Regex 如何删除atom编辑器中每行的前23个字符？

如何删除atom编辑器中每行的前23个字符

谢谢

0,0:10:49.25,0:10:51.00This is incredible, little brother!
0,0:10:52.38,0:10:54.58I never thought it would work so well.
0,0:10:56.92,0:11:01.08Yes, it's more than\N I ever imagined, big brother.
0,0:11:02.96,0:11:06.00My goal was to become\N the strongest man in the world.
0,0:11:08.13,0:11:09.42The strongest man.
0,0:11:10.04,0:11:11.42That was my dream.
0,0:11:14.21,0:11:18.21At last, the ultimate steroid,\N "Biceps King," is complete!
0,0:11:20.13,0:11:21.88Little brother!
0,0:11:21.92,0:11:24.96Drink this and you will obtain\N the power you've been seeking!
0,0:11:26.08,0:11:28.29What? Drink that?
0,0:11:28.79,0:11:32.67Indeed! I added strawberry flavoring,\N to make it go down easier.
0,0:11:33.50,0:11:34.50Brother...
0,0:11:41.21,0:11:43.42Yes, this will do!
0,0:11:45.04,0:11:47.29My brains and your brawn!
0,0:11:47.33,0:11:50.58By combining the greatest of minds\N and the strongest of bodies,
0,0:11:50.79,0:11:55.13we brothers will conquer everything\N on earth
0,0:11:55.17,0:11:56.67and rule as kings!
0,0:12:04.04,0:12:06.08I-Incredible!

文本令人不安

来自atom文档：

ctrl-x（开始录制）

ctrl-x）停止录制

ctrl-x e execute宏

ctrl-x ctrl-e执行宏N次

ctrl-x b执行宏到文件末尾

ctrl-x ctrl-b从文件开头到结尾执行宏

资料来源：

使用指向行首的宏，删除23个字符，然后向下一行。然后执行到文件的末尾

---

另外，你可以按照pushpesh kumar rajwanshi说的做，用空字符串替换

*{23}

。

用空字符串替换

^.{23}

行吗？