微信小程序的网址不能打印出来吗?
上传视频后返回了tempfiles,我想要提取出其中的tempfilepath,但是一直在无法识别,提示我类型错了。
参考代码是https://blog.csdn.net/zhoulib__/article/details/125641526?spm=1001.2014.3001.5506里的,代码问题如下图:
我的代码如下:
Page({
// 链接可能会过期换成自己的就行
data:{
//"audioUrl": "https://dl.stream.qqmusic.qq.com/C4000026gMkr1L9tKN.m4a?guid=7840612806&vkey=4696DBD6111F38EC3D75D0F4FDF769B3736068A795CF585BA18DF3A4BCD8F83D0662B54D70F64934AF74A34C9A354AFF0AA4660CCE4C66F0&uin=&fromtag=120032",
videoUrl: "C:/Users/priesty/Desktop/caogao/test.mp4",
videoInfo: {},
canvasWidth: 0,
canvasHeight: 0,
fps: 0,
duration: 0,
imageList: [],
tempFilePath:"",
},
onLoad() {
console.log("开始"),
this.videoDecoderStart()
},
// async是异步的意思,会返回一个promise对象,而awite会等待这个async的promise完成 并将reslove的结果返回出来
async videoDecoderStart() {
// 自己选择视频
const that = this
/* var {tempFilePath} = await wx.chooseMedia( */
//const tempFilePath = wx.chooseMedia(
wx.chooseMedia(
{
sourceType: ['album'],
success(res){
console.log(res);
//var tempFilePath = res.tempFilePath
that.setData({ //在data维护一组数据
tempFilePath: res.tempFiles.tempFilePath
})
console.log(tempFilePath) //无法识别,为什么?
}
}
)
console.log("选择")
console.log(tempFilePath) //还是无法识别?
// 网络视频
// wx.showLoading({
// title: '加载中...',
// })
// var {
// tempFilePath
// } = await this.getTempPath(this.data.videoUrl)
// wx.hideLoading()
var videoInfo = await wx.getVideoInfo({
src: that.data.tempFilePath //这里一直报错
})
this.setData({
videoInfo: videoInfo,
canvasWidth: videoInfo.width,
canvasHeight: videoInfo.height,
duration: videoInfo.duration,
fps: videoInfo.fps
}, () => {
// 创建视频解析器
this.videoDecoder = wx.createVideoDecoder()
const {
canvas,
context
} = this.initOffscreenCanvas(this.data.canvasWidth, this.data.canvasWidth)
this.videoDecoder.on("start", () => {
this.videoDecoder.seek(0)
this.timer = setInterval(() => {
this.getFrameData(canvas, context)
}, 300);
})
this.videoDecoder.on("seek", () => {})
this.videoDecoder.on("stop", () => {})
this.videoDecoder.start({
source: tempFilePath
})
})
},
问一下该怎么修改?