Python 在特定字符之前删除字节串中的所有内容

Question

I have the following bytestring b'removethis\\x00\\x002020.10.14\\x00\\xf2\\x00^\\n\\x84>\\x01\\x00\\x10\\x01\\x14\\x00\\x00\\x00\\x8d\\xec\\xdc0\\x1bo\\xe7\\x15^\\n\\x84>\\x01\\x00\\x10\\x01\\x04\\x9b_\\x18'

i want everything before the first \\x00 removed so the substring removethis gets removed basically.

So the problem is that i only want everything before the first \\x00 to get removed not any future \\x00 's that may come in the string and im not sure how to do this.

Answer 1

You can build a regex for this.

import re

txt = "b'removethis\x00\x002020.10.14\x00\xf2\x00^\n\x84>\x01\x00\x10\x01\x14\x00\x00\x00\x8d\xec\xdc0\x1bo\xe7\x15^\n\x84>\x01\x00\x10\x01\x04\x9b_\x18'"
x = re.search("\\x00.*", txt)
print(x)

Here we find the first occurance of \\x00 and then take everything to the end of the string as well using .* . The result is your string minus 'removethis'