Ouput json to http.ResponseWriter with template

Question

i've this template:

var ListTemplate = `
{
    "resources": [
        {{ StringsJoin . ", " }}
    ]
  }
`

rendered with:

JoinFunc := template.FuncMap{"StringsJoin": strings.Join}
tmpl := template.Must(template.New("").Funcs(JoinFunc).Parse(ListTemplate))

if i send it to a http.ResponseWriter the output text is escaped.

var list []string
tmpl.Execute(w, list)

how can i write a json this way?

Why not serialise to JSON properly? What is the reason to manually craft it from pieces? — zerkms, Mar 06 '19 at 11:00

icza · Accepted Answer · 2019-03-06T11:18:49.527

4

You shouldn't use Go's template engine (neither text/template nor html/template) to generate JSON output, as the template engine has no knowledge of JSON syntax and rules (escaping).

Instead use the encoding/json package to generate JSON. You may use json.Encoder to write / stream the response directly to an io.Writer, such as http.ResponseWriter.

Example:

type Output struct {
    Resources []string `json:"resources"`
}

obj := Output{
    Resources: []string{"r1", "r2"},
}

enc := json.NewEncoder(w)

if err := enc.Encode(obj); err != nil {
    // Handle error
    fmt.Println(err)
}

Outputs (try it on the Go Playground):

{"resources":["r1","r2"]}

edited Mar 06 '19 at 11:18

answered Mar 06 '19 at 11:01

icza

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thanks, i definitively should do this way. the application is a very simple apl that was not meant to return json at the beginning, just a raw text – George Livanoss Mar 06 '19 at 11:45
is it okay to `json.NewEncoder(w)` for every new request? – Yadu Aug 04 '20 at 10:50

Ouput json to http.ResponseWriter with template

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