Go Lang- Gin : How to extract only the body (and ignore other garbage) from httputil. DumpRequest

Question

我知道你可以从

ioutil.ReadAll(c.Request.Body)

但是使用 httputil.DumpRequest

转储，错误 := httputil.DumpRequest(c.Request, true)

将给出正文内容以及其他值，最后是正文内容。

Content type: application/json IP: 127.0.0.1:36846 header token: Content length: 76 Request Method: POST Request URL: /signup Body: POST /signup HTTP/1.1 Host: 127.0.0.1:8080 Accept: / Accept-Encoding: gzip,deflate Accept-Language: en-US,en;q=0.8 Connection: keep-alive Content-Type: application/json Origin: chrome-extension://hgmloofddffdnphfgcellkdfbfbjeloo User-Agent: Mozilla/5.0 (X11; Linux i686) AppleWebKit/537.36 (KHTML, like Gecko) Chrome/38.0.2125.101 Safari/537.36 {"fname":"aFirstName", "lname":"aLName", "email":"test@test.com", "password":"123"}

有没有一种有效的方法可以只从 httputil.DumpRequest() 获取正文内容？即仅在这种情况下

{"fname":"aFirstName", "lname":"aLName", "email":"test@test.com", "password":"123"}

Answer 1

您不为此使用 httputil.DumpRequest，这是一个调试功能。

假设你想解析 json，你可以这样做:

defer c.Request.Body.Close()
var data yourDataType
if err := json.NewDecoder(c.Request.Body).Decode(&data); err != nil {
    // handle error
}
// handle data