sed 或 awk 或 grep 获取字符串的第一次动态出现

Question

从这个字符串

Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)

我想获得（并在 shell 中打印）“...apache2.service;”之后的第一个匹配项该词可以“启用”或“禁用”

我如何使用 grep 或 sed 或 awk 来获取它？

Answer 1

使用SED:

echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | sed -n 's/.*service; \([^;]*\).*/\1/p'

使用格雷普:

echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | grep -o -P '(?<=service; ).*(?=;)'

使用AWK:

echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | awk -F 'service; |;' '{print $2}'

• service;将打印和之间的字符串;

Answer 2

您可以使用awk

 echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | awk '{print $4}'

或者awk

echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | awk -v RS=" " '/service/{getline;print "",$0}' | tr -d ";"

或者cut

echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | cut -d';' -f 2

或在两个分号之间sed提取disable

 echo "Loaded: loaded (/lib/systemd/system/<name>.service; disabled; vendor preset: enabled)" | sed 's/^[^;]*;//; s/;.*//'