"dialog": {
...
}
and get the entire "dialog"
as a JSON, so I created this struct:
type Dialog struct {
Dialog bson.Raw `json:"dialog" bson:"dialog"`
}
So I query the document like this:
dialog := Dialog{}
query := c.Find(locate).One(&dialog)
and when I print dialog
, I get a bunch of numbers, which I believe are the raw bytes from the query.
The question is: how to unmarshal it into a JSON object?
The only thing I've found about this are https://stackoverflow.com/questions/25276884/marshal-into-a-bson-raw (which doesn't explain how to unmarshal into a json)
Update
Following https://stackoverflow.com/questions/39785289/how-to-marshal-json-string-to-bson-document-in-golang-for-writing-to-mongodb, I did:
fmt.Println(bson.UnmarshalJSON(dialog.Dialog.Data, &a))
which gets me:
json: unknown constant "l"
As you can see I had to extract the Data
from the Raw type, I don't think this is the best way to do it since there's the Kind
field which is not being used. Also, what's this 'l'?
Update 2
I thought I had to Unmarshal into a JSON type in order to work with it but I've found that's better to Unmarshal to a map directly, so here it is:
var a bson.M
fmt.Println(bson.Unmarshal(dialog.Dialog.Data, &a))
fmt.Println(a)
It worked for me
However, I'm still ignoring the Kind
field of the Raw
type. Is there a better way to do it?
答案1
得分: 2
JSON不是一种类型,因此无法将其解组为JSON类型的值。JSON是一种结构化数据的文本表示形式。
bson.Raw
也不等于JSON表示形式,因此不可避免地需要进行某种转换。
你可以将其解组为interface{}
类型的值,然后使用json.Marshal()
来“渲染”JSON表示形式。
如果你希望这个过程是“自动”的,可以创建一个名为JSONStr
的新类型,并通过它实现bson.Setter
和bson.Getter
接口。
以下是示例代码:
type JSONStr string
func (j *JSONStr) SetBSON(raw bson.Raw) (err error) {
var i interface{}
if err = raw.Unmarshal(&i); err != nil {
return
}
data, err := json.Marshal(i)
if err != nil {
return
}
*j = JSONStr(data)
return
}
func (j *JSONStr) GetBSON() (interface{}, error) {
var i interface{}
if err := json.Unmarshal([]byte(*j), &i); err != nil {
return nil, err
}
return i, nil
}
使用这个JSONStr
类型:
type Dialog struct {
Dialog JSONStr `bson:"dialog"`
}
更新:
如果你不需要真正的JSON文本表示形式,只需使用interface{}
类型即可:
type Dialog struct {
Dialog interface{} `bson:"dialog"`
}
然后可以像这样获取JSON文本:
var dialog Dialog
// 加载对话
data, err := json.Marshal(dialog.Dialog)
if err != nil {
// 处理错误
}
fmt.Println(string(data))
(基本上,这就是JSONStr.SetBSON()
方法的作用。)
请注意,interface{}
将“覆盖”所有数据结构。如果你知道它是一个对象,可以使用map。如果你知道它是一个数组,可以使用切片,等等。你还可以使用任何具体类型。
英文:
JSON is not a type, so you can't unmarshal into a value of type JSON. JSON is a textual representation of some structured data.
bson.Raw
is also not equal to JSON representation, so some kind of transformation is inevitable.
What you may do is unmarshal into a value of type interface{}
, and then use json.Marshal()
to "render" the JSON representation.
If you want this to be "automatic", you may create a new type called JSONStr
, and you may implement the bson.Setter
and bson.Getter
interfaces by it.
This is how it could look like:
type JSONStr string
func (j *JSONStr) SetBSON(raw bson.Raw) (err error) {
var i interface{}
if err = raw.Unmarshal(&i); err != nil {
return
}
data, err := json.Marshal(i)
if err != nil {
return
}
*j = JSONStr(data)
return
}
func (j *JSONStr) GetBSON() (interface{}, error) {
var i interface{}
if err := json.Unmarshal([]byte(*j), &i); err != nil {
return nil, err
}
return i, nil
}
And using this JSONStr
type:
type Dialog struct {
Dialog JSONStr `bson:"dialog"`
}
Update:
If you don't really want a JSON text representation, just use type interface{}
:
type Dialog struct {
Dialog interface{} `bson:"dialog"`
}
And obtain the JSON text like this:
var dialog Dialog
// load a dialog
data, err := json.Marshal(dialog.Dialog)
if err != nil {
// handle error
}
fmt.Println(string(data))
(Basically this is what the JSONStr.SetBSON()
method did exactly.)
Note that interface{}
will "cover" all data structures. If you know it's an object, you may use a map. If you know it's an array, you may use a slice, etc. You may also use any concrete type.
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库,让每个人都能够通过互相帮助和分享经验来进步。
评论