如何将python后端与flask和html连接&；css_Python_Html_Css_Flask

如何将python后端与flask和html连接&；css

python html css flask

如何将python后端与flask和html连接&；css,python,html,css,flask,Python,Html,Css,Flask,''' app.py ''' ''' 天气预报 import requests, json import weatherMappingMessage from app import dress from keys import * base_url = "http://api.openweathermap.org/data/2.5/weather?" city_name = complete_url = base_url + "appid=" +

''' app.py

''' ''' 天气预报

import requests, json 
import weatherMappingMessage
from app import dress
from keys import *

base_url = "http://api.openweathermap.org/data/2.5/weather?"
city_name = 

complete_url = base_url + "appid=" + api_key + "&q=" + city_name + "&units=metric" 
response = requests.get(complete_url)

''' HTML文件 '''


了解天气，根据天气穿衣服
搜寻

'''

我需要从HTML获取表单信息（搜索）到后端（城市名称），然后再到烧瓶（城市名称）
如果尝试获取消息，我可以从后端获取消息，但我无法将HTML表单获取到后端进行处理我面临的问题是无法将表单数据从HTML文件获取到后端进行处理

基本上，我需要将cityname添加到后端以获取天气描述。

简短回答：

因为表单提交使用get请求，所以可以使用

request.args

获取查询字符串（）的解析内容：

长答案：

我相信你要求的不仅仅是这段代码。我接受了您提供的代码，并在线添加了缺少的部分（请不要对生产代码执行此操作），还将

cityname

传递到

render\u template

：

import logging
from datetime import datetime

from flask import render_template, request

from app import app, forms


@app.route("/")
def index():
    return render_template("index.html")


@app.route("/dress")
def dress():
    cityname = request.args.get("city_name")

    # missing in example code
    def temperature_condition():
        return 'temp cond'

    # missing in example code
    def clothes():
        return 'clothes'

    feels_temperature = 'feels temp'  # missing in example code
    weather_description = 'weather desc'  # missing in example code

    temp = str(temperature_condition())
    message = str(clothes())
    feels = feels_temperature
    description = weather_description
    return render_template("dress.html", message=message, temp=temp, feels_temperature=feels,
                           weather_description=description, cityname=cityname)  # also pass cityname

我创建了一个简约的

dress.html

：

<html>
    <body>
        <p>message = {{ message }}</p>
        <p>temp = {{ temp }}</p>
        <p>feels_temperature = {{ feels_temperature }}</p>
        <p>weather_description = {{ weather_description }}</p>
        <p>cityname = {{ cityname }}</p>
    </body>
</html>

从城市的

weather\u结果中提取相关数据

，并将其传递到

render\u模板

，就像对其他变量所做的那样。

你好，Harshit soni，欢迎使用SO！几乎很棒的第一个问题，只是一点指导，让它变得更好（然后得到更好的答案）：试着更详细一点。你到底想达到什么目的？您是如何尝试解决您的问题的？您是否收到任何错误消息？如果是，你能包括这些吗？如果你对你的问题付出额外的努力，那么答案会很快出现——而且有额外的价值！看看这个，这个你好，安德鲁，谢谢你的反馈，我试着让它尽可能容易理解，因为我现在可以你好，阿基布，谢谢你的例子，我做了一些改变

cityname = request.args.get("city_name")

import logging
from datetime import datetime

from flask import render_template, request

from app import app, forms


@app.route("/")
def index():
    return render_template("index.html")


@app.route("/dress")
def dress():
    cityname = request.args.get("city_name")

    # missing in example code
    def temperature_condition():
        return 'temp cond'

    # missing in example code
    def clothes():
        return 'clothes'

    feels_temperature = 'feels temp'  # missing in example code
    weather_description = 'weather desc'  # missing in example code

    temp = str(temperature_condition())
    message = str(clothes())
    feels = feels_temperature
    description = weather_description
    return render_template("dress.html", message=message, temp=temp, feels_temperature=feels,
                           weather_description=description, cityname=cityname)  # also pass cityname

<html>
    <body>
        <p>message = {{ message }}</p>
        <p>temp = {{ temp }}</p>
        <p>feels_temperature = {{ feels_temperature }}</p>
        <p>weather_description = {{ weather_description }}</p>
        <p>cityname = {{ cityname }}</p>
    </body>
</html>

import requests, json
import weatherMappingMessage
from app import dress
from keys import *

def weather_for_city(city_name):
    base_url = "http://api.openweathermap.org/data/2.5/weather?"

    complete_url = base_url + "appid=" + api_key + "&q=" + city_name + "&units=metric"
    response = requests.get(complete_url)

    if response.status_code == 200:
        return response.json()  # assumes your API returns a JSON response
    else:
        # perform some error handling here, maybe apply a retry strategy
        pass