Random spawn
#1

I have this random spawn code:

pawn Код:
new Float:RandomUSACoords[10][3] =
{
    (-219.5760,2704.1409,62.5391), // Random Usa Spawn 1
    (-198.5659,2713.6584,62.5391), // Random Usa Spawn 2
    (-247.3498,2678.3420,62.5391), // Random Usa Spawn 3
    (-230.0754,2596.5901,62.5391), // Random Usa Spawn 4
    (-236.6343,2597.4094,78.7681), // Random Usa Spawn 5
    (-250.2850,2602.0071,67.3468), // Random Usa Spawn 6
    (-296.3360,2713.4448,65.5440), // Random Usa Spawn 7
    (-302.6953,2691.2366,62.6875), // Random Usa Spawn 8
    (-166.2483,2720.6833,61.9878), // Random Usa Spawn 9
    (-161.6798,2651.2500,80.1681) // Random Usa Spawn 10
};
OnPlayerSpawn:

pawn Код:
if(gTeam[playerid] == TEAM_USA)
    {
        SetRandomUSAPos(playerid);
    }
pawn Код:
stock SetRandomUSAPos(playerid)
{
    if( gTeam[playerid] == TEAM_USA )
    {
        new rand = random(sizeof(RandomUSACoords));
        SetPlayerPos(playerid, RandomUSACoords[rand][0],RandomUSACoords[rand][1],RandomUSACoords[rand][2]);
    }
    return 1;
}
Can someone tell me why doesn't work? I dont get any error while compiling.

When i spawn, i spawn in a factory near LV bridge, i tried spawning multiple times and i always spawn there.

What's wrong?
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#2

Leave it blank

PHP код:
new Float:RandomUSACoords[][] = 
And use "{}" not "()"
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#3

In that way doesn't work.
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#4

Are you sure that you are from TEAM_USA?
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#5

Ok now works, thanks.
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