我是python的新手,正在尝试读取文件并从中创建字典。
格式如下:

.1.3.6.1.4.1.14823.1.1.27 {
    TYPE = Switch
    VENDOR = Aruba
    MODEL = ArubaS3500-48T
    CERTIFICATION = CERTIFIED
    CONT = Aruba-Switch
    HEALTH = ARUBA-Controller
    VLAN = Dot1q    INSTRUMENTATION:
     Card-Fault            = ArubaController:DeviceID
     CPU/Memory            = ArubaController:DeviceID
     Environment              = ArubaSysExt:DeviceID
     Interface-Fault       = MIB2
     Interface-Performance = MIB2
     Port-Fault            = MIB2
     Port-Performance      = MIB2
}


第一行OID(.1.3.6.1.4.1.14823.1.1.27 {)我希望这是键,其余行是直到}的值

我尝试了几种组合,但无法获得正确的正则表达式来匹配这些组合

有什么帮助吗?

我已经尝试过类似的东西

lines = cache.readlines()

for line in lines:

    searchObj = re.search(r'(^.\d.*{)(.*)$', line)

    if searchObj:
        (oid, cert ) = searchObj.groups()

    results[searchObj(oid)] = ", ".join(line[1:])

    print("searchObj.group() : ", searchObj.group(1))

    print("searchObj.group(1) : ", searchObj.group(2))

最佳答案

您可以尝试以下方法:

import re
data = open('filename.txt').read()
the_key = re.findall("^\n*[\.\d]+", data)
values = [re.split("\s+\=\s+", i) for i in re.findall("[a-zA-Z0-9]+\s*\=\s*[a-zA-Z0-9]+", data)]
final_data = {the_key[0]:dict(values)}


输出:

{'\n.1.3.6.1.4.1.14823.1.1.27': {'VENDOR': 'Aruba', 'CERTIFICATION': 'CERTIFIED', 'Fault': 'MIB2', 'VLAN': 'Dot1q', 'Environment': 'ArubaSysExt', 'HEALTH': 'ARUBA', 'Memory': 'ArubaController', 'Performance': 'MIB2', 'CONT': 'Aruba', 'MODEL': 'ArubaS3500', 'TYPE': 'Switch'}}

关于python - 从文件python创建字典,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46798445/

10-09 06:17
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