yang1zhi 发表于 2018-11-17 15:12
同样是这么看的。
前提条件
Game_Party.prototype.menuActor = function() {
var actor = $gameActors.actor(this._menuActorId);
//if (!this.members().contains(actor)) {
// actor = this.members()[0];
//}
return actor;
};
var actor = $gameActors.actor(2);
$gameParty.setMenuActor(actor);
SceneManager.push(Scene_Status)
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