从第11个单词开始，直到字符'-'；是通过在bash中使用awk命令找到的

我有一个带空格的文本

9   100%   361.7 tr  Back       976.0     0.0    2.8  Get      [FFF] DO SOMETHING HERE - 01

我想把它转换成这个

在这里做点什么

我试过这个：

awk 'match($0, "\.-") {print substr($0, 1, RSTART)}' 

但它给了我类似这样的输出：

9   100%   361.7 tr  Back       976.0     0.0    2.8  Get      [FFF] DO SOMETHING HERE

有没有办法从第11个单词开始，直到找到字符“-”为止

awk '{for(i=1;i<11;i++)$i="";sub(/-.*$/,"");sub(/^ */,"")}7'

事实上，grep可以为您做到这一点：

grep -Po '(\S+\s+){10}\K[^-]*'

你的例子是：

kent$  grep -Po '(\S+\s+){10}\K[^-]*' <<<"9   100%   361.7 tr  Back       976.0     0.0    2.8  Get      [FFF] DO SOMETHING HERE - 01"
DO SOMETHING HERE 

---

kent$grep-Po'（\S+\S+{10}\K[^-]*'或者您可以不使用分叉就完成它

read _ _ _ _ _ _ _ _ _ _ _ word <<< '9   100%   361.7 tr  Back       976.0     0.0    2.8  Get      [FFF] DO SOMETHING HERE - 01'
echo "${word%-*}"

read\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu！第一个和第二个选项工作良好。这就是我要找的！再次感谢！
read _ _ _ _ _ _ _ _ _ _ _ word <<< '9   100%   361.7 tr  Back       976.0     0.0    2.8  Get      [FFF] DO SOMETHING HERE - 01'
echo "${word%-*}"