如何使用grep匹配第一个空格之前的所有字符？

Question

I'd like to use grep to match all characters before the first whitespace.

grep "^[^\s]*" filename.txt

did not work. Instead, all characters before the first s are matched. Is there no \\s available in grep?

Answer 1

您也可以尝试使用perl regex标志P和o标志在输出中仅显示匹配的部分：

grep -oP "^\S+" filename.txt

Answer 2

With a POSIX character class:

grep -o '^[^[:blank:]]*' filename.txt

As for where \\s is available:

• POSIX grep supports only Basic Regular Expressions or, when called grep -E , Extended Regular Expressions, both of which have no \\s
• GNU grep supports \\s as a synonym for [[:space:]]
• BSD grep doesn't seem to support \\s

Alternatively, you could use awk with the field separator explicitly set to a single space so leading blanks aren't ignored:

awk -F ' ' '{ print $1 }'