在bash脚本中使用空格拆分句子 [英] Split a sentence using space in bash script

问题描述

如何使用空格分割一个句子，然后从第二个单词开始打印?

How can I split a sentence using space, and print from second word onwards?

例如，如果我的句子是Hello World Good Morning，那么我想打印如下:

For example, if my sentence is Hello World Good Morning, then I want to print like:

World
Good
Morning

推荐答案

With cut:

$ echo "Hello World Good Morning" | cut -d' ' -f2-
World Good Morning

这告诉 cut 根据 d 分隔符空间剪切"(令人惊讶地)并从第二个字段打印到结尾.

This tells cut to "cut" (surprisingly) based on delimiter space and print from 2nd field up to the end.

使用sed:

$ echo "Hello World Good Morning" | sed 's/^[^ ]* //'
World Good Morning

这会从行首 (^) 获取一个不包含空格的字符块 ([^ ]*)，然后是一个空格并替换它的内容为空.这样，第一个词就被删除了.

This gets, from the beginning of the line (^), a block of characters not containing a space ([^ ]*) and then a space and replaces it with empty content. This way, the first word is deleted.

纯bash:

$ while IFS=" " read -r _ b; do echo "$b"; done <<< "Hello World Good Morning"
World Good Morning

这会将字段分隔符设置为空格并读取虚拟变量 _ 中的第一个块，以及变量 $b 中的其余部分.然后，它打印 $b.

This sets the field separator to the space and reads the first block in a dummy variable _ and the rest in the variable $b. Then, it prints $b.

也在 awk 中，使用这个 Ed Morton 的方法:

Also in awk, using this Ed Morton's approach:

$ echo 'Hello World Good Morning' | awk '{sub(/([^ ]+ +){1}/,"")}1'
World Good Morning

这将 1 块 非空格字符 + 块 空格 替换为空字符串.

This replaces 1 block of not space characters + block of spaces with an empty string.

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