javascript 从数组中获取最大和最小日期
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时间：2020-10-28 12:34:21 来源:igfitidea点击:
Get max and min dates from an array
提问by mpsbhat
I have sample data as:
我的样本数据为：
data = [
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
[10, 622, 782, 783, "2015-05-05"]
]
How can I get the max date value and min date value from the above data? The data may be not in sorted order.
如何从上述数据中获取最大日期值和最小日期值？数据可能没有排序。
回答by folkol
回答by shawn
var data = [
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
[10, 622, 782, 783, "2015-05-05"]
];
var minIdx = 0, maxIdx = 0;
for(var i = 0; i < data.length; i++) {
if(data[i][4] > data[maxIdx][4]) maxIdx = i;
if(data[i][4] < data[minIdx][4]) minIdx = i;
}
alert('Max: ' + maxIdx + ', ' + data[maxIdx][4]);
alert('Min: ' + minIdx + ', ' + data[minIdx][4]);
回答by vamsikrishnamannem
Working example Here
工作示例在这里
var a = [
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
[10, 622, 782, 783, "2015-05-05"]
];
var max = a[0][4];
var min = a[0][4];
for (var i = 0; i < a.length; i++) {
if (a[i][4] > max) {
max = a[i][4];
} else if (a[i][4] < min) {
min = a[i][4];
}
}
console.log(max);
console.log(min);
Reference from here
从这里参考
回答by Jay Patel
Use Math.min()and Math.max()function along with Array.prototypeproperty.
使用Math.min()和Math.max()函数以及Array.prototype属性。
JS: View jsFiddle
JS：查看jsFiddle
data = [
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
[10, 622, 782, 783, "2015-05-05"]
];
dates = []; // Array store only dates
for (i = 0; i < data.length; i++) {
dates.push(new Date(data[i][4]));
}
console.log(dates);
Array.prototype.max = function() { // find max dates
return Math.max.apply(null, this);
};
Array.prototype.min = function() { // find min dates
return Math.min.apply(null, this);
};
alert("Max: "+new Date(dates.max())+"\n\n"+
"Min: "+ new Date(dates.min()));
回答by nehal gala
var data = [
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
[10, 622, 782, 783, "2015-05-05"]
];
var dates = [];
for (var i=0;i<data.length;i++) {
dates.push(data[i][4]);
}
//sort the date
dates.sort(function(a,b){
// Turn your strings into dates, and then subtract them
// to get a value that is either negative, positive, or zero.
return new Date(b.date) - new Date(a.date);
});
// get min and max
var min_date = dates[0];
var max_date = dates.slice(-1)[0]
回答by PHP Worm...
Try this it will help you:
试试这个它会帮助你：
$(document).ready(function(){
var data = [
[10, 622, 782, 783, "2015-05-05"],
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
];
var dates = [];
var max_date='';
var min_date='';
$.each(data, function(k,v){
dates.push(v.pop());
dates.sort(function(a,b){
return new Date(a)- new Date(b);
});
max_date = dates[dates.length-1];
min_date = dates[0];
});
console.log('max_date : '+max_date);
console.log('min_date : '+min_date);
})
回答by PHP Worm...
var data = [
[1, 622, 782, 783, "2015-04-21"],
[2, 622, 782, 783, "2015-04-21"],
[3, 622, 782, 783, "2015-04-22"],
[4, 622, 782, 783, "2015-04-23"],
[5, 622, 782, 783, "2015-04-24"],
[6, 622, 782, 783, "2015-04-28"],
[7, 622, 782, 783, "2015-04-28"],
[8, 622, 782, 783, "2015-04-29"],
[9, 622, 782, 783, "2015-05-04"],
[10, 622, 782, 783, "2015-05-05"]
];
var di = 4,
maxd = mind = data[0][di];
var min = max = new Date(maxd).getTime(),
dt, i = data.length;
while (i--) {
dt = new Date(data[i][di]).getTime();
if (dt > max) {
max = dt;
maxd = data[i][di];
}
if (dt < min) {
min = dt;
mind = data[i][di];
}
}
console.log('min', mind, 'max', maxd);