基于if条件的jQuery DataTable列

我们可以为 jQuery dataTables 列做 if 条件吗?根据变量值,我想决定显示/隐藏哪一列。

我试图通过创建一个对象并通过使用扩展运算符提取列将其传递给 DataTable() 函数来实现这一点,如下所示:

if (offer = "offer1") {
  const offerCols = {
    "columns": [
      {"data": "ioName", "class": "align-middle"},
      {"data": "salesPrice","class": "align-middle"},                    
      {"data": "offerPrice","class": "align-middle"},            
      {"data": "offerQty","class": "align-middle"}, 
    ],
  } 
} else {
  const offerCols = {
    "columns": [
      {"data": "ioName", "class": "align-middle"},
      {"data": "salesPrice","class": "align-middle"},                    
      {"data": "offerPrice","class": "align-middle"},            
      {"data": "offerQty","class": "align-middle"},           
      {"data": "validPeriod","class": "align-middle"}, 
    ], 
  }              
}              

var $table = $("#table").DataTable({  
  "ajax": '/view_offers.php',

  ...offerCols,

  "classes": {
    sLength: "text-left w-auto",
  }
})

但是,这对我不起作用。如何解决以下问题?

stack overflow jQuery DataTable columns based on if condition
原文答案
author avatar

接受的答案

如果您正在寻找用于查看数据的条件列,那么答案是 YES。但我不明白为什么传播运算符。

下面是基于条件的自定义列渲染的代码。

var dataSet = [
    [ "Tiger Nixon", "System Architect", "Edinburgh", "5421", "2011/04/25", "$320,800" ],
    [ "Garrett Winters", "Accountant", "Tokyo", "8422", "2011/07/25", "$170,750" ],
    [ "Ashton Cox", "Junior Technical Author", "San Francisco", "1562", "2009/01/12", "$86,000" ],
    [ "Cedric Kelly", "Senior Javascript Developer", "Edinburgh", "6224", "2012/03/29", "$433,060" ],
    [ "Airi Satou", "Accountant", "Tokyo", "5407", "2008/11/28", "$162,700" ],
    [ "Brielle Williamson", "Integration Specialist", "New York", "4804", "2012/12/02", "$372,000" ],
    [ "Herrod Chandler", "Sales Assistant", "San Francisco", "9608", "2012/08/06", "$137,500" ],
    [ "Rhona Davidson", "Integration Specialist", "Tokyo", "6200", "2010/10/14", "$327,900" ],
    [ "Colleen Hurst", "Javascript Developer", "San Francisco", "2360", "2009/09/15", "$205,500" ],
    [ "Sonya Frost", "Software Engineer", "Edinburgh", "1667", "2008/12/13", "$103,600" ],
    [ "Jena Gaines", "Office Manager", "London", "3814", "2008/12/19", "$90,560" ],
    [ "Quinn Flynn", "Support Lead", "Edinburgh", "9497", "2013/03/03", "$342,000" ],
    [ "Charde Marshall", "Regional Director", "San Francisco", "6741", "2008/10/16", "$470,600" ],
    [ "Haley Kennedy", "Senior Marketing Designer", "London", "3597", "2012/12/18", "$313,500" ],
    [ "Tatyana Fitzpatrick", "Regional Director", "London", "1965", "2010/03/17", "$385,750" ],
    [ "Michael Silva", "Marketing Designer", "London", "1581", "2012/11/27", "$198,500" ],
    [ "Paul Byrd", "Chief Financial Officer (CFO)", "New York", "3059", "2010/06/09", "$725,000" ],
    [ "Gloria Little", "Systems Administrator", "New York", "1721", "2009/04/10", "$237,500" ],
    [ "Bradley Greer", "Software Engineer", "London", "2558", "2012/10/13", "$132,000" ],
    [ "Dai Rios", "Personnel Lead", "Edinburgh", "2290", "2012/09/26", "$217,500" ],
    [ "Jenette Caldwell", "Development Lead", "New York", "1937", "2011/09/03", "$345,000" ],
    [ "Yuri Berry", "Chief Marketing Officer (CMO)", "New York", "6154", "2009/06/25", "$675,000" ],
    [ "Caesar Vance", "Pre-Sales Support", "New York", "8330", "2011/12/12", "$106,450" ],
    [ "Doris Wilder", "Sales Assistant", "Sydney", "3023", "2010/09/20", "$85,600" ],
    [ "Angelica Ramos", "Chief Executive Officer (CEO)", "London", "5797", "2009/10/09", "$1,200,000" ],
    [ "Gavin Joyce", "Developer", "Edinburgh", "8822", "2010/12/22", "$92,575" ],
    [ "Jennifer Chang", "Regional Director", "Singapore", "9239", "2010/11/14", "$357,650" ],
    [ "Brenden Wagner", "Software Engineer", "San Francisco", "1314", "2011/06/07", "$206,850" ],
    [ "Fiona Green", "Chief Operating Officer (COO)", "San Francisco", "2947", "2010/03/11", "$850,000" ],
    [ "Shou Itou", "Regional Marketing", "Tokyo", "8899", "2011/08/14", "$163,000" ],
    [ "Michelle House", "Integration Specialist", "Sydney", "2769", "2011/06/02", "$95,400" ],
    [ "Suki Burks", "Developer", "London", "6832", "2009/10/22", "$114,500" ],
    [ "Prescott Bartlett", "Technical Author", "London", "3606", "2011/05/07", "$145,000" ],
    [ "Gavin Cortez", "Team Leader", "San Francisco", "2860", "2008/10/26", "$235,500" ],
    [ "Martena Mccray", "Post-Sales support", "Edinburgh", "8240", "2011/03/09", "$324,050" ],
    [ "Unity Butler", "Marketing Designer", "San Francisco", "5384", "2009/12/09", "$85,675" ]
];
function getConditionalColumn(isAdmin){
  var columns= [];
  if(isAdmin){
    columns.push({ title: "Name" }) 
      columns.push({ title: "Position" }) 
      columns.push({ title: "Office" })
        columns.push({ title: "Extn" }) 
      columns.push({ title: "Start date" }) 
      columns.push({ title: "Salary" })
  }
  else{
       columns.push({ title: "Name" }) 
      columns.push({ title: "Position" }) 
      columns.push({ title: "Office" })
  }
  return columns;
} 

$(document).ready(function() {

    $('#example').DataTable( {
        data: dataSet,
        columns:getConditionalColumn(false)//toggle here to see conditional column , put true for admin view and false for user view
    });
} );
<link href="https://cdn.datatables.net/1.10.25/css/jquery.dataTables.min.css" rel="stylesheet"/>
<script src="https://code.jquery.com/jquery-3.5.1.js"></script>
<script src="https://cdn.datatables.net/1.10.25/js/jquery.dataTables.min.js"></script>
<table id="example" class="display" width="100%"></table>

答案:

作者头像

columns 中的对象有一个接受 visibleboolean 属性。因此,如果您有其他选项可以在列上设置,那么更简洁的解决方案可能是:

$('#example').DataTable({
    columns: [
        {
            data: 'name',
            title: 'Name'
        },
        {
            data: 'salary',
            title: 'Salary',
            className: 'align-end',
            visible: isAdmin
        }
    ]
});

如果你想指定多个条件,只需有一个返回布尔值的函数:

{
    data: 'salary',
    title: 'Salary',
    className: 'align-end',
    visible: setConditions()
}
function setConditions() {
    if(condition1) return true;

    if(condition2) return true;

    return false;
}