我有一个日志合作伙伴,如下所示,我只想将它分开,它们在[]括号内,包括从一开始的时间。也尝试了下面的代码,但没有运气
def crop_string(line):
try:
#str0 = line.replace(']',"")
#str1 = str0.replace('.000+05:30',"")
str2 = line.split()
#str2 = list(dict.fromkeys(str1))
#x = len(str2)
print(str2[0],str2[5])
pass
except IndexError:
print("exception",str2)
pass
日志
"INFO:DEL:2018-11-24T14:04:49.000+05:30 bltxn03 ChannelGatewayRequestLog INFO :: [ReqOut:][RQID:1958259][STV:1RC][S:EXTGW][RQC:17007][UN:banti mobile centre ][CAT:RET][MSISDN:334455668][USt:Y][UNW:UW][DSMS:1RC 44xxxxxx 11111.0 VG ****][TEMPTID:null][UDH:313231][ST:EXTGW][SRVPRT:190][OINFO:RETAPPUW,MsgReq=true,FT=R ResTyp=RESPONSE][RETMSG:<?xml version="1.0"?><!DOCTYPE COMMAND PUBLIC "-//Ocam//DTD XML Command 1.0//EN" "xml/command.dtd"><COMMAND><TYPE>EXRC121TRFRESP</TYPE><TXNSTATUS>17007</TXNSTATUS><DATE>4/10/2018 14:04:49</DATE><EXTREFNUM>192108828_1548487626</EXTREFNUM><TXNID>UW445566126</TXNID><MESSAGE>Your request cannot be processed at this time, please try again later.</MESSAGE></COMMAND>][FixdInfNtAvail.][TT:860 ms]"
re.findall
应该可以做到这一点。查看下面的代码片段,了解如何将其用于非贪婪正则表达式。 尽管如此,我还是建议您避免这种简单的解决方案:包含括号的字符串数据可能会导致错误的结果相关问题 更多 >
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