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社区首页 >问答首页 >使用flask时连接到postgresql时出错

使用flask时连接到postgresql时出错
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Stack Overflow用户
提问于 2021-06-15 05:06:43
回答 1查看 54关注 0票数 0
代码语言:javascript
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from flask import Flask, jsonify
from flask import request
from slotscal import results
from flask_sqlalchemy import SQLAlchemy
import time
app = Flask(__name__)


app.config['SQLALCHEMY_DATABASE_URI']= 'postgresql://postgres:test123@localhost:5432/flask'

db=SQLAlchemy(app)
代码语言:javascript
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Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/home/hamza/.local/lib/python3.8/site-packages/flask_sqlalchemy/__init__.py", line 1094, in create_all
    self._execute_for_all_tables(app, bind, 'create_all')
  File "/home/hamza/.local/lib/python3.8/site-packages/flask_sqlalchemy/__init__.py", line 1086, in _execute_for_all_tables
    op(bind=self.get_engine(app, bind), **extra)
  File "/home/hamza/.local/lib/python3.8/site-packages/flask_sqlalchemy/__init__.py", line 1017, in get_engine
    return connector.get_engine()
  File "/home/hamza/.local/lib/python3.8/site-packages/flask_sqlalchemy/__init__.py", line 592, in get_engine
    sa_url = make_url(uri)
  File "/home/hamza/.local/lib/python3.8/site-packages/sqlalchemy/engine/url.py", line 694, in make_url
    return _parse_rfc1738_args(name_or_url)
  File "/home/hamza/.local/lib/python3.8/site-packages/sqlalchemy/engine/url.py", line 755, in _parse_rfc1738_args
    raise exc.ArgumentError(
sqlalchemy.exc.ArgumentError: Could not parse rfc1738 URL from string 'postgresql:/test123@localhost/flask'
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Stack Overflow用户

发布于 2021-06-15 09:58:30

连接字符串必须是libpq连接字符串,其语法为is documented in the PostgreSQL documentation

如果您使用URL表单,则必须遵守RFC 1738。特别是,在postgresql:之后需要两个斜杠

代码语言:javascript
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postgresql://test123@localhost/flask
票数 0
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/67977224

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