Python URL下载

Question

以下代码返回none。我该如何解决？我正在使用Python 2.6。

import urllib

URL = "http://download.finance.yahoo.com/d/quotes.csv?s=%s&f=sl1t1v&e=.csv"
symbols = ('GGP', 'JPM', 'AIG', 'AMZN','GGP', 'JPM', 'AIG', 'AMZN')
#symbols = ('GGP')

def fetch_quote(symbols):
    url = URL % '+'.join(symbols)
    fp = urllib.urlopen(url)
    try:
        data = fp.read()
    finally:
        fp.close()

def main():
    data_fp = fetch_quote(symbols)
#    print data_fp
if __name__ =='__main__':
    main()

Answer 1

您必须明确return来自data功能的fetch_quote。像这样：

def fetch_quote(symbols):
    url = URL % '+'.join(symbols)
    fp = urllib.urlopen(url)
    try:
        data = fp.read()
    finally:
        fp.close()
    return data # <======== Return

如果没有明确的return语句，Python会返回None，这就是你所看到的。

Answer 2

您的方法未明确return任何内容，因此returns None