在grep中使用指定的量词检索满意的词汇表

Question

我试着从一个文件中抓取单词：

$ grep -o '\*\*[^*]*\*\*' Principles_20_LifePrinciples.md | grep -v -e "Origin" -e "Etymology"
**circumstance**
**case**
**condition**
**Anxiety**
**anxiety**
**the state of feeling nervous or worried that sth bad is going to happen**
**a worry or fear about sth**
**a strong feeling of wanting to do sth or of wanting sth to happen**

结果，我只想说出一句话：

**circumstance**
**case**
**condition**
**Anxiety**
**anxiety**

具有指定的量化符{,20}的重构代码：

$ grep -E -o '\*\*[^*]{,20}\*\*' Principles_20_LifePrinciples.md

不幸的是，它什么也没回。

如何解决这样的问题？

Answer 1

对于BSD，有关支持的正则表达式的详细信息，请参阅man 7 re_format。它特别指出：

 A bound is `{' followed by an unsigned decimal integer, possibly followed
 by `,' possibly followed by another unsigned decimal integer, always fol-
 lowed by `}'.  The integers must lie between 0 and RE_DUP_MAX (255=)
 inclusive, and if there are two of them, the first may not exceed the
 second.

只有第二个数字可以省略；第一个数字必须给出。

有了这个解决办法：

$ /usr/bin/grep --version
grep (BSD grep) 2.5.1-FreeBSD
$ /usr/bin/grep -Eo '\*\*[^*]{0,20}\*\*' foo
**circumstance**
**case**
**condition**
**Anxiety**
**anxiety**