我需要做这样的事情:当你选择 ex. “登录”,然后在输入文本中显示login
from $scope.logins
与password
JS:
$scope.logins = [{
"login" : "log",
"password" : "pass"
}]
HTML:
<select ng-model="type">
<option value="" disabled selected>Type</option>
<option>login</option>
<option>password</option>
</select>
<input class="form-control" name="type" placeholder="value" ng-model="value" style="margin-bottom:5px;">
感谢您提前的答复。
第一种方法
您可以使用$scope.$watch
为了完成这个
var app = angular.module('mainApp', []);
app.controller('mainController', function($scope) {
$scope.logins = [{
"login" : "log",
"password" : "pass"
}]
$scope.$watch('type',function(val){
$scope.value=$scope.logins[0][val];
});
});
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app="mainApp" ng-controller="mainController" class="col-xs-12">
<select ng-model="type">
<option value="" disabled selected>Type</option>
<option>login</option>
<option>password</option>
</select>
<input class="form-control" name="type" placeholder="value" ng-model="value" style="margin-bottom:5px;">
</div>
第二种方法
您可以根据输入中的类型直接绑定登录对象。 (js中不需要计算)
var app = angular.module('mainApp', []);
app.controller('mainController', function($scope) {
$scope.logins = [{
"login" : "log",
"password" : "pass"
}]
});
<script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.23/angular.min.js"></script>
<div ng-app="mainApp" ng-controller="mainController" class="col-xs-12">
<select ng-model="type">
<option value="" disabled selected>Type</option>
<option>login</option>
<option>password</option>
</select>
<input class="form-control" name="type" placeholder="value" ng-model="logins[0][type]" style="margin-bottom:5px;">
</div>
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