删除 .com python 之后的所有内容
Remove everything after .com python
我在 urls.tmp 文件中获得了包含这 3 个 url 的文件:
https://site1.com.br/wp-content/uploads/2020/06/?SD
https://site2.com.br/wp-content/uploads/tp-datademo/home-4/data/tp-hotel-booking/?SD
https://site3.com.br/wp-content/uploads/revslider/hotel-home/?MD
我想删除每个“com.br/”之后的所有内容。
我试过这个代码:
# open the file
sys.stdout = open("urls.tmp", "w")
# start remove
for i in "urls.tmp":
url_parts = urllib.parse.urlparse(i)
result = '{uri.scheme}://{uri.netloc}/'.format(uri=url_parts)
print(result) #overwrite the file
# close the file
sys.stdout.close()
但是输出给了我这个奇怪的东西:
:///
:///
:///
:///
:///
:///
:///
:///
我是初学者,我做错了什么?
请参阅 答案来解决您的问题。
您可以使用字符串的拆分方法,如:
url = r"https://site1.com.br/wp-content/uploads/2020/06/?SD"
split_by = "com.br/"
new_url = url.split(split_by)[0] + split_by
# this gives you the part before <split_by> and then we can attach it again
new_url == r"https://site1.com.br"
如果您想添加一些额外的检查,您可以查看正则表达式。
您没有要求但可能对初学者有所帮助的事情。
我推荐使用
with open("urls.tmp", "w") as f:
# do something with f
或
import pathlib
urls = pathlib.Path("urls.tmp").read_text()
# which gives you all lines in single string
平淡无奇 open
。如果您想了解更多信息,我建议您查看上下文管理器。
还有 f-strings
自 Python 3.6 在我看来比 "{}".format
.
更容易阅读
可以继续string的find()方法
urllist=[
'https://site1.com.br/wp-content/uploads/2020/06/?SD',
'https://site2.com.br/wp-content/uploads/tp-datademo/home-4/data/tp-hotel-booking/?SD',
'https://site3.com.br/wp-content/uploads/revslider/hotel-home/?MD']
newlist=[]
breaktext='com.br/'
for item in urllist:
position=item.find(breaktext)
newlist.append(item[:position+len(breaktext)])
print (newlist)
您正在遍历 "urls.tmp"
字符串本身,但想逐行遍历打开的文件对象。
所以试试这个:
with open("urls.tmp", "r") as urls_file:
for line in urls_file:
url_parts = urllib.parse.urlparse(line)
result = "{uri.scheme}://{uri.netloc}/".format(uri=url_parts)
print(result)
编辑:作者更新了原始问题,提到源文件内容应该用处理过的 url 重写,这里是示例:
new_urls = []
with open("urls.tmp", "r") as urls_file:
old_urls = urls_file.readlines()
for line in old_urls:
url_parts = urllib.parse.urlparse(line)
proc_url = "{uri.scheme}://{uri.netloc}/\n".format(uri=url_parts)
new_urls.append(proc_url)
with open("urls.tmp", "w") as urls_file:
urls_file.writelines(new_urls)
我在 urls.tmp 文件中获得了包含这 3 个 url 的文件:
https://site1.com.br/wp-content/uploads/2020/06/?SD
https://site2.com.br/wp-content/uploads/tp-datademo/home-4/data/tp-hotel-booking/?SD
https://site3.com.br/wp-content/uploads/revslider/hotel-home/?MD
我想删除每个“com.br/”之后的所有内容。
我试过这个代码:
# open the file
sys.stdout = open("urls.tmp", "w")
# start remove
for i in "urls.tmp":
url_parts = urllib.parse.urlparse(i)
result = '{uri.scheme}://{uri.netloc}/'.format(uri=url_parts)
print(result) #overwrite the file
# close the file
sys.stdout.close()
但是输出给了我这个奇怪的东西:
:///
:///
:///
:///
:///
:///
:///
:///
我是初学者,我做错了什么?
请参阅
您可以使用字符串的拆分方法,如:
url = r"https://site1.com.br/wp-content/uploads/2020/06/?SD"
split_by = "com.br/"
new_url = url.split(split_by)[0] + split_by
# this gives you the part before <split_by> and then we can attach it again
new_url == r"https://site1.com.br"
如果您想添加一些额外的检查,您可以查看正则表达式。
您没有要求但可能对初学者有所帮助的事情。 我推荐使用
with open("urls.tmp", "w") as f:
# do something with f
或
import pathlib
urls = pathlib.Path("urls.tmp").read_text()
# which gives you all lines in single string
平淡无奇 open
。如果您想了解更多信息,我建议您查看上下文管理器。
还有 f-strings
自 Python 3.6 在我看来比 "{}".format
.
可以继续string的find()方法
urllist=[
'https://site1.com.br/wp-content/uploads/2020/06/?SD',
'https://site2.com.br/wp-content/uploads/tp-datademo/home-4/data/tp-hotel-booking/?SD',
'https://site3.com.br/wp-content/uploads/revslider/hotel-home/?MD']
newlist=[]
breaktext='com.br/'
for item in urllist:
position=item.find(breaktext)
newlist.append(item[:position+len(breaktext)])
print (newlist)
您正在遍历 "urls.tmp"
字符串本身,但想逐行遍历打开的文件对象。
所以试试这个:
with open("urls.tmp", "r") as urls_file:
for line in urls_file:
url_parts = urllib.parse.urlparse(line)
result = "{uri.scheme}://{uri.netloc}/".format(uri=url_parts)
print(result)
编辑:作者更新了原始问题,提到源文件内容应该用处理过的 url 重写,这里是示例:
new_urls = []
with open("urls.tmp", "r") as urls_file:
old_urls = urls_file.readlines()
for line in old_urls:
url_parts = urllib.parse.urlparse(line)
proc_url = "{uri.scheme}://{uri.netloc}/\n".format(uri=url_parts)
new_urls.append(proc_url)
with open("urls.tmp", "w") as urls_file:
urls_file.writelines(new_urls)