如何在jquery中对json值进行分组和排序'每个'环

时间:2018-04-25 01:15:48

标签: jquery json ajax

我试图将下面json中的数据从美国代表和参议员拉到下拉菜单中...理想情况下按" Sen"和#34; Rep"然后按字母顺序排列。到目前为止,下面的代码正在努力让所有成员进入下拉列表,但不进行分组或按字母顺序排列。会喜欢任何想法,谢谢!

<script>


$.ajax({
url:'https://theunitedstates.io/congress-legislators/legislators- 
current.json',
dataType: 'json',
type:'get',
cache:false,
success:function(data) {
$(data).each(function(index, value) {

     $("#members").append("<option>" + value.terms[value.terms.length- 
1].type + " " + value.name.first + " " + value.name.last + "</option>");
    });
}

});



</script>

<form>

    <select id="members" style="text-transform:uppercase;">
        <option selected="selected">Select...</option>
    </select>
</form>

1 个答案:

答案 0 :(得分:1)

  

HTML <optgroup>元素会在<select>元素中创建一组选项。

https://developer.mozilla.org/en-US/docs/Web/HTML/Element/optgroup

    <select id="members" style="text-transform:uppercase;">
        <optgroup id="senators" label="Senators"></optgroup>
        <optgroup id="reps" label="US Reps"></optgroup>
    </select>
  

sort()方法对数组中的元素进行排序并返回数组。

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Global_Objects/Array/sort

所以,如果您想按姓氏排序,可以试试这个:

    var reps = [];
    var senators = [];

    $.ajax({
        url: 'https://theunitedstates.io/congress-legislators/legislators-current.json',
        dataType: 'json',
        type: 'get',
        cache: false,
        success: function (data) {

            $(data).each(function (index, value) {

                if (value.terms[value.terms.length - 1].type == 'sen') {
                    // add to senators
                    senators.push(value.name.last + ', ' + value.name.first);
                } else {
                    // add to reps
                    reps.push(value.name.last + ', ' + value.name.first)
                }

            });

            senators.sort();
            reps.sort();

            senators.forEach(function (val) {
                $("#senators").append("<option>" + "SEN. " + " " + val + "</option>");
            });

            reps.forEach(function (val) {
                $("#reps").append("<option>" + "REP. " + val  + "</option>");
            })

        }

    });

JSFiddle: https://jsfiddle.net/eb1ayera/