String Python 2.7:未知url类型:urllib2-BeautifulSoup 导入库 新图书馆:

String Python 2.7:未知url类型:urllib2-BeautifulSoup 导入库 新图书馆:,string,python-2.7,url,beautifulsoup,urllib2,String,Python 2.7,Url,Beautifulsoup,Urllib2,定义变量: i = 1 str_i = str(i) seqPrefix = 'seq_' seq_1 = str('https://anyaddress.com/') quote_page = seqPrefix + str_i #然后,使用Python urllib2获取声明的url的HTML页面 # query the website and return the html to the variable 'page' page = urllib2.urlopen(quote_page)

定义变量:

i = 1
str_i = str(i)
seqPrefix = 'seq_'
seq_1 = str('https://anyaddress.com/')
quote_page = seqPrefix + str_i
#然后,使用Python urllib2获取声明的url的HTML页面

# query the website and return the html to the variable 'page'
page = urllib2.urlopen(quote_page)  


#Finally, parse the page into BeautifulSoup format so we can use BeautifulSoup to work on it.

# parse the html using beautiful soup and store in variable `soup`
soup = BeautifulSoup(page, 'html.parser')
因此,一切都很好……除了:

错误消息:

page=urlib2.urlopen(引用页面) 文件“C:\Python27\lib\urllib2.py”,第154行,在urlopen中 返回opener.open(url、数据、超时) 文件“C:\Python27\lib\urllib2.py”,第423行,打开 协议=请求获取类型() 文件“C:\Python27\lib\urllib2.py”,第285行,get\u类型 提升值错误,“未知url类型:%s”%self.\u原始 ValueError:未知url类型:seq_1

为什么?


txs.

您可以使用局部变量字典vars()


按照您的方式,它试图使用字符串“seq_1”作为URL而不是有效URL的seq_1变量的值来打开URL。

您可以使用局部变量字典vars()


按照您的方式,它试图使用字符串“seq_1”作为URL而不是有效URL的seq_1变量的值来打开URL。

看起来您需要concat
seq_1
&
str_i

Ex:

seq_1 = str('https://anyaddress.com/')
quote_page = seq_1 + str_i
https://anyaddress.com/1
输出:

seq_1 = str('https://anyaddress.com/')
quote_page = seq_1 + str_i
https://anyaddress.com/1

看起来您需要连接
seq_1
&
str_i

Ex:

seq_1 = str('https://anyaddress.com/')
quote_page = seq_1 + str_i
https://anyaddress.com/1
输出:

seq_1 = str('https://anyaddress.com/')
quote_page = seq_1 + str_i
https://anyaddress.com/1