Question

我有一个JSON格式字符串，我希望像参数一样添加到帖子请求中。

我正在尝试执行请求，当我执行请求时，我从Web服务中收到此错误：

#0 /opt/assets/WS/ws.php(48): Slim\Slim::handleErrors(8, 'Trying to get p...', '/opt/assets/WS/...', 48, Array)
#1 [internal function]: {closure}()
#2 /opt/assets/WS/vendor/slim/slim/Slim/Route.php(468): call_user_func_array(Object(Closure), Array)
#3 /opt/assets/WS/vendor/slim/slim/Slim/Slim.php(1357): Slim\Route->dispatch()
#4 /opt/assets/WS/vendor/slim/slim/Slim/Middleware/Flash.php(85): Slim\Slim->call()
#5 /opt/assets/WS/vendor/slim/slim/Slim/Middleware/MethodOverride.php(92): Slim\Middleware\Flash->call()
#6 /opt/assets/WS/vendor/slim/slim/Slim/Middleware/PrettyExceptions.php(67): Slim\Middleware\MethodOverride->call()
#7 /opt/assets/WS/vendor/slim/slim/Slim/Slim.php(1302): Slim\Middleware\PrettyExceptions->call()
#8 /opt/assets/WS/ws.php(59): Slim\Slim->run()
#9 {main}

这就像请求不包含真正的JSON对象。我该如何改造它？我做了一些技巧，但似乎不起作用。那是我的剧本：

assets.controller('AfegirTipusActiusCtrl', function ($scope, $http){

                    // Camps formulari text pla
                 /*   $scope.nomAtribut = "<input type='text' name='firstname'>";
                    $scope.mida = "<input type='number' name='firstname'>";
                    $scope.obligatori = "<input type='checkbox' name='vehicle' value='yes'>";   */

                    // Construeix combo
                    $http.get('http://10.0.203.73/WS/ws.php/getCombo/1').success(function(data) {
                        $scope.options = data;
                    });

                    $scope.atributs = [];

                    $scope.addRow = function(){
                        var newRow = {
                            nomAtribut: "",
                            tipus: "",
                            mida: "",
                            prioritat: "",
                            obligatori: "",
                            observacions: ""
                        }
                        $scope.atributs.push(newRow);
                    }

                   /* $scope.addField = function() {
                        $scope.atributs.push(newRow);
                    };*/

                    $scope.prioritat = $scope.atributs.length;

                    // Envia atributs a WS
                    $scope.sendRow = function(){
                        var d = "{'nomAtribut': 'marca','tipus': 'String','mida': '15','prioritat': '1','obligatori': 'No'}";
                        $http.post("http://10.0.203.73/WS/ws.php/tipusactius/alta", angular.toJson(d)).success(function(data){
                            $scope.status = data;
                        })
                    }
            });

编辑：

我还添加了解析请求的php Slim函数：

 $app->post('/tipusactius/alta', function () use ($app) {
        $json = $app->request->getBody();
        header("Content-Type: application/json");
        $json_decode = json_decode($json); // $json_decode conte l'bjecte format a partir del json del request
        echo $json_decode->{'tipus'};

     //   $sql = "INSERt INTO atributs_actiu (nomAtribut, midaAtribut, tipus_actius_idtipus_actius, atributObligatori, ordre, tipusAtributs_idTipus, observacions
     //                  , combo_idcombo, combo_codi) values ('".$json_decode->{'nomAtribut'}."', '".$json_decode->{'mida'}."', 1, '0', 1, 1, 'atribut de prova', 1, 1)";
        $sql = "INSERt INTO atributs_actiu (nomAtribut, midaAtribut, tipus_actius_idtipus_actius, atributObligatori, ordre, tipusAtributs_idTipus, observacions
                       , combo_idcombo, combo_codi) values ('".$json_decode->{'nomAtribut'}."', '3', 1, '0', 1, 1, 'atribut de prova', 1, 1)";

        $mysqli=getDB();
        $result = $mysqli->query($sql);
 });

解决：就像@vidriduch告诉：

$scope.sendRow = function() {
    var d = {
        'nomAtribut': 'marca',
        'tipus': 'String',
        'mida': '15',
        'prioritat': '1',
        'obligatori': 'No'
    };

    $http.post("http://10.0.203.73/WS/ws.php/tipusactius/alta", angular.toJson(d)).success(function(data) {
        $scope.status = data;
})};

Answer 1

angular.toJson将输入序列化为JSON格式的字符串

https://docs.angularjs.org/api/ng/function/angular.toJson

在你的情况下，你正在序列化字符串......

您的代码可能如下所示：

$scope.sendRow = function() {
    var d = {
        'nomAtribut': 'marca',
        'tipus': 'String',
        'mida': '15',
        'prioritat': '1',
        'obligatori': 'No'
    };

    $http.post("http://10.0.203.73/WS/ws.php/tipusactius/alta", angular.toJson(d)).success(function(data) {
        $scope.status = data;
})};

Answer 2

如果您想将JSON字符串转换为JavaScript对象，则应使用angular.fromJson() method。

另外，请务必将'中的单引号"转换为双引号JSON：

// ' becomes " in your string
var d = '{"nomAtribut": "marca","tipus": "String","mida": "15","prioritat": "1","obligatori": "No"}';

$http.post("http://10.0.203.73/WS/ws.php/tipusactius/alta", angular.fromJson(d))
.success(function(data){
    $scope.status = data;
}

Answer 3

确保您的JSON参数未被任意引用锁定。

将JSON字符串分配给变量并直接传递它将在这里工作。

将JSON格式字符串转换为Angular中的实际对象

3 个答案: