Question

这是我的视图和控制器，我将值发布到php然后是mysql。然后我从数据库中获取/检索值？我没有？

<html>
     <script src="https://ajax.googleapis.com/ajax/libs/jquery/3.2.1/jquery.min.js"></script>
     <script src="https://ajax.googleapis.com/ajax/libs/angularjs/1.2.24/angular.min.js"></script>

<body>
    <div ng-app="samp">

        <div ng-controller="formController" >
            <form ng-submit="submity()">
                <input type="text" ng-model="test.namee">
                <input type="text" ng-model="test.email">
                <input type="submit" value="okay">
            </form>
            ==>{{list}}hyhygtgty
        </div>

    </div>

</body>
</html>
<script type="text/javascript">
var app=angular.module('samp',[]);
app.controller('formController',function($scope,$http)
    {
        $scope.test={};

        $scope.submity=function()
        {
            console.log($scope.test);
            $http(
            {
                method:"POST",
                url:"sample3.php",
                headers: {'Content-Type': 'application/x-www-form-urlencoded'},
                data:{'name':$scope.test.namee,'name1':$scope.test.email},
                success:function(data)
                {
                 alert('sss')
                }
            });
        }

        $scope.myctrl = function(){
            $http(
            {
                method:"GET",
                url:"sample4.php",
                //headers: {'Content-Type': 'application/x-www-form-urlencoded'},
                //data:{'name':$scope.test.namee,'name1':$scope.test.email},
                success:function(data)
                {
                    $scope.list=data;
                    console.log(data);

                }
            });

        }
        $scope.myctrl()

    });
</script>

这是我的php代码

<?php
$conn=mysqli_connect('localhost','root','','boot');
    if (mysqli_connect_errno())
    {
        echo "Failed to connect to MySQL: " . mysqli_connect_error();
    }
    $query=mysqli_query($conn,"select * from boots");
    $data=array();
    while($row = mysqli_fetch_array($query))
    {
        $data[]=$row;
    }

    print json_encode($data);
?>

如何将响应数据核心php发布到angular？

Answer 1

使用then而不是success来抓住http响应。并且success不属于http属性。删除它并使用then来捕捉这样的承诺

$scope.submity = function() {
    $http({
        method: "POST",
        url: "sample3.php",
        headers: {
            'Content-Type': 'application/x-www-form-urlencoded'
        },
        data: {
            'name': $scope.test.namee,
            'name1': $scope.test.email
        },
    }).then(function(response) {
        console.log(response.data);
        alert('sss');
    })
}

如何从php到angular js获取值？

1 个答案: