[英]Extract value from a list formatted like a JSON file in Python
I have a function that returns an image URL from an RSS feed (I'm using feedparser).我有一个从 RSS 提要返回图像 URL 的函数(我使用的是 feedparser)。 The problem is that it returns a strange formatted list.
问题是它返回一个奇怪的格式化列表。
My feed has a key called media_content
.我的提要有一个名为
media_content
的键。 The code used to access it is用于访问它的代码是
import feedparser
NewsFeed = feedparser.parse("https://thegroovecartel.com/feed/")
entry = NewsFeed.entries[0]
post_image = entry.media_content
Now, for some reason post_image is a list of 1 element formattes as:现在,由于某种原因 post_image 是 1 个元素格式的列表:
[{'url': 'https://thegroovecartel.com/i1.wp.com/thegroovecartel.com/wp-content/uploads/2020/10/KAYYA.jpg?fit=300%2C237&ssl=1', 'medium': 'image'}]
Now, I can't understand how to access the actual URL field to have https://thegroovecartel.com/i1.wp.com/thegroovecartel.com/wp-content/uploads/2020/10/KAYYA.jpg?fit=300%2C237&ssl=1
as a string into a variable.现在,我无法理解如何访问实际的 URL 字段以拥有
https://thegroovecartel.com/i1.wp.com/thegroovecartel.com/wp-content/uploads/2020/10/KAYYA.jpg?fit=300%2C237&ssl=1
作为字符串转化为变量。
post_image = [{'url': 'https://thegroovecartel.com/i1.wp.com/thegroovecartel.com/wp-content/uploads/2020/10/KAYYA.jpg?fit=300%2C237&ssl=1', 'medium': 'image'}]
print(post_image[0]['url'])
Result
结果
https://thegroovecartel.com/i1.wp.com/thegroovecartel.com/wp-content/uploads/2020/10/KAYYA.jpg?fit=300%2C237&ssl=1
You've got to treat post_image
as a list, assessing its first element (ie element with index 0), gives you the dictionary that contains your image's attribute您必须将
post_image
视为一个列表,评估其第一个元素(即索引为 0 的元素),为您提供包含图像属性的字典
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