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

Date对象将执行您想要的操作-为每个日期构造一个,然后使用>、<、<=或>=对它们进行比较。

==,!=,==,和!==运算符要求您使用date.getTime(),如中所示

var d1 = new Date();
var d2 = new Date(d1);
var same = d1.getTime() === d2.getTime();
var notSame = d1.getTime() !== d2.getTime();

明确地说,直接检查日期对象是否相等是行不通的

var d1 = new Date();
var d2 = new Date(d1);

console.log(d1 == d2);   // prints false (wrong!) 
console.log(d1 === d2);  // prints false (wrong!)
console.log(d1 != d2);   // prints true  (wrong!)
console.log(d1 !== d2);  // prints true  (wrong!)
console.log(d1.getTime() === d2.getTime()); // prints true (correct)

不过,我建议您使用下拉列表或类似的受约束的日期输入形式,而不是文本框,以免您陷入输入验证地狱。


对于好奇的date.getTime()文档:

返回指定日期的数值,即自1970年1月1日00:00:00 UTC以来的毫秒数。(之前的时间返回负值。)

“一些”发布的代码的改进版本

/* Compare the current date against another date.
 *
 * @param b  {Date} the other date
 * @returns   -1 : if this < b
 *             0 : if this === b
 *             1 : if this > b
 *            NaN : if a or b is an illegal date
*/ 
Date.prototype.compare = function(b) {
  if (b.constructor !== Date) {
    throw "invalid_date";
  }

 return (isFinite(this.valueOf()) && isFinite(b.valueOf()) ? 
          (this>b)-(this<b) : NaN 
        );
};

用法:

  var a = new Date(2011, 1-1, 1);
  var b = new Date(2011, 1-1, 1);
  var c = new Date(2011, 1-1, 31);
  var d = new Date(2011, 1-1, 31);

  assertEquals( 0, a.compare(b));
  assertEquals( 0, b.compare(a));
  assertEquals(-1, a.compare(c));
  assertEquals( 1, c.compare(a));
var curDate=new Date();
var startDate=document.forms[0].m_strStartDate;

var endDate=document.forms[0].m_strEndDate;
var startDateVal=startDate.value.split('-');
var endDateVal=endDate.value.split('-');
var firstDate=new Date();
firstDate.setFullYear(startDateVal[2], (startDateVal[1] - 1), startDateVal[0]);

var secondDate=new Date();
secondDate.setFullYear(endDateVal[2], (endDateVal[1] - 1), endDateVal[0]);
if(firstDate > curDate) {
    alert("Start date cannot be greater than current date!");
    return false;
}
if (firstDate > secondDate) {
    alert("Start date cannot be greater!");
    return false;
}

减去两个日期,得到以毫秒为单位的差值,如果得到0,则是相同的日期

function areSameDate(d1, d2){
    return d1 - d2 === 0
}

这是我在一个项目中所做的,

function CompareDate(tform){
     var startDate = new Date(document.getElementById("START_DATE").value.substring(0,10));
     var endDate = new Date(document.getElementById("END_DATE").value.substring(0,10));

     if(tform.START_DATE.value!=""){
         var estStartDate = tform.START_DATE.value;
         //format for Oracle
         tform.START_DATE.value = estStartDate + " 00:00:00";
     }

     if(tform.END_DATE.value!=""){
         var estEndDate = tform.END_DATE.value;
         //format for Oracle
         tform.END_DATE.value = estEndDate + " 00:00:00";
     }

     if(endDate <= startDate){
         alert("End date cannot be smaller than or equal to Start date, please review you selection.");
         tform.START_DATE.value = document.getElementById("START_DATE").value.substring(0,10);
         tform.END_DATE.value = document.getElementById("END_DATE").value.substring(0,10);
         return false;
     }
}

在表单onsubmit上调用此项。希望这有帮助。