CODEWITHSUNDEEP
jsonmongodbgomgo
user6791424
user6791424

Reputation:

MongoDB Return entire JSON from query using bson.Raw

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 Marshal into a bson.Raw (which doesn't explain how to unmarshal into a json)

Update

Following 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?

Upvotes: 1

Views: 4149

Answers (1)

icza
icza

Reputation: 417592

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.

Upvotes: 2

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