使用Javascript查找本月的第二个和第四个星期二 [英] Finding the second and fourth Tuesday of the month with Javascript

问题描述

我的协会在每个月的第二和第四个星期二有会议，我真的很喜欢一个Javascript代码，所以我不需要每两周更新我们的网站，下一次会议的日期可以计算和显示

我真的很感谢一些帮助，因为我完全没有Javascript，只有在CSS和HTML。

解决方案

这是一个解决方案

HTML

 < ul id =list>< / ul> 
   
 
 
 
 Javascript 
  function getTuesdays（month，year）{
 var d = new Date（year，month，1），
 tuesdays = []; 
 
 d.setDate（d.getDate（）+（9  -  d.getDay（））％7）
 while（d.getMonth（）=== month）{
 tuesdays.push（new Date（d.getTime（）））; 
 d.setDate（d.getDate（）+ 7）; 
} 
 
 return tuesdays; 
} 
 
 var meetingTuesdays = []，
 ul = document.getElementById（list），
 temp，
 li，
i ; 
 
 for（i = 0; i <12; i + = 1）{
 temp = getTuesdays（i，2013）; 
 meetingTuesdays.push（temp [1]）; 
 li = document.createElement（li）; 
 li.textContent = temp [1]; 
 ul.appendChild（li）; 
 
 meetingTuesdays.push（temp [3]）; 
 li = document.createElement（li）; 
 li.textContent = temp [3]; 
 ul.appendChild（li）; 
} 
 
 console.log（meetingTuesdays）; 
  
在 jsfiddle  
 
 
 
更新：为您进一步示范
 
 
 
 Javascript 
  function getTuesdays（month，year）{
 var d = new Date（year，month，1），
 tuesdays = []; 
 
 d.setDate（d.getDate（）+（9  -  d.getDay（））％7）
 while（d.getMonth（）=== month）{
 tuesdays.push（new Date（d.getTime（）））; 
 d.setDate（d.getDate（）+ 7）; 
} 
 
 return tuesdays; 
} 
 
 var today = new Date（），
 theseTuesdays = getTuesdays（today.getMonth（），today.getFullYear（）），
 next; 
 
 theseTuesdays.some（function（tuesday，index）{
 if（index％2 === 1& tuesday> today）{
 next = tuesday; 
 return true; 
} 
 
 return false; 
}）; 
 
 alert（我们的下一次会议是：+ moment（下一个）.format（MMMM Do YYYY））; 
  
在 jsfiddle  
 
My association has meetings on the second and fourth Tuesday of each month and I'd really like a Javascript code so I don't have to update our site every two weeks and the date of the next meeting can be calculated and displayed automatically.

I'd really appreciate some help, as I have absolutely no skils in Javascript, only in CSS and HTML.
解决方案
This is a solution

HTML
<ul id="list"></ul>
Javascript
function getTuesdays(month, year) {
    var d = new Date(year, month, 1),
        tuesdays = [];

    d.setDate(d.getDate() + (9 - d.getDay()) % 7)
    while (d.getMonth() === month) {
        tuesdays.push(new Date(d.getTime()));
        d.setDate(d.getDate() + 7);
    }

    return tuesdays;
}

var meetingTuesdays = [],
    ul = document.getElementById("list"),
    temp,
    li,
    i;

for ( i = 0; i < 12; i += 1) {
    temp = getTuesdays(i, 2013);
    meetingTuesdays.push(temp[1]);
    li = document.createElement("li");
    li.textContent = temp[1];
    ul.appendChild(li);

    meetingTuesdays.push(temp[3]);
    li = document.createElement("li");
    li.textContent = temp[3];
    ul.appendChild(li);
}

console.log(meetingTuesdays);
On jsfiddle

Update: a further demonstration for you

Javascript
function getTuesdays(month, year) {
    var d = new Date(year, month, 1),
        tuesdays = [];

    d.setDate(d.getDate() + (9 - d.getDay()) % 7)
    while (d.getMonth() === month) {
        tuesdays.push(new Date(d.getTime()));
        d.setDate(d.getDate() + 7);
    }

    return tuesdays;
}

var today = new Date(),
    theseTuesdays = getTuesdays(today.getMonth(), today.getFullYear()),
    next;

theseTuesdays.some(function (tuesday, index) {
    if (index % 2 === 1 && tuesday > today) {
        next = tuesday;
        return true;
    }

    return false;
});

alert("Our next meeting is on : " + moment(next).format("MMMM Do YYYY"));
On jsfiddle

                        
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