如何删除单词之间的多个空格？

Question

grep include /etc/nginx/nginx.conf

输出：

include /etc/nginx/modules-enabled/*.conf;
    include             mime.types;
        include /etc/nginx/conf.d/*.conf;
        include /etc/nginx/sites-enabled/*;

期望的输出：

include /etc/nginx/modules-enabled/*.conf;
include mime.types;
include /etc/nginx/conf.d/*.conf;
include /etc/nginx/sites-enabled/*;

Answer 1

awk '/include/ {$1 = $1; print}' < your-file\n

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会那么做。

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通过将某些内容分配给$1，强制awk通过将字段（默认情况下由空格分隔（至少是空格和制表符，可能其他取决于区域设置和 awk 实现））与OFS（默认情况下是空格）来重建记录。

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等价sed的：

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sed -E '/include/!d; s/[[:space:]]+/ /g; s/^ //; s/ $//' < your-file\n

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[[:space:]]其中[[:blank:]]至少包括空格和制表符；[[:space:]]还包括垂直间距字符\xc2\xb9，如果文件具有 MSDOS CRLF 行结尾，则这在这里可能很有用，因为它将删除行末尾的那些虚假 CR。

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^{\xc2\xb9 如垂直制表符、换页符等不应出现在输入中的换行符、记录处理中不会出现的行分隔符，awk依次处理awk每一行的内容}

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