Sau*_*wal 2 regex unix bash sed
我有一个简单的要求,我有输入像L
4,23139,,"XYZ"
Run Code Online (Sandbox Code Playgroud)
在文本文件中,我需要像这样的输出
"4","23139","","ARCMsgEnum".
Run Code Online (Sandbox Code Playgroud)
我编写的命令在某些情况下工作正常:
sed 's/[0-9]*[0-9]/"&"/g' inputTestData.txt | sed s/,,/,\"\",/g
Run Code Online (Sandbox Code Playgroud)
但有一些行的输入为时间戳:
4,23139,,"XYZ","2008-09-04 08:11:33.51 AM"
Run Code Online (Sandbox Code Playgroud)
所以期望的输出是:
"4","23139","","XYZ","2008-09-04 08:11:33.51 AM"
Run Code Online (Sandbox Code Playgroud)
但上面的命令给我的输出如下:
"4","23139","","XYZ",""2008"-"09"-"04" "08":"11":"33"."46" AM"
Run Code Online (Sandbox Code Playgroud)
以下内容可能对您有用:
sed -e 's/"//g' -e 's/[^,]*/"&"/g' inputfile
Run Code Online (Sandbox Code Playgroud)
这将首先去除引号,然后在逗号内的部分中插入引号。
举个例子:
$ echo 4,23139,,\"XYZ\",\"2008-09-04 08:11:33.51 AM\" | sed -e 's/"//g' -e 's/[^,]*/"&"/g'
"4","23139","","XYZ","2008-09-04 08:11:33.51 AM"
Run Code Online (Sandbox Code Playgroud)