首頁猿問按 2 個不同的屬性對數組進行排序/分組

按 2 個不同的屬性對數組進行排序/分組

JavaScript

皈依舞 2022-12-22 15:47:58

我有一個基于其“程序”屬性的分組/排序數組，這很棒。但是現在我需要根據該分組內的不同屬性（可交付）進行排序，這可能嗎？如果是這樣，我怎樣才能做到這一點？這是我的桌子的照片?？纯此侨绾伟闯绦蚪M織的？在程序分組中，我還想根據可交付項目進行排序/分組，因為每個程序中會有兩個以上的方法。此外，如果工作量不大，我也希望使這些行可點擊（展開/折疊），這樣表格在加載后不會有 100 行長。這是我的預期輸出：預期結果+------------+----------------------+-----------+------------+--------------+---------------------+| Program | Deliverable | Date | Approved | Notes | To |+------------+----------------------+-----------+------------+--------------+---------------------+| Program 1 | |+------------+----------------------+-----------+------------+--------------+---------------------+| | Monthly Status Report| |+------------+----------------------+-----------+------------+--------------+---------------------+| | | 05/10/2020| Yes | Example Notes| [email protected] |+------------+----------------------+-----------+------------+--------------+---------------------+| | | 03/30/2020| No | Example Notes| [email protected] |+------------+----------------------+-----------+------------+--------------+---------------------+| | Meeting Minutes | |+------------+----------------------+-----------+------------+--------------+---------------------+

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2 回答

人到中年有點甜

TA貢獻1895條經驗獲得超7個贊

只需對您的結果進行排序。

Object.keys(sortedObj).forEach(key => {

tableContent += "<tr>";

tableContent += "<td>" + key + "</td>";

tableContent += "</tr>";

sortedObj[key].sort((a,b)=>{

if (a.Deliverable > b.Deliverable) return 1;

if (a.Deliverable < b.Deliverable) return -1;

return 0;

}).forEach(obj => {

tableContent += "<tr>";

tableContent += "<td> </td>";

tableContent += "<td>" + obj.To + "</td>";

tableContent += "<td>" + obj.Date + "</td>";

tableContent += "<td>" + obj.Approved + "</td>";

tableContent += "<td>" + obj.Notes + "</td>";

tableContent += "<td>" + obj.Deliverable + "</td>";

tableContent += "</tr>";

});

反對回復 2022-12-22

牛魔王的故事

TA貢獻1830條經驗獲得超3個贊

這是代碼：

var sortedObj = {}

objItems.forEach(item => {

var program = item.Program;

var deliverable = item.Deliverable;

delete (item.Program); //remove this line to keep the program in the item data

delete (item.Deliverable); //remove this line to keep the deliverable in the item data

if(!sortedObj[program]){

sortedObj[program] = {};

}

if(!sortedObj[program][deliverable]){

sortedObj[program][deliverable] = [];

}

sortedObj[program][deliverable].push(item);

});

Object.keys(sortedObj).forEach((program, index) => {

tableContent += "<tr class='breakrow'>";

tableContent += "<td>" + program + "</td>";

tableContent += "</tr>";

Object.keys(sortedObj[program]).forEach((deliverable, index) => {

tableContent += "<tr class='datarow'>";

tableContent += "<td> </td>";

tableContent += "<td>" + deliverable + "</td>";

tableContent += "</tr>";

sortedObj[program][deliverable].forEach(obj => {

tableContent += "<tr class='subdatarow'>";

tableContent += "<td> </td>";

tableContent += "<td>" + obj.To + "</td>";

tableContent += "<td>" + obj.Date + "</td>";

tableContent += "<td>" + obj.Approved + "</td>";

tableContent += "<td>" + obj.Notes + "</td>";

tableContent += "</tr>";

});

$("#deliverables").append(tableContent);

$('.datarow').hide();

$('.subdatarow').hide();

$('#deliverablesTable').on('click', 'tr.breakrow', function(){

console.log('hello');

$(this).nextUntil('tr.breakrow').slideToggle(200);

});

$(document).on('click','#deliverablesTable tr.datarow', function(){

$(this).nextUntil('tr.breakrow, tr.datarow').slideToggle(200);

});

})

反對回復 2022-12-22

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按 2 個不同的屬性對數組進行排序/分組

按 2 個不同的屬性對數組進行排序/分組

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