在记事本++中用相同的开头和结尾替换字符串

Question

Not much of a expert in regular expression here. My IT colleague dump my a text file which containt all our mailing list. But I need to clean it before using it.

I need to find and delete a string of text which always start by "3EA" and finish by "////".

I try several ways, but no cigars.

Here's some example of the string I need to delete :

3AEAEACjkCjm////
3AEAEACjlCjn////
3AEAEACjnCjp////

Thanks

Answer 1

3AE[a-zA-Z]{9}/{4}

Try this.See demo.

https://regex101.com/r/vN3sH3/61

Answer 2

假设在3AE和////之间总是有9个字母字符，在搜索和替换中（在正则表达式模式下）将此模式替换为空白模式：

3AE[A-Za-z]{9}////

Answer 3

You could use the below regex,

3EA\S*?////

Add anchors if necessary.

^3EA\S*?////$