Question

我从一个有10行的文本中找到一行。

desc    = re.findall(r'@description (.*)', comment.strip())

它会返回@description但它还有9个空列表。

print(desc)

返回：

[]
[]
[]
[]
[]
[]
[]
[]
['the desc is here']
[]

那么如何摆脱那些空[]并制作desc=['the desc is here']？

更新

我尝试了列表过滤器，但仍然是相同的返回

评论包含：

/**
 * @param string username   required    the username of the registering user 
 * @param string password   required
 * @param string first_name required
 * @param string last_name  required
 * @param string email      required
 * @package authentication
 * @info user registration
 * @description register a new user into the groupjump platform
 */

更新

注释是一个完整的字符串，所以我将它拆分为这样我就可以逐行阅读

comments = route['comment']
comments = list(filter(None, comments.split('\n')))

实际代码

#!/usr/bin/env python3

import re

routes = []
description = ''
with open('troutes.php', 'r') as f:
    current_comment = ''
    in_comment = False
    for line in f:
        line = line.lstrip()
        if line.startswith('/**'):
            in_comment = True

        if in_comment:
            current_comment += line

        if line.startswith('*/'):
            in_comment = False

        if line.startswith('Route::'):
            matches = re.search(r"Route::([A-Z]+)\('(.*)', '(.*)'\);", line)
            groups = matches.groups()
            routes.append({
                'comment': current_comment,
                'method': groups[0],
                'path': groups[1],
                'handler': groups[2],
            });
            current_comment = '' # reset the comment

for route in routes:

  # get comments
  comments = route['comment']
  comments = list(filter(None, comments.split('\n')))

  for comment in comments:
    params  = re.findall(r'@param (.*)', comment.strip())
    object  = re.findall(r'@package (.*)', comment.strip())
    info    = re.findall(r'@info (.*)', comment.strip())
    desc    = re.search(r'@description (.*)', comment.strip())

    print(comment[15:])

正在阅读的数据：

<?php
/**
 * @param string username   required    the username of the registering user 
 * @param string password   required
 * @param string first_name required
 * @param string last_name  required
 * @param string email      required
 * @package authentication
 * @info user registration
 * @description register a new user into the groupjump platform
 */
Route::POST('v3/register', 'UserController@Register');

/**
 * @param string username   required    the username of the registering user 
 * @param string password   required
 */
Route::GET('v3/login', 'UserController@login');

Answer 1

单个列表的条件是：

if desc:
    print(desc)

这是一个速记版本：

if len(desc) > 0:
    print(desc)

列表是：

desc = [d for d in desc if d]

要仅获取字符串，请执行以下操作：

if desc:
    print(desc[0])

Answer 2

似乎你逐行匹配模式。你为什么不匹配整个评论？

>>> comment = '''/**
...  * @param string username   required    the username of the registering user
...  * @param string password   required
...  * @param string first_name required
...  * @param string last_name  required
...  * @param string email      required
...  * @package authentication
...  * @info user registration
...  * @description register a new user into the groupjump platform
...  */'''
>>>
>>> import re
>>> desc = re.findall(r'@description (.*)', comment)
>>> desc
['register a new user into the groupjump platform']

Answer 3

您可以使用列表推导列表列表中的空字符串过滤列表：

desc = re.findall(r'@description (.*)', comment.strip())
desc = [d for d in desc if len(d[0]) > 0]

另一种解决方案是仅在第一个元素包含某些内容时才打印元素：

desc = re.findall(r'@description (.*)', comment.strip())
for d in desc:
     if len(d) > 0 and d[0]:  # check if there's a first element and if this element isn't empty
           print d

Answer 4

要让你的代码工作，你需要处理一个字符串，如果你有10行像这样：

joined = "\n".join(lines)
for i in re.findall(r'@description (.*)', joined):
    print (i)

Python 3删除空列表

4 个答案: