在“選擇”選項中首先包含 select 值。

Question

Select 首先我希望值來。 我該怎么做？

                $.each(data.departmentList, function (key, value) {
                    $('#searchDepartmentName')
                        .append($("<option></option>")
                            .attr("value", value.mudurlukId)
                            .text(value.mudurlukAdi));
                });
               
                $('#searchDepartmentName').append(optionData);
            },  ```

Answer 1

您可以相應地設置選定的屬性。

 const data = { departmentList: [ {mudurlukId: "1", mudurlukAdi: "test1"}, {mudurlukId: "2", mudurlukAdi: "test2"} ] }; $('#searchDepartmentName').html(data.departmentList.map((v, i) => { const selected = i === 0? " selected": ""; return `<option ${selected} value=${v.mudurlukId}>${v.mudurlukAdi}</option>`; }));

 <head> <script src="https://code.jquery.com/jquery-3.5.1.slim.min.js" integrity="sha256-4+XzXVhsDmqanXGHaHvgh1gMQKX40OUvDEBTu8JcmNs=" crossorigin="anonymous"></script> </head> <body> <select id="searchDepartmentName"> </select> </body>

Answer 2

這將設置要選擇的第一個選項：

$.each(data.departmentList, function (key, value) {
  const select = $('#searchDepartmentName');
  const elem = $("<option></option>")
    .attr("value", value.mudurlukId)
    .text(value.mudurlukAdi);
  
  select.append(elem)
  
  if (elem.is(":first-child")) {
    select.val(value.mudurlukId);
  }
});

在“選擇”選項中首先包含 select 值。

問題描述

2 個解決方案

解決方案1
0 2021-01-11 05:43:36

解決方案2
0 2021-01-11 05:46:16

在“選擇”選項中首先包含 select 值。

問題描述

2 個解決方案

解決方案1 0 2021-01-11 05:43:36

解決方案2 0 2021-01-11 05:46:16

解決方案1
0 2021-01-11 05:43:36

解決方案2
0 2021-01-11 05:46:16