sed 去除特定字符周围的空格 [英] sed Removing whitespace around certain character

问题描述

仅在某些字符周围删除空格的最佳方法是什么.假设破折号 - Some- String- 12345- Here 将变成 Some-String-12345-Here.类似于 sed 's/\ -/-/g;s/-\/-/g' 但我相信一定有更好的方法.

谢谢！

解决方案

如果你的意思是所有的空格，而不仅仅是空格，那么你可以试试 \s:

echo 'Some- String- 12345- Here' |sed 's/\s*-\s*/-/g'

输出:

<前>Some-String-12345-Here

或者使用[:space:]字符类:

echo 'Some- String- 12345- Here' |sed 's/[[:space:]]*-[[:space:]]*/-/g'

不同版本的 sed 可能支持或不支持这些，但 GNU sed 支持.

what would be the best way to remove whitespace only around certain character. Let's say a dash - Some- String- 12345- Here would become Some-String-12345-Here. Something like sed 's/\ -/-/g;s/-\ /-/g' but I am sure there must be a better way.

Thanks!

解决方案

If you mean all whitespace, not just spaces, then you could try \s:

echo 'Some- String- 12345- Here' | sed 's/\s*-\s*/-/g'

Output:

Some-String-12345-Here

Or use the [:space:] character class:

echo 'Some- String- 12345- Here' | sed 's/[[:space:]]*-[[:space:]]*/-/g'

Different versions of sed may or not support these, but GNU sed does.

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