我必须构造一个对象数组。我可以“长手”,但我希望找到一种方法来遍历一些变量,并检查每个变量,将它们“推”到数组中的正确位置。使用相同的“规则”检查多个变量
我有这样的:完美在上面的例子
//this is the starting array...I'm going to update these objects
operationTime = [
{"isActive":false,"timeFrom":null,"timeTill":null},//Monday which is operationTime[0]
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null},
{"isActive":false,"timeFrom":null,"timeTill":null}
];
//I get the below via an API call
var monHours = placeHours.mon_open_close;
var tueHours = placeHours.tue_open_close;
var wedHours = placeHours.wed_open_close;
var thuHours = placeHours.thu_open_close;
var friHours = placeHours.fri_open_close;
var satHours = placeHours.sat_open_close;
var sunHours = placeHours.sun_open_close;
var sunHours = placeHours.sun_open_close;
//here's where I'm stuck.
if (monHours.length>0){
var arr = monHours[0].split("-");
operationTime[0].isActive= true;
operationTime[0].timeFrom= arr[0];
operationTime[0].timeTill= arr[1];
}
else {
operationTime[0].isActive= false;
}
我if/else
使用作品周一,但我不想写这七天使它过于复杂的一周。我怎么能把这个压缩成一个单独的“函数”来测试每个变量并将它推入数组对象的正确位置?
你为什么设置为参考对象'operationTime [2]'为假,如果周一不存在?这是一个错字吗? – Psidom
@Psidom好赶上...一个错字。 – jonmrich