【コメントで質問OKです】
.done,.failの外でreturnすること
var getEventInfo = function(roomID){
console.log(TGVC.sru + "/api/room/event_and_support?room_id=" + roomID)
$.ajax({type: 'GET',url: TGVC.sru + "/api/room/event_and_support?room_id=" + roomID, async: false})
.done(function(re) {
console.log(re);
eventInfo = re;
})
.fail(function() {
// room_info = null;
});
return eventInfo; //.done,.failの外で実行
}