快传号视频批量上传，禁止转载！_快传号视频上传-优快云博客

本文链接：https://blog.youkuaiyun.com/chenzhigaoa/article/details/81944796

本文介绍了如何使用start_upload方法开始视频批量上传，并利用多线程加速上传过程。同时，详细讲解了通过ffmpeg截取视频封面，以及提交上传、保存和发布视频的步骤。

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1 开始上传视频 start_upload方法
这里写图片描述

function start_upload(cookie,filename,size);
//post 参数
var form = 
	{
		'fsize':size,
		'object':filename,
		'parallel':'3',//表示多少个线程同时上传视频
		'insertOnly':'1',
		'channel':'bjyt'
	}
url : https://v.sj.360.cn/upload/video/request
/*
* cookie:用户登录保存cookie
* filename：上传视频文件名，不含路径
* size:文件大小(Byte)
*/

2 多线程上传视频

url:  1步骤返回JSON下的next_block_url + bhash

bhash加密方法：

	function sha1(data) {
		
        function n() {
          this.blocks = [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0],
            this.h0 = 1732584193,
            this.h1 = 4023233417,
            this.h2 = 2562383102,
            this.h3 = 271733878,
            this.h4 = 3285377520,
            this.block = this.start = this.bytes = 0,
            this.finalized = this.hashed = !1,
            this.first = !0
        }
        var r = "object" == typeof e ? e: {}; ! r.JS_SHA1_NO_NODE_JS && "object" == typeof process && process.versions && process.versions.node && (r = global);
        var i = (!r.JS_SHA1_NO_COMMON_JS && "object" == typeof module && module.exports, "function" == typeof t && t.amd, "0123456789abcdef".split("")),
        s = [ - 2147483648, 8388608, 32768, 128],
        o = [24, 16, 8, 0],
        u = ["hex", "array", "digest", "arrayBuffer"],
        a = [],
        f = function(e) {
            return function(t) {
                return (new n()).update(t)[e]()
            }
        };
        return n.prototype.update = function(e) {
            if (!this.finalized) {
                var t = "string" != typeof e;
                t && e.constructor === r.ArrayBuffer && (e = new Uint8Array(e));
                for (var n, i, s = 0,
                u = e.length || 0,
                a = this.blocks; s < u;) {
                    if (this.hashed && (this.hashed = !1, a[0] = this.block, a[16] = a[1] = a[2] = a[3] = a[4] = a[5] = a[6] = a[7] = a[8] = a[9] = a[10] = a[11] = a[12] = a[13] = a[14] = a[15] = 0), t) for (i = this.start; s < u && i < 64; ++s) a[i >> 2] |= e[s] << o[3 & i++];
                    else for (i = this.start; s < u && i < 64; ++s) n = e.charCodeAt(s),
                    n < 128 ? a[i >> 2] |= n << o[3 & i++] : n < 2048 ? (a[i >> 2] |= (192 | n >>