linux - 通过 bash 使用 google 作为字典查找，如何获取第一个定义？

Question

#!/bin/bash
# Command line look up using Google's define feature - command line dictionary

echo "Type in your word:"
read word

/usr/bin/curl -s -A 'Mozilla/4.0'  'http://www.google.com/search?q=define%3A+'$word \
| html2text -ascii -nobs -style compact -width 500 | grep "*"

从 google.com 转储一系列定义，示例如下：

Type in your word:
world
    * universe: everything that exists anywhere; "they study the evolution of the universe"; "the biggest tree in existence"
    * people in general; especially a distinctive group of people with some shared interest; "the Western world"
    * all of your experiences that determine how things appear to you; "his world was shattered"; "we live in different worlds"; "for them demons were as much a part of reality as trees were"

问题是，我不想要所有定义，只想要第一个：

universe: everything that exists anywhere; "they study the evolution of the universe"; "the biggest tree in existence"

如何从输出中抓取该句子？它在两个*之间，可以使用吗？

Answer 1

这将从第一行的开头剥离子弹，打印它并丢弃其余的输出。

sed 's/^ *\* *//; q'

Answer 2

添加这个：

head -n 1 -q | tail -n 1

所以它变成：

#!/bin/bash
# Command line look up using Google's define feature - command line dictionary

echo "Type in your word:"
read word

/usr/bin/curl -s -A 'Mozilla/4.0'  'http://www.google.com/search?q=define%3A+'$word \
| html2text -ascii -nobs -style compact -width 500 | grep "*" | head -n 1 -q | tail -n 1

Answer 3

试试 head 命令