后编辑表单数据，并使用angularjs将其获取到控制器中

Question

我想使用angularjs 将编辑表单数据发布到代码点火器控制器

这是我的angularjs控制器，我在这里收集和发布表单数据

$scope.editlist = function(id){
$http({
method:'post',       
url:'http://localhost/Angular_demo/admin/index.php/welcome/get_edit_data/'+ 
id
}).then(function success(response){
$scope.sid = parseInt(response.data[0]['id']);
$scope.firstname = response.data[0]["first_name"];
$scope.lastname = response.data[0]['last_name'];
$scope.email = response.data[0]['email'];                        
}); 
}
$scope.saveEdit = function(id,firstname,lastname,email){
$http({
method:'post',
data:'id='+ id + '&first_name='+firstname+ 
'&last_name='+lastname+'&email='+email,
url:'http://localhost/Angular_demo/admin/index.php/welcome/update_data'                         
}).then(function success(response){
console.log(response.data);
});
}

我绑定和发布表单数据的视图

<form name="editItem" class="form-horizontal">
<input ng-model="sid" type="hidden" placeholder="Name" name="name" 
class="form-control" />          
<div class="form-group"> 
<label for="inputEmail3" class="col-sm-2 control-label">First 
Name</label>
<div class="col-sm-10"> 
<input type="text" class="form-control" ng-model="firstname"  
id="inputEmail3" placeholder="First name">
</div> 
</div>
<div class="form-group"> 
<label for="inputEmail3" class="col-sm-2 control-label">Last 
Name</label>
<div class="col-sm-10"> 
<input type="text" class="form-control" ng-model="lastname" 
id="inputEmail3" placeholder="Last Name">
</div> 
</div>
<div class="form-group"> 
<label for="inputEmail3" class="col-sm-2 control- 
label">Email</label>
<div class="col-sm-10"> 
<input type="text" class="form-control" ng-model="email" 
id="inputEmail3" placeholder="Email">
</div> 
</div>        
<div class="form-group"> 
<div class="col-sm-offset-2 col-sm-10">
<button type="button" class="btn btn-default" data- 
dismiss="modal">Close</button>
<button type="submit" ng-disabled="editdata.$invalid" 
class="btn btn-primary create-crud" data-dismiss="modal" ng- 
click="saveEdit(sid,firstname,lastname,email)">Submit</button>                  
</div> 
</div>
</form>

这是我的代码点火器控制器，我想在这里获得表单数据

public function update_data(){
$requests = json_decode(file_get_contents('php://input'), TRUE);
$ids= array('id' =>$requests['id']);
echo json_encode($ids);
}

当我提交我的表单数据时，我的控制器中会显示null，所以请帮助如何在控制器中获取表单数据。

Answer 1

$scope.saveEdit = function(id,firstname,lastname,email){
$http({
method:'post',
data:'id='+ id + '&first_name='+firstname+ 
'&last_name='+lastname+'&email='+email,
url:'http://localhost/Angular_demo/admin/index.php/welcome/update_data'                         
}).then(function success(response){
console.log(response.data);
});
}

应该是这样的：-

$scope.saveEdit = function(id,firstname,lastname,email){
$http({
method:'post',
data:{"id": id,"first_name":firstname, 
"last_name":lastname,"email":email}
url:'http://localhost/Angular_demo/admin/index.php/welcome/update_data'                         
}).then(function success(response){
console.log(response.data);
});
}

Answer 2

由于代码点火器代码使用json_decode，$http服务应该使用JavaScript对象来发布数据：

$scope.saveEdit = function(id,firstname,lastname,email) {
var url = 'http://localhost/Angular_demo/admin/index.php/welcome/update_data';                         
var data = {
id: id,
first_name: firstname,
last_name: lastname,
email: email,
};
$http.post(url, data
).then(function success(response){
console.log(response.data);
});
}

AngularJS框架将自动将JavaScript对象编码为JSON字符串。