使用`http.NewRequest(...)`做一个URL编码的POST请求 [英] Make a URL-encoded POST request using `http.NewRequest(...)`
问题描述
我想要发送一个POST请求到一个发送我的数据的API为 application / x-www-form-urlencoded
内容类型。由于我需要管理请求标头,因此我使用 http.NewRequest (method,urlStr string,body io.Reader)
方法创建一个请求。对于这个POST请求,我将我的数据查询添加到URL中,并将主体留空,如下所示:
package main
导入(
字节
fmt
净/ http
净/网址
strconv
)
func main(){
apiUrl:=https://api.com
resource:=/ user /
data:= url.Values {}
data.Set(name,foo)
data.Add(surname,bar)
u,_:= url .ParseRequestURI(apiUrl)
u.Path = resource
u.RawQuery = data.Encode()
urlStr:= fmt.Sprintf(%v,u)//https: //api.com/user/?name=foo&surname=bar
client:=& http.Client {}
r,_:= http.NewRequest(POST ,urlStr,nil)
r.Header.Add(Authorization,auth_token = \XXXXXXX \)
r.Header.Add(Content-Type,application / (data.Encode())))
resp,x-www-form-urlencoded)
r.Header.Add(Content-Length,strconv.Itoa _:= client.Do(r)
fmt.Println(resp.Status)
}
当我回应时,我总是得到 400 BAD REQUEST
。我相信这个问题依赖于我的请求,API不知道我发布了哪个有效载荷。我知道诸如 Request.ParseForm
,但不确定如何在这种情况下使用它。也许我错过了一些头,也许有更好的方式发送有效载荷作为 application / json
类型使用身体
参数?
必须在 body $ c $
http.NewRequest(method,urlStr string,body io.Reader)
方法的参数,作为实现 io.Reader
interface。
基于示例代码:
包主要
导入(
fmt
净/ http
净/网址
strconv
字符串
)
func main(){
apiUrl:=https://api.com
resource:=/ user /
data:= url.Values {}
data.Set(name,foo)
data.Add(surname,bar)
u,_:= url.ParseRequestURI(apiUrl)
u.Path =资源
urlStr:= u.String()//'https://api.com/user/'
client:=& http.Cl客户端{}
r,_:= http.NewRequest(POST,urlStr,strings.NewReader(data.Encode()))// URL编码的有效载荷
r.Header.Add(Authorization ,auth_token = \XXXXXXX \)
r.Header.Add(Content-Type,application / x-www-form-urlencoded)
r.Header。添加(Content-Length,strconv.Itoa(len(data.Encode())))
resp,_:= client.Do(r)
fmt.Println(resp .Status)
}
resp.Status
是 200 OK
这种方式。
I want to make a POST request to an API sending my data as a application/x-www-form-urlencoded
content type. Due to the fact that I need to manage the request headers, I'm using the http.NewRequest(method, urlStr string, body io.Reader)
method to create a request. For this POST request I append my data query to the URL and leave the body empty, something like this:
package main
import (
"bytes"
"fmt"
"net/http"
"net/url"
"strconv"
)
func main() {
apiUrl := "https://api.com"
resource := "/user/"
data := url.Values{}
data.Set("name", "foo")
data.Add("surname", "bar")
u, _ := url.ParseRequestURI(apiUrl)
u.Path = resource
u.RawQuery = data.Encode()
urlStr := fmt.Sprintf("%v", u) // "https://api.com/user/?name=foo&surname=bar"
client := &http.Client{}
r, _ := http.NewRequest("POST", urlStr, nil)
r.Header.Add("Authorization", "auth_token=\"XXXXXXX\"")
r.Header.Add("Content-Type", "application/x-www-form-urlencoded")
r.Header.Add("Content-Length", strconv.Itoa(len(data.Encode())))
resp, _ := client.Do(r)
fmt.Println(resp.Status)
}
As I response, I get always a 400 BAD REQUEST
. I believe the problem relies on my request and the API does not understand which payload I am posting. I'm aware of methods like Request.ParseForm
, not really sure how to use it in this context though. Maybe am I missing some further Header, maybe is there a better way to send payload as a application/json
type using the body
parameter?
URL-encoded payload must be provided on the body
parameter of the http.NewRequest(method, urlStr string, body io.Reader)
method, as a type that implements io.Reader
interface.
Based on the sample code:
package main
import (
"fmt"
"net/http"
"net/url"
"strconv"
"strings"
)
func main() {
apiUrl := "https://api.com"
resource := "/user/"
data := url.Values{}
data.Set("name", "foo")
data.Add("surname", "bar")
u, _ := url.ParseRequestURI(apiUrl)
u.Path = resource
urlStr := u.String() // 'https://api.com/user/'
client := &http.Client{}
r, _ := http.NewRequest("POST", urlStr, strings.NewReader(data.Encode())) // URL-encoded payload
r.Header.Add("Authorization", "auth_token=\"XXXXXXX\"")
r.Header.Add("Content-Type", "application/x-www-form-urlencoded")
r.Header.Add("Content-Length", strconv.Itoa(len(data.Encode())))
resp, _ := client.Do(r)
fmt.Println(resp.Status)
}
resp.Status
is 200 OK
this way.
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