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I am trying to do a sed substitution over two lines like this:

sed -r '/user-type/{N;s/user-type: (.*)\n.*gth: .*([0-9]+).*$/\1 \2/}'

I got this syntax from an answer here, which explains I first match the first line of the pattern and then add the next line to the pattern space with N to make the substitution. This has worked for me on previous occasions, but this time the weirdest thing happens.

For input:

user-type: admin
password minumim length: 8
user-type: auth
password minumim length: 8

I get:

 8min
 8th

When I expected:

admin 8
auth 8

It looks like the second saved match is printed from the beggining of the line and replaces what was previously printed.

Any idea?

Community
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user177388
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1 Answers1

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Your command works for me:

$ sed -r '/user-type/{N;s/user-type: (.*)\n.*gth: .*([0-9]+).*$/\1 \2/}' file.txt
admin 8
auth 8

If, however, if convert the input to DOS format, I get the output that you see:

$ sed -r '/user-type/{N;s/user-type: (.*)\n.*gth: .*([0-9]+).*$/\1 \2/}' <(sed 's/$/\r/' <file.txt)
 8min
 8th

One solution is to convert your input file to unix-style line-endings.

Another solution is to adjust the sed command to be tolerant of \r characters:

$ sed -r '/user-type/{N;s/user-type: (.*)\n.*gth: .*([0-9]+).*$/\1 \2/; s/\r//}' <(sed 's/$/\r/' <file.txt)
admin 8
auth 8

Or:

$ sed -r '/user-type/{N;s/user-type: ([^\r]*).?\n.*gth: .*([0-9]+).*$/\1 \2/}' <(sed 's/$/\r/' <file.txt)
admin 8
auth 8
John1024
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