Question

我希望从serice页面到php页面有多重价值。我想传递数据＆＃39;，＆＃39; albimg＆＃39;，＆＃39; albvideo＆＃39;到php page.now我发现错误，如图所示。

var deferred = $q.defer();
            data.pagename = "create_album";
            $http.post('js/data/album.php', {action:{"data":data,"albimg":albimg,"albvideo":albvideo}})
                    .success(function (data, status, headers, config)
                    {
                        console.log(status + ' - ' + action);
                        deferred.resolve(action);
                    })
                    .error(function (action, status, headers, config)
                    {
                        deferred.reject(action);
                        console.log('error');
                    });

            return deferred.promise;
php page:
$postdata = file_get_contents("php://input",true);
$request = json_decode($postdata);
$now = date('Y-m-d H:i:s');
echo $sql="INSERT INTO `$prefix.album` (CONTENT_VALUES,CreatedTime)VALUES('$postdata','$now')";

Answer 1

你可以使用param属性：像这样;

var data = {"data":data,"albimg":albimg,"albvideo":albvideo};

$http({ url: "js/data/album.php",    method: "POST",    params: data    })

Answer 2

看看这个例子：

 function PhoneListCtrl($scope, phones) {
      $scope.phones = phones;
      $scope.orderProp = 'age';
    }

    PhoneListCtrl.resolve = {
      phones: function(Phone, $q) {
        var deferred = $q.defer();



 deferred.reject();
    Phone.query(function(successData) {
            deferred.resolve(successData); 
    }, function(errorData) {
            deferred.reject();
    });
    return deferred.promise;
  },
  delay: function($q, $defer) {
    var delay = $q.defer();
    $defer(delay.resolve, 1000);
    return delay.promise;
  }
}

我认为您忘了在函数中添加$defer。你真的需要异步吗？ Beacouse如果您使用$q它需要它。但是如果你只想将数据发送到php文件最简单的方法来使用这样的东西：

angular.module('httpExample', [])
.controller('FetchController', ['$scope', '$http', '$templateCache',
  function($scope, $http, $templateCache) {
    $scope.method = 'Post';
    $scope.url = 'http-hello.php';

    $scope.fetch = function() {
      $scope.code = null;
      $scope.response = null;

      $http({method: $scope.method, url: $scope.url, cache: $templateCache}).
        then(function(response) {
          $scope.status = response.status;
          $scope.data = response.data;
        }, function(response) {
          $scope.data = response.data || "Request failed";
          $scope.status = response.status;
      });
    };

    $scope.updateModel = function(method, url) {
      $scope.method = method;
      $scope.url = url;
    };
  }]);

Answer 3

成功回调中没有定义动作参数。

您的代码

private void nextpageintent(String photoUri){
    Bundle bundle = new Bundle();
    bundle.putString("photo", photoUri);
    picture picture = new picture();
    picture .setArguments(bundle);
}

应该是

.success(function (data, status, headers, config) // No Action here
    {
        console.log(status + ' - ' + action); // This line gives error
        deferred.resolve(action);
    })

将多个值从js页面传递到php页面

3 个答案: