获取spotify链接返回python

时间:2015-11-21 19:22:22

标签: json python-3.x

我尝试在Python中使用此功能,以便从spotify获取链接:

import os
import urllib.parse
import urllib.request
import json
from tkinter import *
from tkinter import filedialog


def makeRequestToSpotify(songname):
    encodedname = urllib.parse.quote_plus(songname)
    url = "https://api.spotify.com/v1/search?query=" + encodedname+"&type=track"                                                                                                                             "
    content = urllib.request.urlopen(url)
    data = json.loads(content.readall().decode('utf-8'))
    songLink = data["tracks"]["items"][0]["externalurls"]["spotify"]
    print(songLink)
def getMusicFilesFromFolder(afolder):
    extensions = [".mp3",".wma",".aac",".m4a"]
    filenames = []
    for x in os.walk(afolder):
        for file in x[2]:
            file_name, filext = os.path.splitext(file)
            for ext in extensions:
                if filext == ext:
                    filenames.append(file_name)
                    break
    for file in filenames:
        makeRequestToSpotify(file)
    print(filenames)

def buttonClick():
    file_path = filedialog.askdirectory()
    getMusicFilesFromFolder(file_path)
#User Interface
root = Tk()
root.title("Music Repository with Spotify")
root.geometry("600x600")
app = Frame(root)
app.grid()
chooseFileButton = Button(app, command=buttonClick, text = "Choose a Folder")
chooseFileButton.grid()

root.mainloop()

这给了我一个index out of range - 错误。

我希望从此程序中获取该文件夹中每首歌曲的链接

0 个答案:

没有答案