tfp = open(filename,' wb')
OSError:[Errno 22}参数无效:'已下载/ misc / jquery.js?v = 1.4.4'
任何人都可以帮我解决这个错误吗?我认为它与jquery.js?v=1.4.4
无效有关。我是python的新手;如果我遗漏了一些明显的东西,我道歉。
以下是代码:
import os
from urllib.request import urlretrieve
from urllib.request import urlopen
from bs4 import BeautifulSoup
downloadDirectory = "downloaded"
baseUrl = "http://pythonscraping.com"
def getAbsoluteURL(baseUrl, source):
if source.startswith("http://www."):
url = "http://"+source[11:]
elif source.startswith("http://"):
url = source
elif source.startswith("www."):
url = source[4:]
url = "http://"+source
else:
url = baseUrl+"/"+source
if baseUrl not in url:
return None
return url
def getDownloadPath(baseUrl, absoluteUrl, downloadDirectory):
path = absoluteUrl.replace("www.", "")
path = path.replace(baseUrl, "")
path = downloadDirectory+path
directory = os.path.dirname(path)
if not os.path.exists(directory):
os.makedirs(directory)
return path
html = urlopen("http://www.pythonscraping.com")
bsObj = BeautifulSoup(html, "html.parser")
downloadList = bsObj.findAll(src=True)
for download in downloadList:
fileUrl = getAbsoluteURL(baseUrl, download["src"])
if fileUrl is not None:
print(fileUrl)
urlretrieve(fileUrl, getDownloadPath(baseUrl, fileUrl, downloadDirectory))
答案 0 :(得分:1)
对于函数urlretrieve(url, filename, reporthook, data)
,
您为filename
参数提供的参数必须是操作系统上的有效文件名。
在这种情况下,当你运行
urlretrieve(fileUrl, getDownloadPath(baseUrl, fileUrl, downloadDirectory))
您为url
提供的参数是“http://pythonscraping.com/misc/jquery.js?v=1.4.4”,您为filename
提供的参数是“downloads / misc / jquery.js?v = 1.4.4”。
“jquery.js?v = 1.4.4”我认为这不是一个有效的文件名。
解决方案:在getDownloadPath
功能中,将return path
更改为
return path.partition('?')[0]
答案 1 :(得分:0)
已下载/ misc / jquery.js?v = 1.4.4不是有效的文件名。 我认为这是一个更好的解决方案:
import requests
from bs4 import BeautifulSoup
download_directory = "downloaded"
base_url = "http://www.pythonscraping.com/"
# Use Requests instead urllib
def get_files_url(base_url):
# Return a list of tag elements that contain src attrs
html = requests.get(base_url)
soup = BeautifulSoup(html.text, "lxml")
return soup.find_all(src=True)
def get_file_name(url):
# Return the last part after the last "/" as file name
# Eg: return a.png as file name if url=http://pythonscraping.com/a.png
# Remove characters not valid in file name
file_name = url.split("/")[-1]
remove_list = "?><\/:\"*|"
for ch in remove_list:
if ch in file_name:
file_name = file_name.replace(ch, "")
return download_directory + "/" + file_name
def get_formatted_url(url):
if not url.startswith("http://"):
return base_url + url
elif base_url not in url:
return None
else:
return url
links = get_files_url(base_url)
for link in links:
url = link["src"]
url = get_formatted_url(url)
if url is None:
continue
print(url)
result = requests.get(url, stream=True)
file_name = get_file_name(url)
print(file_name)
with open(file_name, 'wb') as f:
for chunk in result.iter_content(10):
f.write(chunk)