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How to check party members using code?


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Posted (edited)
set .@partyid,getcharid(1);
	set .@mob_base_exp,(( getmonsterinfo( killedrid,MOB_BASEEXP ) / 100 ) * .rotd_exp_bonus );
	set .@mob_job_exp,(( getmonsterinfo( killedrid,MOB_JOBEXP ) / 100 ) * .rotd_exp_bonus );
	if( .@partyid ){
		set .@aid,getcharid(3);
		set .@baselevel,BaseLevel;
		set .@map$,strcharinfo(3);
		getpartymember .@partyid,1;
		getpartymember .@partyid,2;
		while( .@i < $@partymembercount ){
			if( isloggedin( $@partymemberaid[.@i],$@partymembercid[.@i] ) )
				if( attachrid( $@partymemberaid[.@i] ) )
					if( strcharinfo(3) == .@map$ && ( BaseLevel - .@baselevel ) <= .party_level_range && ( BaseLevel - .@baselevel ) >= ~.party_level_range ){	
						set BaseExp,( BaseExp + .@mob_base_exp );
						set JobExp,( JobExp + .@mob_job_exp );
						dispbottom "[ROTD Party] : "+.@mob_base_exp+" EXP";
					}
			set .@i,.@i + 1;
		}
		attachrid( .@aid );
	}else{
		set BaseExp,( BaseExp + .@mob_base_exp );
		set JobExp,( JobExp + .@mob_job_exp );
		dispbottom "[ROTD SOLO] : "+.@mob_base_exp+" EXP";
	}
	if( rand( 100 ) < .rotd_item_rate && .rotd_zeny ) 
		set Zeny,Zeny + rand( .rotd_zeny );
	if( rand( 100 ) < .rotd_item_rate && .rotd_item_amount ) 
		getitem .rotd_item_id,rand( .rotd_item_amount ),.@aid;

Trying to find out how to get how many party members are in a party, then divide the experience rates;

Was trying to do   if ( $@partymembercount > 2 ){ set .@mob_job_exp,(( getmonsterinfo( killedrid,MOB_JOBEXP ) / 100 ) * .rotd_exp_bonus / 2 ); }

...But I guess $@partymembercount isn't the code to get how many members in a party tho  /hum Anyone can help me a bit in this?.. Thanks~!

Edited by Talon

2 answers to this question

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Posted

 

Check the wiki. There is everything you need to know about party related commands and informations: http://rathena.org/wiki/Getpartymember

 

first you have to use this command to store the informations in global variables:

getpartymember <party id>{,<type>};

then just use this global variable for your script:

$@partymembercount

Okay, thanks for replied. I'll have a look when I'm free =)

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