Home > Backend Development > Python Tutorial > How Can I Convert SQLAlchemy Row Objects to Python Dictionaries?

How Can I Convert SQLAlchemy Row Objects to Python Dictionaries?

Barbara Streisand
Release: 2024-11-19 01:15:03
Original
785 people have browsed it

How Can I Convert SQLAlchemy Row Objects to Python Dictionaries?

Converting SQLAlchemy Row Objects to Python Dictionaries

When working with SQLAlchemy, you may encounter situations where you need to convert row objects into Python dictionaries. This can be particularly useful for iterating over column name-value pairs or passing row data to external applications.

Using SQLAlchemy's Internal Dictionary

One approach to converting a SQLAlchemy row object to a dictionary is by accessing its internal dict attribute. This attribute holds a dictionary-like representation of the object's attributes. To use this approach, simply iterate over the dict attribute:

for u in session.query(User).all():
    print(u.__dict__)
Copy after login

This will output a dictionary containing the column names and values for each row.

Using SQLAlchemy's Marshal Library

Alternatively, SQLAlchemy provides the marshal library, which offers methods for converting SQLAlchemy objects to different formats, including dictionaries. The following code demonstrates how to use marshal.to_dict() to convert User objects to dictionaries:

from sqlalchemy.ext import marshal

for u in session.query(User).all():
    print(marshal.to_dict(u))
Copy after login

This method will produce a dictionary with a key for each column name and a value for each column value.

Note: The use of marshal.to_dict() requires SQLAlchemy version 0.6 or higher. If you are using an earlier version, you can use the dict attribute instead.

The above is the detailed content of How Can I Convert SQLAlchemy Row Objects to Python Dictionaries?. For more information, please follow other related articles on the PHP Chinese website!

source:php.cn
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Latest Articles by Author
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template