使用sed匹配行尾，NOT后面紧跟某个字符

Question

I have a set of lines followed by '\\' I want to print up to the line not containing the '\\' character at the end of the line. Note that xx# yy# change per file and to be re-usable, I want my script to print lines from match1 to '\\' end of line.

match1 xx1 xx1 xx1 xx1 \
yy1 yy1 yy1 yy1 \
zz\<1\> zz1 zz1 zz1 
xx2 xx2 xx2 xx2 \
...

output should be:

match1 xx1 xx1 xx1 xx1 \
yy1 yy1 yy1 yy1 \
zz\<1\> zz1 zz1 zz1 

My best effort came up with:

sed -n '/match1/,/\\/p' file

but does not work. Any help or suggestions are appreciated.

Answer 1

The following command will do what you want:

sed -n '/match1/,/[^\\]$/p' file

Using the ^ at the start of a character class negates the match.

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Following to glenn jackman's great comment the above solution has a little problem. Since [^\\\\] requires at least one character it would not work if the first line which not ends with a \\ is an empty line. Let me add his solution which should be the preferred one:

sed -n '/match1/,/\(^\|[^\\]\)$/p' file