为什么不检查此条件？

如何解决为什么不检查此条件？

我目前正在做一个小项目，我想通过谷歌日历 API 检查房间是否繁忙。 为此，我使用此功能：

//checking for change of all values. Then console.log values on change and executing request if busy.
function avalabilityCheck() {
    [...inputs].forEach(input => {
            input.addEventListener('change',function () {
                    if (date.value !== "" && startTime.value !== "" && endTime.value !== ""
                    ) {let isBusy = true;
                    //looping through all rooms in compartment
                        for (let key in comp_1) {
                            if (comp_1.hasOwnProperty(key)) {
                                let calendarID = comp_1[key];
                                let roomName = key;
                                //console.log(value);
                                //user input that goes into the freebusy query
                                let requestBody = {
                                    timeMin: date.value + "T" + startTime.value + ":00.000Z",timeMax: date.value + "T" + endTime.value + ":00.000Z",items: [
                                        {
                                            id: calendarID
                                        }
                                    ],timeZone: "GMT+01:00"
                                };


                                //make request to gcalendar if rooms are free. Giving back array on what times room is busy.
                                var freeRequest = gapi.client.calendar.freebusy.query(requestBody);

                                //executing request.
                                freeRequest.execute(function (resp) {
                                    var responSEObject = JSON.stringify(resp);
                                    console.log(responSEObject);
                                    if (resp.calendars[calendarID].busy.length < 1) {
                                        console.log(`${roomName} is free`);
                                    } else { isBusy = false;
                                    console.log("room is Busy");}

                                })

                          }
                        }

                    console.log("finito");
                    if (isBusy === false) {
                        console.log("working?");
                    }
                    else{console.log("not working");}
                    } else {
                        console.log("change date pls");
                    }
                }
            )
        }
    )
}

现在let isBusy = true;，当房间很忙时，我希望 isBusy 为假：

if (resp.calendars[calendarID].busy.length < 1) {
                                        console.log(`${roomName} is free`);
                                    } else { isBusy = false;
                                    console.log("room is Busy");} 

所以当我运行应用程序时，控制台应该给出：

console.log("finito");
if (isBusy === false) {
console.log("working?");
 }
else{console.log("not working");}

但它只向控制台提供“finito”，显然没有检查 isBusy 的状态。 有没有人看到这里的问题是什么？ 谢谢！