我有一個項目,我使用查詢參數傳遞到後端搜索。如何檢查查詢參數是否爲空的AngularJS形式
它們使用AngularJS中的$ http.get方法傳遞。
一些參數不是搜索所必需的,所以我希望它們在空的時候不在url中。
如何獲得此功能?
下面是我的代碼:
<!DOCTYPE html>
<html lang="en" ng-app = "searchApp">
<script type="text/javascript">
var searchApp = angular.module("searchApp", []);
searchApp.controller("searchListingsCtrl", ['$scope', '$http', function($scope, $http) {
$scope.searchListings = function(){
if ($scope.checkIn == '') {}
var searchListingObj = {
checkIn : $scope.checkIn,
checkOut : $scope.checkOut,
country : $scope.country,
city : $scope.city,
state : $scope.state,
accommodates : $scope.accommodates,
}
var res = $http.get('http://www.localhost:8080/messenger/webapi/listings', searchListingObj);
res.success(function(data, status, headers, config) {
$scope.message = data;
});
res.error(function(data, status, headers, config) {
alert("failure message: " + JSON.stringify({data: data}));
});
};
}]);
</script>
<body ng-controller="searchListingsCtrl">
<form action="/listings" name="listings" ng-submit="searchListings()">
<input type="text" name="neighborhood" placeholder="Neighborhood:" ng-model="state">
<input type="text" name="town" placeholder="Town:" ng-model="city">
<input type="text" name="country" placeholder="Country:" ng-model="country">
People:<select class="peopleSelect" placeholder="People:" ng-model="accommodates"> </select>
<input type="text" id="arrival" name="arrival" value="Arrival" ng-model="checkIn">
<input type="text" id="departure" name="depart" value="Departure" ng-model="checkOut">
<input type="submit" name="submit" value="Search" id="Search_submit">
</form>
</body>
無法理解您的問題。你能否詳細說明一下? –
我重新編寫了它,請再次檢查問題,然後詢問我的更多信息。 –
你的代碼現在是什麼樣子? –