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Porti automate (nu merg)
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Question
amaru
Vroiam sa fac un tutorial "cum sa faci porti automate" si ca sa nu ma fac de cacao am zis sa incerc intai. ;D
La compilare nu da nicio eroare, warn nimic, dar in joc nu merge.
#define FILTERSCRIPT #include <a_samp> #define SCM SendClientMessage #define COLOR_BLUE 0x2641FEAA forward inchidebariera(playerid); forward inchidepoarta(playerid); forward deschidepoarta(playerid); forward PlayerToPoint(Float:radi, playerid, Float:x, Float:y, Float:z); new bariera; new poarta; public OnFilterScriptInit() { bariera = CreateObject(968, 1544.695557, -1630.902100, 13.169105, 0.0000, 268.0403, 270.0000); poarta = CreateObject(980, 1588.003174, -1638.080078, 14.826468, 0.0000, 0.0000, 0.0000); return 1; } public inchidebariera(playerid) { SetObjectRot(bariera, 0.0000, 268.0403, 270.0000); return 1; } public inchidepoarta(playerid) { MoveObject(poarta, 1588.003174, -1638.080078, 14.826468, 5); return 1; } public PlayerToPoint(Float:radi, playerid, Float:x, Float:y, Float:z) { if(IsPlayerConnected(playerid)) { new Float:oldposx, Float:oldposy, Float:oldposz; new Float:tempposx, Float:tempposy, Float:tempposz; GetPlayerPos(playerid, oldposx, oldposy, oldposz); tempposx = (oldposx -x); tempposy = (oldposy -y); tempposz = (oldposz -z); //printf("DEBUG: X:%f Y:%f Z:%f",posx,posy,posz); if (((tempposx < radi) && (tempposx > -radi)) && ((tempposy < radi) && (tempposy > -radi)) && ((tempposz < radi) && (tempposz > -radi))) { return 1; } } return 0; } public deschidepoarta(playerid) { if (PlayerToPoint(15, playerid,1539.1755,-1627.4850,13.3828)) { SetObjectRot(bariera,0.0000, 0.0000, 270.0000); SetTimerEx("inchidebariera", 5 * 1000, 0, "i", playerid); SCM(playerid, COLOR_BLUE,"Bariera e deschisa si se va inchide in 5 secunde."); } else if (PlayerToPoint(15, playerid,1589.053344,-1638.123168,14.122960)) { MoveObject(poarta,1587.995850, -1638.034668, 9.452288,5); SetTimerEx("inchidepoarta", 5 * 1000, 0, "i", playerid); SCM(playerid, COLOR_BLUE,"Poarta e deschisa si se va inchide in 5 secunde."); } return 1; }2 answers to this question
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