在bourne shell中从字符串中提取n个单词

Question

我想从一个字符串中提取特定数量的单词。

例如:s1= "Hello: I am new to this forum"

我如何从字符串中提取前4个单词"Hello: I am new" ?

Answer 1

你可以这样做;

#!/bin/bash
s1="Hello: I am new to this forum"
echo $s1 | cut -d' ' -f-4

-d' ':使用空格作为字段分隔符-f-4:字段4及

之前

man cut;

   -d, --delimiter=DELIM
          use DELIM instead of TAB for field delimiter
   -f, --fields=LIST
          select only these fields;  also print any line that contains no delimiter character, unless the -s option is specified

with awk solution;

#!/bin/bash
s1="Hello: I am new to this forum"
echo $s1 | awk  '{printf "%s %s %s %sn", $1, $2, $3, $4}' 

或只回显;

#!/bin/bash
s1="Hello: I am new to this forum"
echo ${s1/% to*/}

with sed;

#!/bin/bash
s1="Hello: I am new to this forum"
echo $s1 | sed -E 's/(([^ ]+ ){4}).*/1/'