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.
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'
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