在 unix 文件中查找以“#”开头的所有单词

Question

输入文件看起来像这样：-

mary #had a little #lamb its #fleece #was white as snow

输出应该类似于

#had #lamb #fleece #was

Answer 1

使用sed

sed -r 's/(^| )[^#][^ ]+//g'

测试

echo "mary #had a little #lamb its #fleece #was white as snow" | sed -r 's/(^| )[^#][^ ]+//g'

#有一只#lamb #fleece #was

使用grep

grep -o "#[^ ]*"

测试

$ echo "mary #had a little #lamb its #fleece #was white as snow" | grep -o "#[^ ]*"
#had
#lamb
#fleece
#was

Answer 2

如果您不介意前导 space 并且不想匹配中间可能包含 # 的单词，例如 lit#tle，您也可以尝试以下操作：

grep -o -P "(\s|^)#\w+"

示例：

echo "#mary #had a lit#tle #lamb its #fleece #was white as snow" | grep -o -P "(\s|^)#\w+"
#mary
 #had
 #lamb
 #fleece
 #was