Pattern match a variable using Lua

Question

Evening all,

I'm looking at passing text from a monitoring system as a variable and then using the regex/LUA to look for duplicate usernames

Example line of text is:

Unathorised Change Profile Entries: (Audit trail entry) USERNAME - USERNAME

Any advice would be greatly appreciated.

Check [`if(string.match(str, "(%w+) %- %1$")) then print("Matching!") end`](http://ideone.com/RHThad). — Wiktor Stribiżew, Jan 31 '17 at 23:28
Almost there! Looks great! Thank you. I'm passing the variable text from some alert text, here's my test: — greenage, Feb 01 '17 at 17:06
a = alarm.get ("VD27075360-92047") str = a if(string.match(str, "(%w+) %- %1$")) then print("Matching!") end — greenage, Feb 01 '17 at 17:06
You may adjust the pattern, say, instead of `%w` use `%a` to match just letters. If the spaces are not obligatory, use `*` quantifier after each of them. — Wiktor Stribiżew, Feb 01 '17 at 17:07
Geting this: Error in line 5: bad argument #1 to 'match' (string expected, got table) — greenage, Feb 01 '17 at 17:08
Here's what I have at the moment: a = alarm.get ("VD27075360-92047") str = a if(string.match(str, "(%w+) %- %1$")) then print("Matching!") end — greenage, Feb 01 '17 at 17:29
You are stll not passing a *string*, see [Print value from a lua table if pattern matches](http://stackoverflow.com/questions/25252653/print-value-from-a-lua-table-if-pattern-matches). — Wiktor Stribiżew, Feb 01 '17 at 17:50
Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/134608/discussion-between-greenage-and-wiktor-stribizew). — greenage, Feb 01 '17 at 18:22

score 0 · Answer 1 · answered Feb 01 '17 at 21:14

0

Big thank you for the help above.

Here's my final working code:

a = alarm.get ()

str = a

if (string.match(a.message, "(%a+) %- %1$")) then print("Matching!")

else print("No match") end

answered Feb 01 '17 at 21:14

greenage

1 Answers1