鑑於數據:
var pagenumber=[1,2,3,4,5]; //only numbers
var zipcode=[77889,99887,33667,11122,44559]; // only numbers
var streetname=['Hillroad', 'Hillroad', 'Baghdad Street', 'Hongway', 'Chinatown']; //only letters
var roadnumber=[55,27,1,13,16]; //only numbers
第一,您需要使您的數據更容易管理
var data = pagenumber.map(function(itemValue, index) {
return {
pagenumber:itemValue, // == pagenumber[index]
zipcode:zipcode[index],
streetname:streetname[index],
roadnumber:roadnumber[index]
};
});
然後進行排序從另一個答案
data.sort(function(a, b) {
if (a.zipzode != b.zipcode) {
// numeric
return a.zipcode - b.zipcode;
}
if (a.streetname != b.streetname) {
// alpha
return a.streetname < b.streetname ? -1 : a.streetname > b.streetname ? 1 : 0;
}
if (a.roadnumber % 2 != b.roadnumber % 2) {
// even before odd
return b.roadnumber % 2 - a.roadnumber % 2;
}
// numeric
return a.roadnumber - b.roadnumber;
});
借款,可以簡化爲
data.sort(function(a, b) {
return (a.zipcode - b.zipcode) || (a.streetname < b.streetname ? -1 : a.streetname > b.streetname ? 1 : 0) || (b.roadnumber % 2 - a.roadnumber % 2) || (a.roadnumber - b.roadnumber);
});
就個人而言,我不使用的中間步驟,如果我能避免它...所以下面是等同於BOT地圖和排序在一個鏈接的命令
var sortedData = pagenumber.map(function(itemValue, index) {
return {
pagenumber:itemValue,
zipcode:zipcode[index],
streetname:streetname[index],
roadnumber:roadnumber[index]
};
}).sort(function(a, b) {
return (a.zipcode - b.zipcode) || (a.streetname < b.streetname ? -1 : a.streetname > b.streetname ? 1 : 0) || (b.roadnumber % 2 - a.roadnumber % 2) || (a.roadnumber - b.roadnumber);
});
你在排序任何這些數組方面迄今爲止做了什麼?請告訴我們 – HenryDev
我沒有排序它。不知道如何。但我會! – Heresh
如果每個數組中的每個索引引用單個實體...即索引0 = 1,77889,Hillroad,55等,所以當對一個數組進行排序時,您還需要重新排序其他數組,存儲我可以想到的數據! –