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考慮以下幾點：如何按關聯數組值排序（複雜數組結構）？
，我在一個HTML表格表示數組數據，如：
1）我怎麼會通過b1或b3數組排序？
我曾嘗試：
var o = {
"orgs": {
"655": {
"data": {
"cons": 30,
"b3ports": 0,
"b9": 2,
"b1": 25,
"b2": 14,
"b3": 10,
"ports": 0,
"rica": 30
},
"depth": 1,
"agents": [207072],
"orgunit_id": "TEAM00655",
"name": "TEAM00655: Jabba - Mooi River (Muhammad Jaffar)"
},
"853": {
"data": {
"cons": 356,
"b3ports": 1,
"b9": 8,
"b1": 283,
"b2": 122,
"b3": 77,
"ports": 1,
"rica": 356
},
"depth": 2,
"agents": [208162],
"orgunit_id": "TEAM00853",
"name": "TEAM00853: Jabba - Mooiriver (Bongiwe Gwala)"
},
"921": {
"data": {
"cons": 22,
"b3ports": 0,
"b9": 2,
"b1": 20,
"b2": 7,
"b3": 5,
"ports": 0,
"rica": 22
},
"depth": 1,
"agents": [210171, 212842],
"orgunit_id": "TEAM00921",
"name": "TEAM00921: Jabba - Nolwazi Zungu"
},
},
"agents": {
"207072": {
"name": "Bongiwe Gwala",
"oid": 655,
"depth": 1,
"aid": "A0207072",
"orgunit_id": "TEAM00655",
"data": {
"cons": 30,
"b3ports": 0,
"b9": 2,
"b1": 25,
"b2": 14,
"b3": 10,
"ports": 0,
"rica": 30
},
"aname": "A0207072: Bongiwe Gwala",
"oname": "TEAM00655: Jabba - Mooi River (Muhammad Jaffar)"
},
"208162": {
"name": "Nkosikhona MADLALA",
"oid": 853,
"depth": 2,
"aid": "A0208162",
"orgunit_id": "TEAM00853",
"data": {
"cons": 356,
"b3ports": 1,
"b9": 8,
"b1": 283,
"b2": 122,
"b3": 77,
"ports": 1,
"rica": 356
},
"aname": "A0208162: Nkosikhona MADLALA",
"oname": "TEAM00853: Jabba - Mooiriver (Bongiwe Gwala)"
},
"212842": {
"name": "SANELE KHUMALO",
"oid": 921,
"depth": 1,
"aid": "A0212842",
"orgunit_id": "TEAM00921",
"data": {
"cons": 22,
"b3ports": 0,
"b9": 2,
"b1": 20,
"b2": 7,
"b3": 5,
"ports": 0,
"rica": 22
},
"aname": "A0212842: SANELE KHUMALO",
"oname": "TEAM00921: Jabba - Nolwazi Zungu"
},
},
"orglist": [853, 655, 921],
}
function sort_data(data, sortby, asc) {
console.log(data);
if (asc == "asc") {
data.sort(function(a, b) {
a.sortby - b.sortby;
});
} else {
data.sort(function(a, b) {
a.sortby + b.sortby;
});
}
// update_data;
}
var a = sort_data(o, "b1", "asc");
console.log(a);<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>UPDATE：
我得到了排序，以感謝@NicolasAlbert。
現在我需要通過agents以及。它需要通過orgs先訂購由agents我曾嘗試：
if (asc == "desc") {
data.orglist.sort(function(a, b) {
a_org = data.orgs[a].data[sortby];
b_org = data.orgs[b].data[sortby];
a_agent = data.agents[a].data[sortby];
b_agent = data.agents[b].data[sortby];
// return d - c;
return b_org - a_org && a_agent - b_agent;
// data.agents[a].data[sortby] - data.agents[b].data[sortby]
});
} else {
data.orglist.sort(function(a, b) {
a_org = data.orgs[a].data[sortby];
b_org = data.orgs[b].data[sortby];
a_agent = data.agents[a].data[sortby];
b_agent = data.agents[b].data[sortby];
return a_org - b_org && a_agent - b_agent;
});
}
這不起作用怎麼過......
另一個更新
我已經修改我的代碼做：
data.orglist.sort(function(a, b) {
a_org = data.orgs[a].data[sortby];
b_org = data.orgs[b].data[sortby];
agents = data.orgs[b].agents.sort(function(a, b){
a_agent = data.agents[a].data[sortby];
b_agent = data.agents[b].data[sortby];
return b_agent - a_agent
});
return b_org - a_org && agents;
});
但是，這兩種排序和orgsagents在同一時間。
最後更新：
我得到它的工作，整理兩個orgs和agents，我不得不創建兩個排序功能：
function sort_org(data, sortby, order) {
/*
Sort the orgs
*/
var a_org, b_org;
if (order == "desc") {
data.orglist.sort(function(a, b) {
a_org = data.orgs[a].data[sortby];
b_org = data.orgs[b].data[sortby];
return b_org - a_org;
});
} else {
data.orglist.sort(function(a, b) {
a_org = data.orgs[a].data[sortby];
b_org = data.orgs[b].data[sortby];
return a_org - b_org;
});
}
}
function sort_agent(data, sortby, order) {
/*
Sort the agents
*/
var a_agent, b_agent;
if (order == "desc") {
for (var orgid in data.orglist){
data.orgs[data.orglist[orgid]].agents.sort(function(a, b){
a_agent = data.agents[a].data[sortby];
b_agent = data.agents[b].data[sortby];
return b_agent - a_agent
})
}
} else {
for (var orgid in data.orglist){
data.orgs[data.orglist[orgid]].agents.sort(function(a, b){
a_agent = data.agents[a].data[sortby];
b_agent = data.agents[b].data[sortby];
return a_agent - b_agent
})
}
}
}
然後我就打電話功能連續...即
sort_org(o, "b1", "asc");
sort_agent(o, "b1", "asc");
我希望這可以幫助別人......
謝謝您的回答...我只是reallised，我做了一個錯字，當我修改數組：'orgslist'，'agents'和'orgs'是數組中的鍵。 我已更新我的代碼... – Renier 2014-09-25 08:53:33
呃，實際上，順序可以修改，但只有瀏覽器引擎本身，據我所知。換句話說，屬性的順序沒有在規範中定義，儘管我發現對象屬性通常保持與它們定義的順序相同 - 儘管這不能保證。 – Agamemnus 2014-09-25 08:56:22
@Renier我用你的新數據格式更新了我的answser。 – 2014-09-25 09:00:05