res.json不是函数-nodejs

问题描述 投票:0回答:1

我在使用res.json时遇到此错误,以获取我的函数的返回信息,这是如下所述在其他地方使用的代码,运行正常。

async function getPlaylist(req, res, playlistId) {

  try {

    // const calltoken = await getToken()

    // const token = calltoken.data.access_token

    // console.log(token)

    const config = {
      headers: {
        'Authorization': 'Bearer ' + 'BQA0K9bKgBVn8xTp-yTsoaKs5VfS7EyjMIL03OEOy05wq08ZmLkNfqbbnsL_hFT1AV2FGN5tAQdeDV1X224', //token,
        'Content-Type': 'application/json',
        'Accept': 'application/json'
      }
    }

    const url = 'https://api.spotify.com/v1/playlists/1j2L3DjzZ3SdN64r83Sblj?si=cuvOrPONSO6caE9XD6smEg'

    await axios.get(url, config)
      .then(function (response) {

        var playlist = response

        var items = playlist.data.tracks.items

        // console.log(items)

        const playlistfull = []

        items.forEach(index => {

          var playlistdata = {
            name: index.track.name,
            artists: index.track.album.artists[0].name,
            album: index.track.album.name,
            url: index.track.external_urls.spotify
          }

          playlistfull.push(playlistdata)

        })

        return res.json(playlistfull)

      })

  } catch (error) {

    return console.log(error)

  }
}

我在使用res.json时遇到此错误,以返回我的函数,这是如下所述在其他地方使用的代码,运行正常。异步函数getPlaylist(req,res,playlistId){...

javascript node.js
1个回答
0
投票

您必须在NodeJS中使用节点的resparameter赋予function

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