如何将文件分成几个部分并使用机器人框架将它们放入字典中

我有一个这样的文件:

Name: John
Class: II
Age:  8
Interest: Sports

Name: Emma
Class: III
Hobby: Dance

所以我想读取这个文件并将内容放入一个名为key的字典中.这些部分的行数不尽相同.如何使用Robot Framework关键字实现此字典.

这是你需要的吗？

*** Settings ***
Library    OperatingSystem
Library    String
Library    Collections

*** Test Cases ***
Split File By Names
    ${my_dict}    Create Dictionary
    ${data}    Get File    
    @{lines}    Split To Lines    ${data}
    Remove Values From List    ${lines}    ${EMPTY}
    :FOR    ${line}    IN    @{lines}
    \    ${key}    ${value}    Split String    ${line}    :
    \    ${name}    Set Variable If    '${key}' == 'Name'    ${value.strip()}    ${name}
    \    Run Keyword If    '${key}' == 'Name'    Set To Dictionary    ${my_dict}    ${name}=@{EMPTY}
    \    Run Keyword If    '${key}' <> 'Name'    Append To List    ${my_dict.${name}}    ${line}
    Log    ${my_dict}

无论如何,解析文件的RF方式很糟糕.我宁愿去蟒蛇.

#!/usr/bin/python
# -*- coding: utf-8 -*-


class ParseFile:

    def __init__(self):
        self.my_dict = {}

    def parse_file_to_dict(self):
        with open('') as f:
            lines = f.read().splitlines()
            for line in (l for l in lines if l != ""):
                key, value = line.split(":", 1)
                if key == "Name":
                    name = value
                    self.my_dict[name] = []
                else:
                    self.my_dict[name].append(line)
        return self.my_dict

......然后在RF中调用它.

*** Settings ***
Library    ParseFile.py

*** Test Cases ***
Do It In Python
    ${my_dict}    Parse File To Dict
    Log    ${my_dict}

请注意,这两种方式都严格依赖于您提供的数据结构.也就是说,如果"名称"不在每个部分的第一行,它将无法工作,需要对数据进行更多处理.