如果日期少于或多于 15 分钟（日落或日出）

Question

我正在尝试处理日期，现在是日落或日出。

因为我不确定如何使用 If Else 语句 关于 Date

我得到了日落时间和日出时间的数据。

但想知道如果当前日期是日落或日出如何使用 +-15 minutes

示例：

当前日期：2015-12-17T17:45:00 （或介于 17:45 到 18:15 之间）

日落：2015-12-17T18:00:00

结果=日落（因为+-15 minutes）

到目前为止，这是我的代码：

var SunData = {'set':'2015-12-17T09:14:56', 'rise':'2015-12-17T15:10:52'};

var date = new Date();
    date = date.getTime();

var sun_rise = new Date(SunData['rise']);
    sun_rise = sun_rise.getTime();

var sun_set = new Date(SunData['set']);
    sun_set = sun_set.getTime();

var sunRiseSetDiff = 900000; // 15 minutes

// Here
var result;
if(false) {
    result = 'sun-rise';
} else if (false) {
    result = 'sun-set';
} else {
    result = 'none';
}

https://jsfiddle.net/gd1s1o0y/

你能帮我解决一下 If Else 语句 吗？谢谢

Answer 1

你是说

var result = 'none';
if(Math.abs(date - sun_rise) <= sunRiseSetDiff) {
    result = 'sun-rise';
}
else if(Math.abs(date - sun_set) <= sunRiseSetDiff) {
    result = 'sun-set';
}

Fiddle

Answer 2

勾选这个fiddle

var SunData = {'set':'2015-12-17T09:14:56', 'rise':'2015-12-17T15:10:52'};

var date = new Date();
        date = date.getTime();

var sun_rise = new Date(SunData['rise']);
        sun_rise = sun_rise.getTime();

var sun_set = new Date(SunData['set']);
        sun_set = sun_set.getTime();

var sunRiseSetDiff = 900000; // 15 minutes

// Here
var sunsetResult = Math.abs( date - sun_set ) > sunRiseSetDiff ? false: true;
var sunriseResult = Math.abs( date - sun_rise ) > sunRiseSetDiff ? false: true;

if(sunriseResult) {
    result = 'sun-rise';
} else if (sunsetResult) {
    result = 'sun-rise';
} else {
    result = 'none';
}

alert(result);

Answer 3

只是用注释重写你的函数。

// Always parse date strings manually, you can use a library if you want
// but a bespoke function is pretty simple if the format is guaranteed
// and is a valid date
function qParse(s) {
  var b = s.split(/\D/);
  return new Date(b[0], b[1]-1, b[2], b[3]||0, b[4]||0, b[5]||0, b[6]||0);
}
    
// The OP seemed to have sunrise and sunset transposed
var sunData = {'rise':'2015-12-17T09:14:56', 'set':'2015-12-17T15:10:52'};

// There is no need for getTime, date objects can be used in simple
// arithmetic directly
var now = new Date();
var buffer = 9e5; // 15 mintues

// Initialise result to a default string
var result = 'Not near sunrise or sunset';
    
// Change value if near sunset
if (Math.abs(now - qParse(sunData.set)) <= buffer) {
  result = "Close to sunset"; 

// Otherwise, change value if near sunrise
} else if (Math.abs(now - qParse(sunData.rise)) <= buffer) {
  result = "Close to sunrise"; 
}
    
// Show result
document.write(result);

如果日期少于或多于 15 分钟（日落或日出）

If date less or more than 15 minuites (If Sunset or Sunrise)

javascript

conditional

datetime

date