如何通过PyMongo仅输出聚合查询结果中的书名字段内容

Hey there! No worries, this is super straightforward. Your books variable is a standard Python dictionary, so you just need to access the value tied to the _id key (which holds the book name you want).

Here's what you can do:

• If you just need to print the name directly (including any trailing spaces from the original result):

  print(books['_id'])
  

• Since your result shows 'Harry Potter ' with an extra trailing space, to get clean Harry Potter without unwanted whitespace, add the strip() method to trim any leading/trailing spaces:

  print(books['_id'].strip())
  

That's it! This will pull only the book name from your aggregation result and print it exactly as you requested.

内容的提问来源于stack exchange，提问作者Will.S89