英文:
Make a URL-encoded POST request using `http.NewRequest(...)`
问题
我想向一个API发送POST请求,将我的数据作为application/x-www-form-urlencoded
的内容类型发送。由于我需要管理请求头,所以我使用http.NewRequest(method, urlStr string, body io.Reader)
方法创建请求。对于这个POST请求,我将我的数据查询附加到URL上,将请求体留空,就像这样:
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)
}
作为响应,我始终得到400 BAD REQUEST
。我认为问题出在我的请求上,API无法理解我正在发送的有效载荷。我知道有像Request.ParseForm
这样的方法,但不确定如何在这个上下文中使用它。也许我漏掉了一些其他的头部信息,也许有更好的方法可以使用body
参数将有效载荷作为application/json
类型发送吗?
英文:
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?
答案1
得分: 247
URL编码的有效载荷必须作为实现了io.Reader
接口的类型,提供给http.NewRequest(method, urlStr string, body io.Reader)
方法的body
参数。
根据示例代码:
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.Set("surname", "bar")
u, _ := url.ParseRequestURI(apiUrl)
u.Path = resource
urlStr := u.String() // "https://api.com/user/"
client := &http.Client{}
r, _ := http.NewRequest(http.MethodPost, 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")
resp, _ := client.Do(r)
fmt.Println(resp.Status)
}
这样,resp.Status
将会是200 OK
。
英文:
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.Set("surname", "bar")
u, _ := url.ParseRequestURI(apiUrl)
u.Path = resource
urlStr := u.String() // "https://api.com/user/"
client := &http.Client{}
r, _ := http.NewRequest(http.MethodPost, 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")
resp, _ := client.Do(r)
fmt.Println(resp.Status)
}
resp.Status
is 200 OK
this way.
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