如何将文件保存到服务器中？

Question

我在服务器中收到以下请求，我正在使用该请求(->请求:params)进行提取：

{"_parts":[["video",{"_data":{"size":2971246,"blobId":"D002459C-47C5-4403-ABC6-A2DE6A46230A","offset":0,"type":"video/quicktime","name":"DCDE604A-954F-4B49-A1F9-1BCC2C2F37BC.mov","__collector":null}}],["key","VAL"]]}

它包含一个带有名称和blobId的文件"video“。但是，我想访问文件的数据并将其保存到文件中。到目前为止，我已经尝试了以下几种方法：

(defn upload-shot-video [req]
  (prn "uploading video")
  (prn "video is! " (-> req :multipart-params))
  (prn "video is " (-> req :params))
  (prn "video before is " (slurp (-> req :body)))
  (.reset (-> req :body))
  (prn "req full" (-> req))
  (prn "video after is " (-> req :body))
  (prn "video is! " (-> req :multipart-params))
  (clojure.java.io/copy (-> req :body) (clojure.java.io/file "./resources/public/video.mov"))

  (let [filename (str (rand-str 100) ".mov")]
    (s3/put-object
     :bucket-name "humboi-videos"
     :key filename
     :file "./resources/public/video.mov"
     :access-control-list {:grant-permission ["AllUsers" "Read"]})
    (db/add-video {:name (-> req :params :name)
                   :uri (str "https://humboi-videos.s3-us-west-1.amazonaws.com/" filename)}))
  (r/response {:res "okay!"}))

其中我试图将(-> req :body)保存到文件(这是一个输入流)中。这肯定是不正确的。通过将数据保存到服务器上的文件中，将服务器收到的这个文件保存到文件中的正确方法是什么？如何从请求中提取这些数据？

Answer 1

如果您使用的是Ring，则需要使用wrap-multipart-params中间件。

(ns controller
  (:require [ring.middleware.params :refer [wrap-params]]
            [ring.middleware.multipart-params :refer [wrap-multipart-params]])

(defn upload-shot-video [req]
  (let [uploaded-file (-> req :params "file" :tempfile) ;; here is a  java.io.File instance of your file
    (save-file uploaded-file)
    {:status 201 :body "Upload complete"}))

(def app
  (-> upload-shot-video
      wrap-params
      wrap-multipart-params))