regex

linux主机巡检脚本(内存，磁盘，cpu)

由 試著忘記壹切 提交于 2021-02-08 19:44:11
#!/bin/bash #author by acrossyao #date: 2021-02-08 #放假巡检脚本 echo "---------------------------------------「OS系统巡检信息」---------------------------------------" #OS_IP=`hostname -i | awk '{print $1}'` OS_IP="" IPLIST=`hostname -i` for elem in $IPLIST do regex="\b(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[1-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[0-9])\.(25[0-5]|2[0-4][0-9]|1[0-9][0-9]|[1-9][0-9]|[1-9])\b" ckStep2=`echo $elem | egrep $regex | wc -l` if [ $ckStep2 -eq 0 ] then aa=1 else OS_IP+=$elem", " fi done OS_HOSTNAME=`hostname` OS

Pylint ignore-patterns Not Working

由 我与影子孤独终老i 提交于 2021-02-08 19:41:25
问题 I'm using an rc file that has: ignore-patterns=".*local.*" I expect this to ignore all files with the word local in the name. So, tmplocal.py , tmp.local.py , tmp_local.py , local_tmp.py , etc. When running Pylint (1.7.2), these files are not ignored. Any suggestions? 回答1: What ended up working for me was not including quotes. My current config looks something like this: ignore-patterns=one.py,.*local.* $ pylint --version pylint 1.7.2, astroid 1.5.3 Python 3.5.2 (default, Nov 23 2017, 16:37

Pylint ignore-patterns Not Working

由 醉酒当歌 提交于 2021-02-08 19:41:08
问题 I'm using an rc file that has: ignore-patterns=".*local.*" I expect this to ignore all files with the word local in the name. So, tmplocal.py , tmp.local.py , tmp_local.py , local_tmp.py , etc. When running Pylint (1.7.2), these files are not ignored. Any suggestions? 回答1: What ended up working for me was not including quotes. My current config looks something like this: ignore-patterns=one.py,.*local.* $ pylint --version pylint 1.7.2, astroid 1.5.3 Python 3.5.2 (default, Nov 23 2017, 16:37

Javascript regex to remove string inside bracket

由 荒凉一梦 提交于 2021-02-08 19:40:08
问题 I'd like to remove character between { and } . Example : input_string = "i like apple {nobody knows}"; expected result : "i like aple" 回答1: You can use var out = input_string.replace(/{[^}]*}/,'') If you want to remove more than one occurrence, use var out = input_string.replace(/{[^}]*}/g,'') To remove things between /* and */ , this one should work : var out = input_string.replace(/(?!<\")\/\*[^\*]+\*\/(?!\")/g,'') 来源： https://stackoverflow.com/questions/14648375/javascript-regex-to-remove

Need to extract filename from URL [duplicate]

由 ▼魔方 西西 提交于 2021-02-08 16:23:38
问题 This question already has answers here : Get last part of url using a regex (2 answers) Closed 6 years ago . I am trying to extract a file name with some specific extension from a URL. For example, I have a URL like "https://abc.xyz.com/path/somefilename.xy". I need to extract "somefilename.xy" from the above URL, and nothing else. basically I need to write that code in my java program I am not that good in regular expressions, so can somebody help me in that. 回答1: You could also do it

Need to extract filename from URL [duplicate]

由 别说谁变了你拦得住时间么 提交于 2021-02-08 16:22:06
问题 This question already has answers here : Get last part of url using a regex (2 answers) Closed 6 years ago . I am trying to extract a file name with some specific extension from a URL. For example, I have a URL like "https://abc.xyz.com/path/somefilename.xy". I need to extract "somefilename.xy" from the above URL, and nothing else. basically I need to write that code in my java program I am not that good in regular expressions, so can somebody help me in that. 回答1: You could also do it

Need to extract filename from URL [duplicate]

由 不羁岁月 提交于 2021-02-08 16:20:52
问题 This question already has answers here : Get last part of url using a regex (2 answers) Closed 6 years ago . I am trying to extract a file name with some specific extension from a URL. For example, I have a URL like "https://abc.xyz.com/path/somefilename.xy". I need to extract "somefilename.xy" from the above URL, and nothing else. basically I need to write that code in my java program I am not that good in regular expressions, so can somebody help me in that. 回答1: You could also do it

Need to extract filename from URL [duplicate]

由 前提是你 提交于 2021-02-08 16:11:04
问题 This question already has answers here : Get last part of url using a regex (2 answers) Closed 6 years ago . I am trying to extract a file name with some specific extension from a URL. For example, I have a URL like "https://abc.xyz.com/path/somefilename.xy". I need to extract "somefilename.xy" from the above URL, and nothing else. basically I need to write that code in my java program I am not that good in regular expressions, so can somebody help me in that. 回答1: You could also do it

Need to extract filename from URL [duplicate]

由 北城余情 提交于 2021-02-08 16:10:48
问题 This question already has answers here : Get last part of url using a regex (2 answers) Closed 6 years ago . I am trying to extract a file name with some specific extension from a URL. For example, I have a URL like "https://abc.xyz.com/path/somefilename.xy". I need to extract "somefilename.xy" from the above URL, and nothing else. basically I need to write that code in my java program I am not that good in regular expressions, so can somebody help me in that. 回答1: You could also do it

Why is my Perl in-place script exiting with a zero exit code even though it's failing?

由 こ雲淡風輕ζ 提交于 2021-02-08 15:19:20
问题 I have a one-liner Perl search and replace that looks roughly like this: perl -p -i -e 's/foo/bar/' non-existent-file.txt Because the file doesn't exist (which isn't intentional, but this is part of an automated build script, so I want to protect against that), Perl exits with this error: Can't open non-existent-file.txt: No such file or directory. However, the exit code is still zero: echo $? 0 Am I doing something wrong? Should I be modifying my script, or the way I'm invoking Perl? I was
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