bash - 删除行并更新订单

Question

我有一个文件，其中包含以数字开头的行，例如

1 This is the first line
2 this is the second line
3 this is the third line
4 This is the fourth line

我想要做的是删除一行，例如第 2 行并更新编号，以便文件如下所示，我想在 bash 脚本中执行此操作。

1 This is the first line
2 this is the third line
3 This is the fourth line

谢谢

Answer 1

IMO 可能会更容易一些awk：

awk '!/regex/ {$1=++x; print}' inputFile

在/.../您可以将regex发生在需要删除的行上。

测试：

$ cat inputFile
1 This is the first line
2 this is the second line
3 this is the third line
4 This is the fourth line

$ awk '!/second/ {$1=++x; print}' inputFile
1 This is the first line
2 this is the third line
3 This is the fourth line

$ awk '!/third/ {$1=++x; print}' inputFile
1 This is the first line
2 this is the second line
3 This is the fourth line

$ awk '!/first/ {$1=++x; print}' inputFile
1 this is the second line
2 this is the third line
3 This is the fourth line

注意：由于我们正在重新构建该$1字段，任何空白序列都将被删除。

Answer 2

您可以使用这组命令：

grep -v '^2 ' file | cut -d' ' -f2- | nl -w1 -s' '

1. 使用grepwith-v选项可以删除第 2 行。
2. cut程序剪切第一列，即行号。
3. 最后，我们只需要重新编号行，所以我们使用nl.