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declare: -g: invalid option #11
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Hello @cowbe0x004, thank you for your bug report and specially thank you for providing such a good information about the problem. I see that the version of bash that you are using does not support https://raw.githubusercontent.com/diego-treitos/linux-smart-enumeration/fix%2311/lse.sh |
yep, that works. Thank you for the fast fix. |
I am glad it is fixed. I added the fix to the main branch, so it is already available in version 1.5. Thanks for your help! |
Thank you for the great script! Not sure if there's alternative for declare, but I'm getting these errors in older machines.
$ bash --version
GNU bash, version 3.1.17(1)-release (i486-pc-linux-gnu)
$ declare -g
bash: line 13: declare: -g: invalid option
declare: usage: declare [-afFirtx] [-p] [name[=value] ...]
===================================================================( sudo )=====
[!] sud000 Can we sudo without a password?................................. nope
[!] sud010 Can we list sudo commands without a password?..................../lse.sh: line 328: declare: -g: invalid option
declare: usage: declare [-afFirtx] [-p] [name[=value] ...]
nope
[] sud040 Can we read /etc/sudoers?....................................... nope
[] sud050 Do we know if any other users used sudo?........................ nope
============================================================( file system )=====
[] fst000 Writable files outside user's home.............................../lse.sh: line 328: declare: -g: invalid option
declare: usage: declare [-afFirtx] [-p] [name[=value] ...]
nope
[] fst010 Binaries with setuid bit........................................./lse.sh: line 328: declare: -g: invalid option
declare: usage: declare [-afFirtx] [-p] [name[=value] ...]
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