Javascript 如何在selectize plugin from controller中选择多个值
如何添加所选的几个选项 我正在使用这个插件: 我有一个对象,如下图所示: 这是我的客户脚本Javascript 如何在selectize plugin from controller中选择多个值,javascript,arrays,angularjs,object,selectize.js,Javascript,Arrays,Angularjs,Object,Selectize.js,如何添加所选的几个选项 我正在使用这个插件: 我有一个对象,如下图所示: 这是我的客户脚本 //skills $scope.$selectSkill = $('#selectSkill').selectize({ valueField: 'id', labelField: 'name', searchField: 'name', placeholder: "Select Skills", options: $scope.data.skills,
//skills
$scope.$selectSkill = $('#selectSkill').selectize({
valueField: 'id',
labelField: 'name',
searchField: 'name',
placeholder: "Select Skills",
options: $scope.data.skills,
create: false,
sortField: {
field: 'name',
direction: 'asc'
}
});
这是我的HTML
<select id="selectSkill"
name="selectSkill"
ng-model="selectSkill"
multiple required> </select>
我希望这两个被选中,如下图所示
我该怎么做
我试过这样的方法:
var $select = $("#selectSkill").selectize();
var yourDefaultIds = [slectedSkills.skills]; //(those 2 vlaues)
selectize.setValue(yourDefaultIds);;
这应该行得通,我已经测试过了。
将这两个值转换为数组,然后设置,而不是设置为字符串
var $select = $("#selectSkill").selectize();
var yourDefaultIds = [slectedSkills.skills]; //(those 2 vlaues)
selectize.setValue(yourDefaultIds);;
//populate skills starts here
var templateSkills = finTempObj.skills;
var str_array_skills = templateSkills.split(',');
var $select = $("#selectSkill").selectize();
var selectize = $select[0].selectize;
selectize.setValue(str_array_skills);
selectize.refreshOptions();
//populate skills ends here