Linux 通过bash使用google作为字典查询，如何获取第一个定义？

从google.com转储一系列定义示例如下：

#!/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 "*"

问题是，我不想要所有的定义，只想要第一个：

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"

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如何从输出中抓取那个句子？它介于两个*之间，可以使用吗？

这将从第一行开始剥离项目符号，打印它并丢弃剩余的输出

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

---

尝试使用head命令添加以下内容：

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

因此，它变成：

head -n 1 -q | tail -n 1

---

这并没有像上面那样删除文本格式，但它确实隔离了用于命令后格式设置的文本。

#!/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