如何以正确的格式存储从 POST XML 请求收到的 XML？ Python 请求库

发布于 2025-01-14 17:30:53 字数 1442 浏览 1 评论 0原文

我正在使用 Python requests 库向网站发送一个 XML 文件，并收到一堆 XML 代码（以字节格式），如下所示：

b'<?xml version="1.0" encoding="UTF-8"?>\n<GetCategorySpecificsResponse xmlns="urn:ebay:apis:eBLBaseComponents"><Timestamp>2022-03-15T09:54:41.461Z</Timestamp><Ack>Success</Ack><Version>1219</Version><Build>E1219_CORE_APICATALOG_19146446_R1</Build><Recommendations><CategoryID>19006</CategoryID><NameRecommendation>.....

但是，如何才能以正确的格式和所有正确的缩进获取上面的 xml ？我想将上面的字符串存储在另一个文件中，但是对于当前的字符串，它只是一条永远向右侧延伸的长线，这对我来说并不是真正有用......

下面是我的代码（以 r.content 作为上面的 xml）：

import requests

xml_file = XML_FILE

headers = {'Content-Type':'text/xml'}

with open(XML_FILE) as xml:
    r = requests.post(WEBSITE_URL, data=xml, headers=headers)

print(r.content)

new_file = open(ANOTHER_FILE)
new_file.write(str(r.content))
new_file.close()

我想存储在 new_file 中的 xml 示例：

<?xml version="1.0" encoding="UTF-8"?>
<GetCategorySpecificsResponse
  xmlns="urn:ebay:apis:eBLBaseComponents">
  <Timestamp>2022-03-15T08:30:01.877Z</Timestamp>
  <Ack>Success</Ack>
  <Version>1219</Version>
  <Build>E1219_CORE_APICATALOG_19146446_R1</Build>
  <Recommendations>
    <CategoryID>19006</CategoryID>
.....
</GetCategorySpecificsResponse>

谢谢！

原文

I'm sending an XML file to a website with Python requests library and received back a bunch of XML code (in format of bytes) like below:

b'<?xml version="1.0" encoding="UTF-8"?>\n<GetCategorySpecificsResponse xmlns="urn:ebay:apis:eBLBaseComponents"><Timestamp>2022-03-15T09:54:41.461Z</Timestamp><Ack>Success</Ack><Version>1219</Version><Build>E1219_CORE_APICATALOG_19146446_R1</Build><Recommendations><CategoryID>19006</CategoryID><NameRecommendation>.....

However, how can I get the xml above in its correct format and with all the correct indentations? I want to store the string above in another file, but with the current string, it's just a long line going forever toward the right side that is not really useful to me...

Below is my code (with the r.content as the xml above):

import requests

xml_file = XML_FILE

headers = {'Content-Type':'text/xml'}

with open(XML_FILE) as xml:
    r = requests.post(WEBSITE_URL, data=xml, headers=headers)

print(r.content)

new_file = open(ANOTHER_FILE)
new_file.write(str(r.content))
new_file.close()

Example of the xml I want to store in new_file:

<?xml version="1.0" encoding="UTF-8"?>
<GetCategorySpecificsResponse
  xmlns="urn:ebay:apis:eBLBaseComponents">
  <Timestamp>2022-03-15T08:30:01.877Z</Timestamp>
  <Ack>Success</Ack>
  <Version>1219</Version>
  <Build>E1219_CORE_APICATALOG_19146446_R1</Build>
  <Recommendations>
    <CategoryID>19006</CategoryID>
.....
</GetCategorySpecificsResponse>

Thank you!

分享到QQ

分享到微博

如果你对这篇内容有疑问，欢迎到本站社区发帖提问参与讨论，获取更多帮助，或者扫码二维码加入 Web 技术交流群。

发布评论

需要登录才能够评论，你可以免费注册一个本站的账号。

岁月如刀 2025-01-21 17:30:53

一种方法是通过解析器传递响应并保存到文件。例如，这样的事情应该有效：

from bs4 import BeautifulSoup as bs
soup= bs(r.text,"lxml")
with open("file.xml", "w", encoding='utf-8') as file:
    file.write(str(soup.prettify()))

One way to do it is to pass the response through a parser and save to file. For example, something like this should work:

from bs4 import BeautifulSoup as bs
soup= bs(r.text,"lxml")
with open("file.xml", "w", encoding='utf-8') as file:
    file.write(str(soup.prettify()))

回复收藏 0 原文

~没有更多了~