CODEWITHSUNDEEP
curlfile-uploadgo
vs.
vs.

Reputation: 103

How can I upload a file using golang code?

With golang code I have to transfer file to remote service using their API. Their requirement is that request MUST NOT use multipart/form-data. I tried this curl command:

curl -i -X PUT -F [email protected] -H "Content-Type: text/plain"  https://url.of.endpoint.com

it doesn't work since it simulates form, but this command:

curl -i -X PUT -T text.txt -H "Content-Type: text/plain"  https://url.of.endpoint.com

works perfectly.

How can I translate this curl command to golang code?

Upvotes: 8

Views: 11745

Answers (2)

Arjan
Arjan

Reputation: 21465

You have to create a "PUT" request and set its request body to the contents of the file:

package main

import (
    "log"
    "net/http"
    "os"
)

func main() {
    data, err := os.Open("text.txt")
    if err != nil {
        log.Fatal(err)
    }
    defer data.Close()
    req, err := http.NewRequest("PUT", "http://localhost:8080/test.txt", data)
    if err != nil {
        log.Fatal(err)
    }
    req.Header.Set("Content-Type", "text/plain")

    client := &http.Client{}
    res, err := client.Do(req)
    if err != nil {
        log.Fatal(err)
    }
    defer res.Body.Close()
}

Upvotes: 12

KoRoN
KoRoN

Reputation: 375

It would be like this:

package main

import (
    "fmt"
    "net/http"
    "net/http/httputil"
    "os"
)

func main() {
    // Open a file.
    f, err := os.Open("text.txt")
    if err != nil {
        panic(err)
    }
    defer f.Close()

    // Post a file to URL.
    resp, err := http.Post("https://url.of.endpoint.com", "text/plain", f)
    if err != nil {
        panic(err)
    }
    defer resp.Body.Close()

    // Dump response to debug.
    b, err := httputil.DumpResponse(resp, true)
    if err != nil {
        panic(err)
    }
    fmt.Println(b)
}

Upvotes: 2

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