Scripting Helpers is winding down operations and is now read-only. More info→
Log in to vote
0
Table does not working with table.find condition?
Hi, i have make a list of a boss in my server, when a player attack a boss i want to check if the specific boss is in the list, i have make this:
1 | local list = { |
2 | "Fox", |
3 | "Yeti", |
4 | "Crocodile", |
5 | "Pharaoh" |
6 | } |
When the player attack the boss i have this condition:
1 | if table.find(list, HitParent) then |
2 | print("Rangeboss") |
HitParent = the name of the boss when the player attack, so for exemple HitParent = Fox
But the print is’nt fire ^^ maybe my condition is not exactly, somebody have a solution for check the name of the boss in the list ?
1
It's probably not the table.find part, maybe it's something else with your code? Try posting the whole thing so we can check if there's an error somewhere that's causing it to not print. killerbrenden 1537 — 5y
3 answers
Log in to vote
1
Use a for loop to search if the item is in the array. For example, you can do something like this:
1 | for i,v in pairs (list) do |
2 | if v == hitParent then |
3 | print("Hit boss") |
4 | end |
5 | end |
Hope this helped!
0
Not Working, the print isn't fire :/ Thorngard 4 — 5y
0
for i,v in pairs (list) do if v == hitParent then print("Hit boss") else print("Not on the list") end end AlexTheCreator 461 — 5y
Log in to vote
0
for this to work, HitParent must be a string value and you can do this:
1 | for i,v in pairs(list) do |
2 | if v == HitParent then |
3 | print"hello" |
4 | end |
5 | end |
0
also table.find() does not exist in lua floppybeann 0 — 5y
Log in to vote
0
if list[HitParent] then print(HitParent) end