如何仅替换与 sed 中多个模式匹配的行？

Question

$ echo -e 'CH12\nCH23au' | sed '/^CH/s=^=<b>='
<b>CH12
<b>CH23au

我知道我可以匹配以CHby开头的行，^CH但如何匹配多个模式？

例子：

输入：

CH12
CH23au

输出：

CH12
<b>CH23au

如何只放在<b>有^CH和au在线的地方？

Answer 1

If the CH must occur at the beginning of the line, the order of CH and au is fixed, so you can look for ^CH.*au.

$ echo -e 'CH12\nCH23au' | sed '/^CH.*au/s=^=<b>='
CH12
<b>CH23au
$

If the order of the two patterns is not fixed, one could do something like

sed -e '/pattern1/{;/pattern2/s/old/new/;}'

but the perl solution

perl -pe 'if (/pattern1/ && /pattern2/) {s/old/new/;}'

is probably more readable.

Answer 2

另一种方法：

sed -e '/^CH/!b' -e '/au/!b' -e 's/^/<b>/'

b（如果没有给出任何标签参数，则分支到最后）就像continue或next在其他语言中。所以上面是这样的：

for (; line = readline(); print line) { # The implicit loop in sed
  if (!/^CH/) continue;
  if (!/au/) continue;
  line =~ s/^/<b>/
}