Question

我要求以DD-MM-YYYYY格式填写甄选名单,并附上以下文字。它以90天的日期选定名单。

现在客户的要求发生了变化,他只想填满星期六、星期一或星期三的日期。我无法解决这个问题,我试图修改文字,但并没有工作。

以下文字正在发挥作用。赞赏在这方面提供的任何帮助。

<script>
$(function(){
    function pad(n){return n<10 ?  0 +n : n}
    var date = new Date();
    var selectElement = $( <select name="dddDate" class="ddDate" > ), optionElement;
    for (var count =0; count < 90; count++){
        formattedDate = pad(date.getUTCDate()) +  -  + pad(date.getUTCMonth()+1) +  -  +  date.getUTCFullYear();
        optionElement = $( <option> )
        optionElement.attr( value ,formattedDate);
        optionElement.text(formattedDate);
        selectElement.append(optionElement);
        date.setDate(date.getDate() +1);
    }
    $( #ddDate ).append(selectElement);
});
</script>

<body>
<div id="ddDate"> </div>
</body>

Answer 1

Date.getDay( ): return the day-of-the-week field of a Date. Return values are between 0 (Sunday) and 6 (Saturday). http://jsfiddle.net/UvDEa/

    $(function(){
    function pad(n){return n<10 ?  0 +n : n}
    var date = new Date();
    var selectElement = $( <select name="dddDate" class="ddDate" > ), optionElement;
    for (var count =0; count < 90; count++){
        var day = date.getUTCDay();
        if (day == 6 || day == 1 || day == 3) {
          formattedDate = pad(date.getUTCDate()) +  -  + pad(date.getUTCMonth()+1) +  -  +  date.getUTCFullYear();
          optionElement = $( <option> )
          optionElement.attr( value ,formattedDate);
          optionElement.text(formattedDate);
          selectElement.append(optionElement);  
        }
        date.setDate(date.getDate() +1);
    }
    $( #ddDate ).append(selectElement);
    });

Answer 2

可为此使用<代码>。

if ($.inArray(date.getUTCDay(), [6, 1, 3]) > -1) {
    // Include this date
}

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Answer 3

仅加一,如果对“天空”功能进行这种检查。

if( formattedDate.getDay() == 1 || formattedDate.getDay() == 3 || formattedDate.getDay() == 6)
{
    optionElement = $( <option> );
    optionElement.attr( value ,formattedDate);
    optionElement.text(formattedDate);
    selectElement.append(optionElement);
}

EDIT:

$(function(){
function pad(n){return n<10 ?  0 +n : n}
var date = new Date();
var selectElement = $( <select name="dddDate" class="ddDate" > ), optionElement;
for (var count =0; count < 90; count++){
    formattedDate = pad(date.getUTCDate()) +  -  + pad(date.getUTCMonth()+1) +  -  +  date.getUTCFullYear();
if( formattedDate.getDay() == 1 || formattedDate.getDay() == 3 || formattedDate.getDay() == 6)
{
    optionElement = $( <option> );
    optionElement.attr( value ,formattedDate);
    optionElement.text(formattedDate);
    selectElement.append(optionElement);
 }
    date.setDate(date.getDate() +1);
}
$( #ddDate ).append(selectElement);

iii

Answer 4

这应当行之有效。

$(function(){
function pad(n){return n<10 ?  0 +n : n}
var date = new Date();
var selectElement = $( <select name="dddDate" class="ddDate" > ), optionElement;
for (var count =0; count < 90; count++){
    formattedDate = pad(date.getUTCDate()) +  -  + pad(date.getUTCMonth()+1) +  -  +  date.getUTCFullYear();
if( date.getUTCDay() == 1 || date.getUTCDay() == 3 || date.getUTCDay() == 6)
{
    optionElement = $( <option> );
    optionElement.attr( value ,formattedDate);
    optionElement.text(formattedDate);
    selectElement.append(optionElement);
 }
    date.setDate(date.getDate() +1);
}
$( #ddDate ).append(selectElement);

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