Question

我正在使用Spring + AngularJS。如何在angularjs中发送后端下拉值？

HTML：

<div class="col-md-4">
    <select  ng-model="algoList">
        <option ng-repeat="algo in algoList" value="{{algo}}">{{algo}}</option>
    </select>
    <button class="btn btn-primary" ng-click="applyAlgo()">
        <span class="glyphicon glyphicon-right"></span> Apply
    </button>
</div>

JS：

$scope.algoName = "";
$scope.algoList= ["DC_CleaningOperations","FilerOperation"];
$scope.applyAlgo = function() {
var response =  $http.post('/idw-dv/getAlgo', $scope.algoName);
response.success(function(data, status, headers, config) {
    swal("Following algorithm is applied successfully. :",data,"success");
    console.log(data);
});
response.error(function(data, status, headers, config) {
    swal( "Exception details: " + JSON.stringify({data: data}));
});

控制器：

@RequestMapping(value = "/getAlgo", method = RequestMethod.POST)
@ResponseBody
public String getAlgorithm(@RequestBody String algoName) {
    return algoName;
}

Answer 1

所以这就是问题，你为select字段定义的模型是错误的，最终会引发你的问题。

如果您将ng-model更改为“algoName”，则下拉列表中的所选值现在将存储在algoName中。

此值现在将作为String传输到Spring Controller。

希望这有效......

    <div class="col-md-4">
            <select  ng-model="algoName">
                    <option ng-repeat="algo in algoList" value="{{algo}}">{{algo}}</option>
            </select>

            <button class="btn btn-primary" ng-click="applyAlgo()">
                    <span class="glyphicon glyphicon-right"></span> Apply
            </button>
    </div>

如何在angularjs中发送后端下拉值？

1 个答案: