Kendo UI下拉列表未显示正确的值

问题描述 投票：0回答：1

我有这样的情况，我的下拉列表据说它应该根据我的userPropertyID=3显示GUEST HOUSE但不知何故它仍然没有在下拉列表中显示正确的值。需要你的帮助，谢谢你的时间。

  var userPropertyID = 3;

  var propertyTBL = [
  {"propertyID":"1","propertyName":"HOTEL"},
  {"propertyID":"2","propertyName":"RESORT"},
  {"propertyID":"3","propertyName":"GUEST HOUSE"},
  {"propertyID":"4","propertyName":"HOME"},
  {"propertyID":"5","propertyName":"MOTEL"}];

  
  var dropdownlist = $("#dropdownlist").kendoDropDownList({
    dataTextField: "propertyName",
    dataValueField: "propertyID",
    dataSource: propertyTBL
  });
  dropdownlist.select(userPropertyID);
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>Untitled</title>

	<link rel="stylesheet" href="https://kendo.cdn.telerik.com/2019.1.115/styles/kendo.common.min.css" />
    <link rel="stylesheet" href="https://kendo.cdn.telerik.com/2019.1.115/styles/kendo.silver.min.css" />
    <link rel="stylesheet" href="https://kendo.cdn.telerik.com/2019.1.115/styles/kendo.silver.mobile.min.css" />

    <script src="https://kendo.cdn.telerik.com/2019.1.115/js/jquery.min.js"></script>
    <script src="https://kendo.cdn.telerik.com/2019.1.115/js/kendo.all.min.js"></script>
    <style>*{font-family:Arial;font-size:11px;}</style>
<body>
Property Name:<input id="dropdownlist"/>

</body>
</html>
kendo-ui kendo-grid kendo-dropdown kendo-datasource
1个回答
2
投票

发生这种情况是因为您没有将正确的kendo小部件分配给变量。所以它不知道剑道的功能。

正确的方法是这样（通过使用.data（'kendoDropDownList'））

var dropdownlist = $("#dropdownlist").data('kendoDropDownList');

所以在你的情况下，你需要改变你的代码部分。

var dropdownlist = $("#dropdownlist").kendoDropDownList({
    dataTextField: "propertyName",
    dataValueField: "propertyID",
    dataSource: propertyTBL
}).data('kendoDropDownList');

但请记住，索引从0开始（我在你的代码中看到，在索引3处，你再次将值归因于索引0，所以不要对此感到困惑）。

现在调用.value（）函数将起作用

dropdownlist.select(userPropertyID - 1);

最后，如果您在创建小部件后立即在实际应用程序中设置值，则可以使用value属性进行设置

var dropdownlist = $("#dropdownlist").kendoDropDownList({
    dataTextField: "propertyName",
    dataValueField: "propertyID",
    dataSource: propertyTBL,
    value: userPropertyID
}).data('kendoDropDownList');

编辑：要完成它，这是您修改后的代码段

  var userPropertyID = 3;

  var propertyTBL = [
  {"propertyID":"1","propertyName":"HOTEL"},
  {"propertyID":"2","propertyName":"RESORT"},
  {"propertyID":"3","propertyName":"GUEST HOUSE"},
  {"propertyID":"4","propertyName":"HOME"},
  {"propertyID":"5","propertyName":"MOTEL"}];

  
  var dropdownlist = $("#dropdownlist").kendoDropDownList({
    dataTextField: "propertyName",
    dataValueField: "propertyID",
    dataSource: propertyTBL
  }).data('kendoDropDownList');
  dropdownlist.select(userPropertyID - 1);
<!DOCTYPE html>
<html>
<head>
  <meta charset="utf-8">
  <title>Untitled</title>

	<link rel="stylesheet" href="https://kendo.cdn.telerik.com/2019.1.115/styles/kendo.common.min.css" />
    <link rel="stylesheet" href="https://kendo.cdn.telerik.com/2019.1.115/styles/kendo.silver.min.css" />
    <link rel="stylesheet" href="https://kendo.cdn.telerik.com/2019.1.115/styles/kendo.silver.mobile.min.css" />

    <script src="https://kendo.cdn.telerik.com/2019.1.115/js/jquery.min.js"></script>
    <script src="https://kendo.cdn.telerik.com/2019.1.115/js/kendo.all.min.js"></script>
    <style>*{font-family:Arial;font-size:11px;}</style>
<body>
Property Name:<input id="dropdownlist"/>

</body>
</html>
最新问题
© www.soinside.com 2019 - 2024. All rights reserved.