搜索某个人时是否可以显示与该人有关的所有资料。并且还必须选择多个人。
这是我的示例代码.....
<!doctype html>
<html ng-app="plunker">
<head>
<script src="http://ajax.googleapis.com/ajax/libs/angularjs/1.0.5/angular.js"></script>
<script src="http://angular-ui.github.com/bootstrap/ui-bootstrap-tpls-0.2.0.js"></script>
<script src="example.js"></script>
<link href="//netdna.bootstrapcdn.com/twitter-bootstrap/2.3.1/css/bootstrap-combined.min.css" rel="stylesheet">
</head>
<body>
<div class='container-fluid' ng-controller="TypeaheadCtrl">
<input type="text" ng-model="selectedStuff" typeahead="stuff as stuff.desc for stuff in stuffs | filter:{name: $viewValue, desc: $viewValue}"/>
<span>{{selectedStuff.name}}</span>
<span>{{selectedStuff.desc}}</span>
</div>
</body>
</html>
和js代码.....
angular.module('plunker', ['ui.bootstrap']);
function TypeaheadCtrl($scope) {
$scope.stuffs= [
{
"name":"thing1",
"desc":"this is the first thing",
"title": "this is the first title"
},
{
"name":"thing2",
"desc":"this is the second thing",
"title": "this is the second title"
}
]
}
在这里，当我搜索时，它仅显示名称或desc或title，但是我需要显示整个三个属性而不是一个属性。
我需要选择多个类似的标签。
请帮助解决此问题。
提前致谢
最佳答案
我认为您需要将html更改为
<input type="text" ng-model="selectedStuff"
typeahead="stuff as stuff.name + ' ' +stuff.desc + ' ' + stuff.title for stuff in stuffs | filter:{name: $viewValue, desc: $viewValue}"/>
关于javascript - Angular ui-bootstrap提前选择多个，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/29789542/