如何使用Python生成html目录列表 [英] How to generate an html directory list using Python
问题描述
def list_files(startpath):
for root,dirs,files in os .walk(startpath):
level = root.replace(startpath,'').count(os.sep)
if level <= 1:
print('< li>格式(os.path.basename(root)))
else:
print('< li> {}'。format(os.path.basename(root )))
for file in:
last_file = len(files)-1
if f == files [last_file]:
print('< li> {} < / ul>'。format(f))
elif f == files [0] and level-1>格式(f))
print('< ul>< li> {}< / li>'
print('< li> {} < / li>'。format(f))
print('< / li>< / ul>)
如果只有根目录,一级子目录和文件,它似乎工作得很好。但是,添加另一层次的子目录会导致出现问题(因为在我认为的结尾,关闭标签没有足够的时间输入)。但是我很难把握住它。
如果不能这样做,有没有更简单的方法来做到这一点?我正在使用Flask,但是我对模板非常不熟悉,所以也许我错过了一些东西。 目录树生成及其呈现为html。
要生成树,您可以使用一个简单的递归函数:
def make_tree(path):
tree = dict(name = os.path.basename(path),children = [])
try:lst = os。 listdir(path)
除了OSError:
pass #ignore错误
else:
在lst中的名字:
fn = os.path.join(path,name)
如果os.path.isdir(fn):
tree ['children']。append(make_tree(fn))
else:$ b $ tree ['children']。append (dict(name = name))
返回树
使用jinja2循环递归
特性:
<!doctype html>
< title>路径:{{tree.name}}< / title>
< h1> {{tree.name}}< / h1>
< ul>
{% - for tree.children递归%}
< li> {{item.name}}
{% - if item.children - %}
< ; ul> {{loop(item.children)}}< / ul>
{% - endif%}< / li>
{% - endfor%}
< / ul>
将html放入 templates / dirtree.html
文件。
要测试它,请运行以下代码并访问 http:// localhost:8888 /
:
path = os.path.expanduser(u'〜')
return render_template('dirtree.html',tree = make_tree(path))
$ b $ if if __name __ ==__ main__:
app.run(host ='localhost',port = 8888,debug = True)
pre>
I am having some problems using Python to generate an html document. I am attempting to create an HTML list of a directory tree. This is what I have so far:
def list_files(startpath):
for root, dirs, files in os.walk(startpath):
level = root.replace(startpath, '').count(os.sep)
if level <= 1:
print('<li>{}<ul>'.format(os.path.basename(root)))
else:
print('<li>{}'.format(os.path.basename(root)))
for f in files:
last_file = len(files)-1
if f == files[last_file]:
print('<li>{}</li></ul>'.format(f))
elif f == files[0] and level-1 > 0:
print('<ul><li>{}</li>'.format(f))
else:
print('<li>{}</li>'.format(f))
print('</li></ul>')
It seems to work well if there is only the root directory, one level of sub-directories and files. However, adding another level of sub-directories causes there to be problems (because the close tag isn't input enough times at the end I think). But I'm having a hard time getting my head around it.
If it can't be done this way, is there an easier way to do it? I'm using Flask but I'm very inexperienced with templates so perhaps I'm missing something.
You could separate the directory tree generation and its rendering as html.
To generate the tree you could use a simple recursive function:
def make_tree(path):
tree = dict(name=os.path.basename(path), children=[])
try: lst = os.listdir(path)
except OSError:
pass #ignore errors
else:
for name in lst:
fn = os.path.join(path, name)
if os.path.isdir(fn):
tree['children'].append(make_tree(fn))
else:
tree['children'].append(dict(name=name))
return tree
To render it as html you could use jinja2's loop recursive
feature:
<!doctype html>
<title>Path: {{ tree.name }}</title>
<h1>{{ tree.name }}</h1>
<ul>
{%- for item in tree.children recursive %}
<li>{{ item.name }}
{%- if item.children -%}
<ul>{{ loop(item.children) }}</ul>
{%- endif %}</li>
{%- endfor %}
</ul>
Put the html into templates/dirtree.html
file.
To test it, run the following code and visit http://localhost:8888/
:
import os
from flask import Flask, render_template
app = Flask(__name__)
@app.route('/')
def dirtree():
path = os.path.expanduser(u'~')
return render_template('dirtree.html', tree=make_tree(path))
if __name__=="__main__":
app.run(host='localhost', port=8888, debug=True)
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