bash 使用 sed 获取字符串的第一个单词
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Get first word of string with sed
提问by Vladimir Voitekhovski
How can i cut with sed the following property to have only MAC?
我怎样才能用 sed 剪切以下属性MAC？
MAC evbyminsd58df
I did this but it works in the other side:
我这样做了，但它在另一边工作：
sed -e 's/^.\{1\}//'
回答by fedorqui 'SO stop harming'
Just remove everything from the space:
只需从空间中删除所有内容：
$ echo "MAC evbyminsd58df" | sed 's/ .*//'
MAC
As you mention cut, you can use cutselecting the first field based on space as separator:
正如您提到的那样，您可以使用cut基于空间选择第一个字段作为分隔符：
$ echo "MAC evbyminsd58df" | cut -d" " -f1
MAC
With pure Bash, either of these:
使用纯 Bash，以下任一方式：
$ read a _ <<< "MAC evbyminsd58df"
$ echo "$a"
MAC
$ echo "MAC evbyminsd58df" | { read a _; echo "$a"; }
MAC
回答by Arjun Mathew Dan
with cut(space as delimiter, select first field):
with cut（空格作为分隔符，选择第一个字段）：
echo "MAC evbyminsd58df" | cut -d " " -f 1
with awk(select print first field, space is default delimiter):
with awk（选择打印第一个字段，空格是默认分隔符）：
echo "MAC evbyminsd58df" | awk '{print }'
回答by Avinash Raj
Use grep like below,
像下面这样使用grep，
grep -o '^[^ ]\+' file
OR
或者
grep -o '^[^[:space:]]\+' file
^Asserts that we are at the start. [^[:space:]]Negated POSIX character class which matches any character but not of a space , zero or more times.
^断言我们处于开始阶段。[^[:space:]]否定 POSIX 字符类，匹配任何字符但不匹配空格，零次或多次。
