unix - sed 使用通配符替换

Question

如何用 sed 进行下面的替换？

输入

group_0 group_10 n_name_0 n_name_10 n_name_20 n_name_5 n_name_40 team_20 team_1

所需输出

group_0 group_10 n_name n_name n_name n_name n_name n_name team_20 team_1

我尝试使用 sed -i 's/n_name*/n_name/g' 但它会删除 n_name 之后的所有内容

Answer 1

sed -i 's:\(n_name\)_[[:digit:]]*:\1:g'

Answer 2

解决方案非常简单。在http://www.unix.com/shell-programming-scripting/31583-wildcards-sed.html上找到了这个

$commandSed ="sed -r 's/n_name_[0-9]*/un_cell/g' test.txt>out.txt";
system($commandSed);

Answer 3

根据您的输入数据工作：

sed -r 's/_[0-9]+//g'

请参阅下面的行：

kent$   echo "group0 group1 n_name_0 n_name_10 n_name_20 n_name_5 n_name_40 team0 team1"|sed -r 's/_[0-9]+//g'
group0 group1 n_name n_name n_name n_name n_name team0 team1

更新

为您的新输入更新

sed -r 's/(n_name)_[0-9]+/\1/g'

测试：

kent$ echo "group_0 group_10 n_name_0 n_name_10 n_name_20 n_name_5 n_name_40 team_20 team_1"|sed -r 's/(n_name)_[0-9]+/\1/g' 
group_0 group_10 n_name n_name n_name n_name n_name team_20 team_1

更新

我假设您想在您的 shell 脚本中使用该行。所以看下面的测试：

kent$  ls
test.txt

kent$  cat test.txt 
group_0 group_10 n_name_0 n_name_10 n_name_20 n_name_5 n_name_40 team_20 team_1

kent$  commandSed=$(sed -r 's/(n_name)_[0-9]+/\1/g' test.txt > out.txt)

kent$  ls
out.txt  test.txt

kent$  cat out.txt
group_0 group_10 n_name n_name n_name n_name n_name team_20 team_1

实际上 commandSed var 在这里没有任何意义。

如果你这样做：

kent$  commandSed=$(sed -r 's/(n_name)_[0-9]+/\1/g' test.txt)   

（不重定向到新文件）

kent$  echo $commandSed 
group_0 group_10 n_name n_name n_name n_name n_name team_20 team_1

如果你想在新文件和 commandSed 变量中都有输出， tee 是你的朋友：

kent$  commandSed=$(sed -r 's/(n_name)_[0-9]+/\1/g' test.txt|tee out.txt)  

kent$  echo $commandSed                                                  
group_0 group_10 n_name n_name n_name n_name n_name team_20 team_1

kent$  cat out.txt 
group_0 group_10 n_name n_name n_name n_name n_name team_20 team_1