如何在Linux中使用SED regex从一行中提取多个字符串并将其写入文件？

Question

I have an XML file with multiple lines like below ( I only care about the lines that start with SOURCE)

SOURCE BUSINESSNAME ="" DATABASETYPE ="Oracle" DBDNAME ="OrclExp11g" DESCRIPTION ="" NAME ="EMPLOYEES" OBJECTVERSION ="1" 

SOURCE BUSINESSNAME ="" DATABASETYPE ="Oracle" DBDNAME ="OrclExp11g" DESCRIPTION ="" NAME ="HR" OBJECTVERSION ="1"

In every line that starts with SOURCE I need to get 3 strings and write them to another file like below.

Oracle,OrclExp11g,EMPLOYEES

Oracle,OrclExp11g,HR

sed -n -e '/SOURCE /p' InputFile.XML | sed -r 's/.* NAME \=\"(.+)\" OBJECTVERSION \=\".*/\1/'  > $Source_List.Out

I am new to using SED but so far I was able to get out only one string out using SED. I really appreciate if anyone can help me how to get 3 strings out. Thanks so much in advance!

Answer 1

As you guessed sed is your friend, you could replace matched regex using \\1 , \\2 and so on.

$ sed -nE '/SOURCE/{s/^.*DATABASETYPE ="([^"]*)".*DBDNAME ="([^"]*)".*NAME ="([^"]*)".*$/\1,\2,\3/;p}' file >outputfile

Output

$ cat outputfile
Oracle,OrclExp11g,EMPLOYEES
Oracle,OrclExp11g,HR

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Notes

• -E enable extended regex.
• -n with sed suppresses the normal output. Only the lines that you would print with p will be printed.