如果已添加到tr中，则附加tr但不附加

Question

var data = [{
  "org_name": "General",
  "suborg_name": "General1"
}, {
  "org_name": "General",
  "suborg_name": "General2"
}, {
  "org_name": "General",
  "suborg_name": "General3"
}, {
  "org_name": "Fund",
  "suborg_name": "Fund1"
}, {
  "org_name": "Fund",
  "suborg_name": "Fund2"
}, {
  "org_name": "Fund",
  "suborg_name": "Fund3"
}]


var tr;
$.each(data, function(i, v) {
  tr = $('<tr class="org_name"/>');
  tr.append("<td id=''>" + v.org_name + "</td>");
  tr.append("<td id=''>" + v.suborg_name + "</td>");

  $("table").append(tr);
})

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
</table>

我想要的输出是：

General General1
        General2
        General3
Fund    Fund1
        Fund2
        Fund3

这怎么可能？

Answer 1

你可以简单地在循环中建立一个支票。也许与另一个数组/对象进行比较。

＆＃13;

＆＃13;

var data = [{
  "org_name": "General",
  "suborg_name": "General1"
}, {
  "org_name": "General",
  "suborg_name": "General2"
}, {
  "org_name": "General",
  "suborg_name": "General3"
}, {
  "org_name": "Fund",
  "suborg_name": "Fund1"
}, {
  "org_name": "Fund",
  "suborg_name": "Fund2"
}, {
  "org_name": "Fund",
  "suborg_name": "Fund3"
}];

var added = [];

$.each(data, function(i, v) {
  var name = " ";

  if( added.indexOf(v.org_name) == -1 ) {
      name = v.org_name;
      added.push(v.org_name);
  }
  
  var tr = $('<tr class="org_name"/>');
  tr.append("<td id=''>" + name + "</td>");
  tr.append("<td id=''>" + v.suborg_name + "</td>");

  $("table").append(tr);
})

＆＃13;

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
</table>

＆＃13;

＆＃13;

＆＃13;

Answer 2

点击此处jsfiddle

var tr;
var orgName=' ';
$.each(data, function(i, v) {
    if(orgName!=v.org_name){
  tr = $('<tr class="org_name"/>');
    tr.append("<td id=''>" + v.org_name + "</td>");
    tr.append("<td id=''>" + v.suborg_name + "</td>");
}else{
  tr = $('<tr class="org_name"/>');
    tr.append("<td id=''> </td>");
    tr.append("<td id=''>" + v.suborg_name + "</td>");
}
orgName = v.org_name;
  $("table").append(tr);
})

Answer 3

查找是否存在具有组织名称的现有tr。如果存在，则将空字符串放在第一个td中，并在相同的org tr之后追加。如果没有，则在第一个td中添加组织名称并附加到表。

var data = [{
      "org_name": "General",
      "suborg_name": "General1"
    }, {
      "org_name": "General",
      "suborg_name": "General2"
    }, {
      "org_name": "General",
      "suborg_name": "General3"
    }, {
      "org_name": "Fund",
      "suborg_name": "Fund1"
    }, {
      "org_name": "Fund",
      "suborg_name": "Fund2"
    }, {
      "org_name": "Fund",
      "suborg_name": "Fund3"
    }]


    var tr;

    $.each(data, function(i, v) {
      
      var rowExist = $('table').find('tr.' + v.org_name).length;
      tr = $('<tr class="org_name ' + v.org_name + '"/>');
      var orgName = '';
      if (rowExist == 0)
      {
        orgName = v.org_name;
        $('table').append(tr);
      }
      else 
      {
        $('table').find('tr.' + v.org_name).last().after(tr);
      }
      tr.append("<td id=''>" + orgName  + "</td>");
      tr.append("<td id=''>" + v.suborg_name + "</td>");  
    });

<script src="https://ajax.googleapis.com/ajax/libs/jquery/2.1.1/jquery.min.js"></script>
<table>
</table>