bash 为什么 sed 不替换所有出现的事件？
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Why does sed not replace all occurrences?
提问by Nishant George Agrwal
If I run this code in bash:
如果我在 bash 中运行此代码：
echo dog dog dos | sed -r 's:dog:log:'
it gives output:
它给出了输出：
log dog dos
How can I make it replace all occurrences of dog?
我怎样才能让它取代所有出现的狗？
回答by Bruno Reis
You should add the gmodifier so that sed performs a globalsubstitution of the contents of the pattern buffer:
您应该添加g修饰符，以便 sed 执行模式缓冲区内容的全局替换：
echo dog dog dos | sed -e 's:dog:log:g'
For a fantastic documentation on sed, check http://www.grymtheitroade.com/Unix/Sed.html. This globalflag is explained here: http://www.grymtheitroade.com/Unix/Sed.html#uh-6
有关 sed 的精彩文档，请查看http://www.grymtheitroade.com/Unix/Sed.html。这里解释了这个全局标志：http: //www.grymtheitroade.com/Unix/Sed.html#uh-6
The official documentation for GNU sedis available at http://www.gnu.org/software/sed/manual/
官方文档GNU sed可在http://www.gnu.org/software/sed/manual/ 获得
回答by alestanis
You have to put a gat the end, it stands for "global":
你必须g在最后放一个，它代表“全局”：
echo dog dog dos | sed -r 's:dog:log:g'
^
