[英]Extract the Version number from the output using regex python?
I want the version number from the output and for both the output I want one code only我想要 output 和 output 的版本号,我只想要一个代码
Edition: Read the output starts with Cisco till Version then extract the Version Number版本:阅读 output 从 Cisco 开始直到版本然后提取版本号
For example:Read the line like this Cisco IOS Software, s2t54 Software (s2t54-ADVIPSERVICESK9-M), Version 15.5(1)SY2, the ouput Version例如:读这样一行 Cisco IOS Software, s2t54 Software (s2t54-ADVIPSERVICESK9-M), Version 15.5(1)SY2, 输出版本
Output: 15.5(1)SY2 Output:15.5(1)SY2
Output 1: ''' Cisco IOS Software, s2t54 Software (s2t54-ADVIPSERVICESK9-M), Version 15.5(1)SY2, RELEASE SOFTWARE (fc6) ROM: System Bootstrap, Version 12.2(50r)SYS3, RELEASE SOFTWARE (fc1) CPU: MPC8572_E, Version: 2.2, (0x80E80022) CORE: E500, Version: 3.0, (0x80210030) ''' Output 1: ''' Cisco IOS Software, s2t54 Software (s2t54-ADVIPSERVICESK9-M), Version 15.5(1)SY2, RELEASE SOFTWARE (fc6) ROM: System Bootstrap, Version 12.2(50r)SYS3, RELEASE SOFTWARE (fc1) CPU :MPC8572_E,版本:2.2,(0x80E80022)核心:E500,版本:3.0,(0x80210030)'''
Output 2: Cisco IOS Software, IOS-XE Software, Catalyst 4500 L3 Switch Software (cat4500es8-UNIVERSALK9-M), Version 03.08.07.E RELEASE SOFTWARE (fc2) licensed under the GNU General Public License ("GPL") Version 2.0. Output 2: Cisco IOS Software, IOS-XE Software, Catalyst 4500 L3 Switch Software (cat4500es8-UNIVERSALK9-M), Version 03.08.07.E RELEASE SOFTWARE (fc2) licensed under the GNU General Public License ("GPL") Version 2.0 . The software code licensed under GPL Version 2.0 is free software that comes GPL code under the terms of GPL Version 2.0.在 GPL 2.0 版下许可的软件代码是根据 GPL 2.0 版条款获得 GPL 代码的免费软件。
I tried ths code:我试过这个代码:
r = re.findall(r'Version\s*(([\w]+))', str)
r[0]
it gives output:它给出了 output:
15.5 15.5
03.08.07.E 03.08.07.E
expected output:预期 output:
15.5(1)SY2 15.5(1)SY2
03.08.07.E 03.08.07.E
I think this is the expression you are looking for:我认为这是您正在寻找的表达方式:
Version\s*(.+?)[\s|,]
It matches anything after "Version" until it finds a comma or a blank space它匹配“版本”之后的任何内容,直到找到逗号或空格
Try Version\s+([^,\s]+).+
尝试Version\s+([^,\s]+).+
Explanation:解释:
Version
- matches Version
literally Version
- 从字面上匹配Version
\s+
- matches one or more of white spaces \s+
- 匹配一个或多个空格
([^,\s]+)
- match one or more characters other from whitespace \s
or comma ,
and store it inside first capturing group ([^,\s]+)
- 匹配除空格\s
或逗号之外的一个或多个字符,
并将其存储在第一个捕获组中
.+
- match one or more of any characters (except newline), this is to consume rest of the output and prevent from matching multiple versions in one output .+
- 匹配任意字符中的一个或多个(换行符除外),这是为了消耗output的rest,防止在一个output中匹配多个版本
I think the best is;我认为最好的是;
a = r'Version\s*(.+?)[\s|,|[]' a = r'版本\s*(.+?)[\s|,|[]'
example output:例如 output:
6.4.2 6.4.2
03.06.06E 03.06.06E
03.16.07b.S 03.16.07b.S
12.0(18b) 12.0(18b)
12.2(33.3.8)SB13 12.2(33.3.8)SB13
12.2(33)SXI2a 12.2(33)SXI2a
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