Question

我使用angular将数据发布到PHP JSON webservice但无法正确获取

这是我的.http方法代码

    app.controller('AddCustomer',['$http','$scope',function($http,$scope){
$scope.freshRequest = true;
$scope.addCustomer=function(){
    var data_to_send={};
    data_to_send.name=$scope.name;
    alert(data_to_send.name);
    $http({
        method: 'POST',
        url: '../service/add_new_Customer.php', 
        data: data_to_send,
        headers: {'Content-Type': 'application/x-www-form-urlencoded'}
        }).success(function(returnedData){
        $scope.freshRequest = false;
        $scope.response = returnedData.response;
        $scope.accountNumber= returnedData.accountNumber;
        $scope.updatedBalance= returnedData.updatedBalance;
        $scope.transactionSuccessful = returnedData.transactionSuccessful;          
    }).error(function (data, status, headers, config) {
        $scope.status = status + ' ' + headers;
        console.log($scope.status)});



};}]);

下面是我用来调用addCustomer（）方法的htm表单

<form class="form-horizontal" novalidate ng-controller="AddCustomer as add" ng-submit="addCustomer()">
<input id="name" name="name" type="text" placeholder="" class="form-control input-md" required="" ng-model="name">
<input id="fname" name="fname" type="text" placeholder="" class="form-control input-md" required="" ng-model="fname">
 <button id="button1id" type="submit" name="button1id" class="btn btn-primary">Add</button>

当我正在执行$ _REQUEST [＆＃39; name＆＃39;]或$ _POST [＆＃39; name＆＃39;]时，我的php为字段名称变为null;

下面是php代码

<?php
class Summary{
   public $response= "Account Could not be added invalid data";
   public $transactionSuccessful= "";
   public $updatedBalance= "";
   public $accountNumber= "";
}
$e = new Summary();
$final_res =json_encode($jsonObj) ;
if( $_REQUEST['name'] ){
$e->response =  $_REQUEST['name'];
}
else{
$e->response =  "Account Could not be added invalid data";
}
 $e->transactionSuccessful= true;
$e->updatedBalance=  $_POST['name'];
$e->accountNumber=  $_POST['name'];  
echo json_encode($e);

Answer 1

首先，您需要将发布的数据解码为json，然后才能访问该对象的属性。 file_get_contents("php://input")允许您阅读原始发布的数据。这就是你如何得名。

$postdata = file_get_contents("php://input");
$request = json_decode($postdata);
@$name = $request->name;

无法从php web服务上的http.post获取数据

1 个答案: