问题描述
我在使用 (-> req :params) 提取的服务器中收到以下请求:
{"_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“视频”。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!"}))
我试图将 (-> req :body) 保存到文件中(这是一个输入流)。这一定是不正确的。通过将数据保存到服务器上的文件中,将服务器接收到的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?
解决方法
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))