简单的烧瓶应用程序,从.html文件中读取其内容。外部样式表被阻止? [英] Simple flask application that reads its content from a .html file. External style sheet being blocked?
问题描述
文件夹中的文件可以在这里。这3个文件都在同一个文件夹中。
您的代码不是使用Flask提供的文件,它只是读取文件并将其发送到浏览器 - 这就是为什么URL不起作用。
您需要从方法中呈现文件。
首先在 .py
文件的同一目录中创建一个模板
文件夹并移动你的html文件到这个文件夹中。创建另一个名为 static
的文件夹,并将样式表放在该文件夹中。
您应该有
/flaskwp1.py
/模板
webcode.html
/静态
webcodestyle.css
$ / code>
然后,调整您的代码:
<$ p $ () - > from flask import Flask,render_template
$ b app = Flask('flaskwp1')
#webcode = open('webcode.html')。不需要
@ app.route('/')
def webprint():
return render_template('webcode.html')
if __name__ =='__main__':
app.run(host ='0.0.0.0',port = 3000)
编辑 webcode.html
:
< link rel =stylesheet
type =text / css
href =/ static / webcodestyle.css/>
I made a very simple flask application that reads its content from a .html file. The application works except for the style. Strangely my inline css code works but not the external style sheet. I've checked the syntax, it should work. Does flask somehow prevent the .css file from being read?
The files in the folder can be viewed here. Those 3 files are all in the same folder.
Your code is not serving files using Flask, it is simply reading a file and sending it to the browser - which is why URLs are not working.
You need to render the file from within the method.
First make a templates
folder in the same directory as your .py
file and move your html file into this folder. Create another folder called static
and put your stylesheet in that folder.
You should have
/flaskwp1.py
/templates
webcode.html
/static
webcodestyle.css
Then, adjust your code thusly:
from flask import Flask, render_template
app = Flask('flaskwp1')
# webcode = open('webcode.html').read() - not needed
@app.route('/')
def webprint():
return render_template('webcode.html')
if __name__ == '__main__':
app.run(host = '0.0.0.0', port = 3000)
Edit webcode.html
:
<link rel="stylesheet"
type="text/css"
href="/static/webcodestyle.css"/>
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