Awk-将后续行的长度写入行 [英] Awk - Writing length of subsequent line into line

问题描述

假定一个包含交替行的文本文件.具体而言，每个线对的第1行以#"开头，​​而每个线对的下一行包含字母数字字符串.

Assume a text file that comprises alternating lines. Specifically, line 1 of each line pair starts with a "#", whereas the subsequent line of each line pair contains an alphanumeric string.

$ cat file
#Foo
1234567
#Bar
1234
#Baz
123456789

如何自动将第2行的长度(以及关键字)附加到每对线对的第1行?我相信awk是执行此操作的正确选择.

How do I automatically append the length of line 2 (as well as a keyword) to line 1 of each line pair? I believe that awk is the right choice for such an operation.

$ awk 'desired code' file
#Foo_Length7
1234567
#Bar_Length4
1234
#Baz_Length9
123456789

这是我的尝试，但是我不知道用什么替换length($0):

Here's my try, but I can't figure out what to substitute the length($0) with:

awk '{if ($1~/^#/) print $0"_Length"length($0); else print $0}' file

推荐答案

$ awk '!(NR%2){print prev "_Length" length($0) ORS $0} {prev=$0}' file
#Foo_Length7
1234567
#Bar_Length4
1234
#Baz_Length9
123456789

如果愿意，可以将!(NR%2)替换为!/^#/或类似名称.

You can replace !(NR%2) with !/^#/ or similar if you prefer.

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