有人能提出一种方法来比较两个大于、小于和过去不使用JavaScript的日期的值吗?值将来自文本框。
当前回答
嗨,这是我比较日期的代码。在我的情况下,我正在检查是否允许选择过去的日期。
var myPickupDate = <pick up date> ;
var isPastPickupDateSelected = false;
var currentDate = new Date();
if(currentDate.getFullYear() <= myPickupDate.getFullYear()){
if(currentDate.getMonth()+1 <= myPickupDate.getMonth()+1 || currentDate.getFullYear() < myPickupDate.getFullYear()){
if(currentDate.getDate() <= myPickupDate.getDate() || currentDate.getMonth()+1 < myPickupDate.getMonth()+1 || currentDate.getFullYear() < myPickupDate.getFullYear()){
isPastPickupDateSelected = false;
return;
}
}
}
console.log("cannot select past pickup date");
isPastPickupDateSelected = true;
其他回答
日期比较:
var str1 = document.getElementById("Fromdate").value;
var str2 = document.getElementById("Todate").value;
var dt1 = parseInt(str1.substring(0,2),10);
var mon1 = parseInt(str1.substring(3,5),10);
var yr1 = parseInt(str1.substring(6,10),10);
var dt2 = parseInt(str2.substring(0,2),10);
var mon2 = parseInt(str2.substring(3,5),10);
var yr2 = parseInt(str2.substring(6,10),10);
var date1 = new Date(yr1, mon1, dt1);
var date2 = new Date(yr2, mon2, dt2);
if(date2 < date1)
{
alert("To date cannot be greater than from date");
return false;
}
else
{
alert("Submitting ...");
document.form1.submit();
}
function datesEqual(a, b)
{
return (!(a>b || b>a))
}
from_date ='10-07-2012';
to_date = '05-05-2012';
var fromdate = from_date.split('-');
from_date = new Date();
from_date.setFullYear(fromdate[2],fromdate[1]-1,fromdate[0]);
var todate = to_date.split('-');
to_date = new Date();
to_date.setFullYear(todate[2],todate[1]-1,todate[0]);
if (from_date > to_date )
{
alert("Invalid Date Range!\nStart Date cannot be after End Date!")
return false;
}
使用此代码使用javascript比较日期。
谢谢D.吉普
在javascript中比较日期的最简单方法是首先将其转换为Date对象,然后比较这些日期对象。
下面是一个具有三个功能的对象:
日期.比较(a,b)返回一个数字:-如果a<b,则为1如果a=b,则为0如果a>b,则为1如果a或b是非法日期,则为NaNdates.inRange(d,开始,结束)返回布尔值或NaN:如果d在开始和结束之间(含),则为true如果d在开始之前或结束之后,则为false。如果一个或多个日期是非法的,则为NaN。日期转换由其他函数用于将其输入转换为日期对象。输入可以是日期对象:输入按原样返回。数组:解释为[年,月,日]。注:月份为0-11。数字:解释为自1970年1月1日以来的毫秒数(时间戳)字符串:支持多种不同的格式,如“YYYY/MM/DD”、“MM/DD/YYYY”、“Jan 31 2009”等。对象:解释为具有年、月和日期属性的对象。注:月份为0-11。
.
// Source: http://stackoverflow.com/questions/497790
var dates = {
convert:function(d) {
// Converts the date in d to a date-object. The input can be:
// a date object: returned without modification
// an array : Interpreted as [year,month,day]. NOTE: month is 0-11.
// a number : Interpreted as number of milliseconds
// since 1 Jan 1970 (a timestamp)
// a string : Any format supported by the javascript engine, like
// "YYYY/MM/DD", "MM/DD/YYYY", "Jan 31 2009" etc.
// an object : Interpreted as an object with year, month and date
// attributes. **NOTE** month is 0-11.
return (
d.constructor === Date ? d :
d.constructor === Array ? new Date(d[0],d[1],d[2]) :
d.constructor === Number ? new Date(d) :
d.constructor === String ? new Date(d) :
typeof d === "object" ? new Date(d.year,d.month,d.date) :
NaN
);
},
compare:function(a,b) {
// Compare two dates (could be of any type supported by the convert
// function above) and returns:
// -1 : if a < b
// 0 : if a = b
// 1 : if a > b
// NaN : if a or b is an illegal date
// NOTE: The code inside isFinite does an assignment (=).
return (
isFinite(a=this.convert(a).valueOf()) &&
isFinite(b=this.convert(b).valueOf()) ?
(a>b)-(a<b) :
NaN
);
},
inRange:function(d,start,end) {
// Checks if date in d is between dates in start and end.
// Returns a boolean or NaN:
// true : if d is between start and end (inclusive)
// false : if d is before start or after end
// NaN : if one or more of the dates is illegal.
// NOTE: The code inside isFinite does an assignment (=).
return (
isFinite(d=this.convert(d).valueOf()) &&
isFinite(start=this.convert(start).valueOf()) &&
isFinite(end=this.convert(end).valueOf()) ?
start <= d && d <= end :
NaN
);
}
}
If you are using **REACT OR REACT NATIVE**, use this and it will work (Working like charm)
如果两个日期相同,则返回TRUE,否则返回FALSE
const compareDate = (dateVal1, dateVal2) => {
if (dateVal1.valueOf() === dateVal2.valueOf()){
return true;
}
else { return false;}
}