如何只用sed替换一行中的最后一场比赛? [英] How to replace only last match in a line with sed?

问题描述

使用sed，我可以使用

sed 's/pattern/replacement/'

所有匹配使用

sed 's/pattern/replacement/g'

如何只替换最后匹配项，而不管它之前有多少匹配项?

How do I replace only the last match, regardless of how many matches there are before it?

推荐答案

复制粘贴我在别处发布的内容:

Copy pasting from something I've posted elsewhere:

$ # replacing last occurrence
$ # can also use sed -E 's/:([^:]*)$/-\1/'
$ echo 'foo:123:bar:baz' | sed -E 's/(.*):/\1-/'
foo:123:bar-baz
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):/\1-/'
456:foo:123:bar:789-baz
$ echo 'foo and bar and baz land good' | sed -E 's/(.*)and/\1XYZ/'
foo and bar and baz lXYZ good
$ # use word boundaries as necessary - GNU sed
$ echo 'foo and bar and baz land good' | sed -E 's/(.*)\band\b/\1XYZ/'
foo and bar XYZ baz land good

$ # replacing last but one
$ echo 'foo:123:bar:baz' | sed -E 's/(.*):(.*:)/\1-\2/'
foo:123-bar:baz
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):(.*:)/\1-\2/'
456:foo:123:bar-789:baz

$ # replacing last but two
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):((.*:){2})/\1-\2/'
456:foo:123-bar:789:baz
$ # replacing last but three
$ echo '456:foo:123:bar:789:baz' | sed -E 's/(.*):((.*:){3})/\1-\2/'
456:foo-123:bar:789:baz

进一步阅读:

• 单词边界的错误行为在带有限定符的组内使用 - 例如: echo 'it line with it here's it too' |sed -E 's/with(.*\bit\b){2}/XYZ/' 失败
• 贪婪与不情愿与占有量词
• 参考 - 这个正则表达式是什么意思?
• sed 手册:反向引用和子表达式

• Buggy behavior if word boundaries is used inside a group with quanitifiers - for example: echo 'it line with it here sit too' | sed -E 's/with(.*\bit\b){2}/XYZ/' fails
• Greedy vs. Reluctant vs. Possessive Quantifiers
• Reference - What does this regex mean?
• sed manual: Back-references and Subexpressions

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