我对表单中的所有字段都有字形眼。如果用户单击 glyphicon-eye-open,那么它将更改为 glyphicon-eye-close,然后我将该特定字段名称推送到数组中。
在我的 JSON 响应中,我获取了隐藏字段值,但如何使用该值并调用精确的 glyphicon-eye。
JSON 响应:
{
"response": {
"status": {
"code": "0",
"message": "Success"
},
"service": {
"servicetype": "4",
"functiontype": "1005"
},
"data": {
"session_id": "372",
"roles": [
{
"hiddenfields": [
{
"fname": "firstname",
"fblink": "fblink",
"country": "country",
"martialStatus": "martialStatus"
}
]
}
]
}
}
}
控制器 :
$scope.user = {
fname: "firstname",
lname: "lastname",
dob: "dob",
gender: "gender",
country: "country",
state: "state",
city: "city",
pincode: "pincode",
martialStatus: "martialStatus",
emailId: "emailId",
mobile: "mobile",
qualification: "qualification",
fblink: "fblink"
};
$scope.allow = {};
$scope.users = [];
$scope.push = function(){
$scope.users = [];
var user = {},
allow = $scope.allow;
console.log(allow);
Object.keys(allow).forEach(function(key){
allow[key] ? user[key] = $scope.user[key] : null;
});
$scope.users.push(user);
}
HTML :
<a class="menu-toggle" class="btn btn-default" ng-model="allow.fname"><i class="glyphicon" ng-class="{'glyphicon-eye-open':allow.fname, 'glyphicon-eye-close':!allow.fname}" ng-click="push(allow.fname = allow.fname?false:true)"></i></a>
如果字段值在数组中,那么我需要显示 glyphicon-eye-close。