首页 > 解决方案 > Flask API 为特定的 URl 提供 URL not found 错误

问题描述

我有这样的 API:

from flask import Flask
import db

app = Flask(__name__)

@app.route('/home')
def home():
    return "flask mongodb atlas!"
if __name__ == '__main__':
    app.run(port=8000)

@app.route('/text')
def text():
    return "This is a text"

当我尝试在本地主机(http://127.0.0.1:8000)中运行它并测试/homeurl 时,它给了我正确的结果,但是当我尝试时/text,它给了我在站点中找不到 URL 错误并在 cmd 中出现以下错误:

 * Serving Flask app 'app' (lazy loading)
 * Environment: production
   WARNING: This is a development server. Do not use it in a production deployment.
   Use a production WSGI server instead.
 * Debug mode: off
 * Running on http://127.0.0.1:8000/ (Press CTRL+C to quit)
127.0.0.1 - - [29/Oct/2021 02:21:37] "GET /text HTTP/1.1" 404 -
127.0.0.1 - - [29/Oct/2021 02:22:06] "GET /text HTTP/1.1" 404 -
127.0.0.1 - - [29/Oct/2021 02:26:22] "GET /home HTTP/1.1" 200 -

我认为没有任何语法错误,但我不知道问题出在哪里

标签: pythonapiflask

解决方案


尝试

from flask import Flask
import db

app = Flask(__name__)

@app.route('/home')
def home():
    return "flask mongodb atlas!"

@app.route('/text')
def text():
    return "This is a text"

if __name__ == '__main__':
    app.run(port=8000)

您需要在启动之前定义所有应用程序


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