SED替代使用通配符

Question

我怎樣才能開展與下面的sed替換？

輸入

group_0 group_10 n_name_0 n_name_10 n_name_20 n_name_5 n_name_40 team_20 team_1 

所需輸出

group_0 group_10 n_name n_name n_name n_name n_name n_name team_20 team_1 

我嘗試使用SED -i的/ n_name */n_name/G'，但它n_name後刪除一切

Answer 1

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

Answer 2

作品根據您輸入的數據：

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

看到下面的一行：

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

更新的新輸入

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

測試：

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

我假設你想在你的shell腳本中使用該行。所以看下面的測試：

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 

其實commandSed var在這裏沒有任何意義。

如果你這樣做：

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

（不重定向到新的文件）

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

，如果你想擁有輸出在新的文件和commandSed變量都，發球是你的朋友：

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 

Answer 3

解決方法很簡單。發現在http://www.unix.com/shell-programming-scripting/31583-wildcards-sed.html

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