30.11.2017, 04:22
"castka" is already an integer.
You can't use strval on it (since strval need a string as a parameter).
You should read error message carefully.
You can't use strval on it (since strval need a string as a parameter).
PHP код:
PlayerInfo[playerid][BankMoney]+=strval(castka);
GivePlayerMoney(playerid,-strval(castka));