匹配后从多行字符串中提取定界符的整个字符串右

时间:2019-02-26 13:52:20

标签: python regex string

从多行字符串中,我尝试在匹配后从=号右边提取整个字符串。但是,仅提取字符串的一部分。 如何解决此问题?我也愿意接受其他实现的搜索/提取操作。

import re
s = '''jaguar.vintage.aircards = 2
jaguar.vintage.hw.sdb.size = 512.1 GB
jaguar.vintage.hw.tm.firmware = SWI9X15C_05.05.16.02 r21040 carmd-fwbuild1 2014/03/17 23:49:48
jaguar.vintage.hw.tm.hardware = 1.0
jaguar.vintage.hw.tm.iccid = 8901260591783960689
jaguar.vintage.hw.tm.imei = 359225051166726
jaguar.vintage.hw.tm.imsi = 310260598396068
jaguar.vintage.hw.tm.model = MC7354
jaguar.vintage.hw.wifi1.mac = 00:30:1a:4e:06:7a
jaguar.vintage.hw.wifi2.mac = 00:30:1a:4e:06:79
jaguar.vintage.part = P34110-002
jaguar.vintage.product = P34101
jaguar.vintage.psoc = 0.1.16
jaguar.vintage.serial = 34110002T0021
jaguar.vintage.slavepsoc1 = 0.1.5
jaguar.vintage.sw.app.release = 4.0.0.41387-201902131138git367fbda8e
'''
# print(s)
# release = (s.split('jaguar.vintage.sw.app.release =')[1]).strip()
# print(release)
#part_number = jaguar.vintage.part = P34110-002
pnumsrch = r"jaguar.vintage.part =.*?(?=\w)(\w+)"
part_number = re.findall(pnumsrch, s)
print(part_number[0])

# release_number = jaguar.vintage.sw.app.release = 4.0.0.41387-201902131138git367fbda8e
relnumsrch = r"jaguar.vintage.sw.app.release =.*?(?=\w)(\w+)"
rel_number = re.findall(relnumsrch, s)
print(rel_number[0])
  

实际:
  P34110
  4

     

预期:
  P34110-002
  4.0.0.41387-201902131138git367fbda8e

2 个答案:

答案 0 :(得分:0)

只需捕捉不是换行Demo的所有内容:

pat = re.compile(r'jaguar\.vintage\.part = ([^\n]+)')
pat2 = re.compile(r'jaguar\.vintage\.sw\.app\.release = ([^\n]+)')

>>> pat.findall(s)

['P34110-002']
>>> pat2.findall(s)

['4.0.0.41387-201902131138git367fbda8e']

您还应该按照模式逃避句点。

如@WiktorStribiżewjust . is good enough for the [^\n] portion所述:

pat = re.compile(r'jaguar\.vintage\.part = (.+)')
pat2 = re.compile(r'jaguar\.vintage\.sw\.app\.release = (.+)')

答案 1 :(得分:0)

由于默认情况下.不匹配换行符,因此您可以简单地使用.*来匹配其余行:

pnumsrch = r"jaguar.vintage.part = (.*)"

和:

relnumsrch = r"jaguar.vintage.sw.app.release = (.*)"