Question

我正在尝试将我的angularjs应用程序连接到一个简单的php脚本，它只是从sqlite3数据库返回一个sql查询。

这是我的PHP脚本：

<?php
 date_default_timezone_set('UTC');
try {
        $objDb = new PDO('sqlite:../dbase/shopper');
        $objDb -> setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

        $sql = "SELECT item.* ,
                        t.name AS type_name
                        FROM items item
                        INNER JOIN types t
                        ON t.id = item.type
                        ORDER BY item.date ASC";

        $result = $objDb->query($sql);

        if(!$result) {
                throw new  PDOException("The result returned no object");
        }
        $result->setFetchMode(PDO::FETCH_ASSOC);
        $items = $result->fetchAll();

        $sql = "SELECT *
                        FROM types
                        ORDER BY id";
        $result = $objDb->query($sql);
        echo var_dump($result);
        if(!$result) {
                throw new  PDOException("The result returned no object");
        }
        $result->setFetchMode(PDO::FETCH_ASSOC);
        $types = $result->fetchAll();

        echo json_encode(array(
                'error' => false,
                'items' => $items,
                'types' => $types
        ), JSON_HEX_TAG | JSON_HEX_APOS |JSON_HEX_QUOT |JSON_HEX_AMP );

} catch (PDOException $e) {
        echo json_encode(array(
                'error' => true,
                'message' => $e->getMessage()
        ),JSON_HEX_TAG | JSON_HEX_APOS |JSON_HEX_QUOT |JSON_HEX_AMP );
}

当我检查php文件地址时，我可以得到结果：

object(PDOStatement)#3 (1) { ["queryString"]=> string(41) " SELECT * FROM types ORDER BY id " }
{"error":false,"items":[{"id":"1","item":"Butter","qty":"1","type":"1","done":"0","date":"2014-10-06 02:45:51","type_name":"Qty"}],"types":[{"id":"1","name":"Qty"},{"id":"2","name":"Kg"}]}

如果我使用angularjs，我会得到未定义的结果。

.controller('ShoppingListController', function($scope, $http, $log ){ 
    $scope.types = [];
    $scope.items = [];

    $scope.item = '';
    $scope.qty = '';
    $scope.types = '';

     $scope.select = function( ) {

        $http({method: 'GET',url: 'mod/select.php'})
            .success(function(data){
                    console.log(data)
                    $scope.items = data.items;


                if(data.types) {

                    $scope.types = data.types;
                    $scope.type = $scope.types[0].id;

                }
            })
            .error(function(data,status,header){
                throw new Error('Something went wrong with selecting record');
            });
     };

     $scope.select();

});

console.log（数据）显示：

object(PDOStatement)#3 (1) {
  ["queryString"]=>
  string(41) " SELECT * 
            FROM types 
            ORDER BY id "
}
{"error":false,"items":[{"id":"1","item":"Butter","qty":"1","type":"1","done":"0","date":"2014-10-06 02:45:51","type_name":"Qty"}],"types":[{"id":"1","name":"Qty"},{"id":"2","name":"Kg"}]}

我该如何解决这个问题？

Answer 1

您必须返回JSON响应。从数据库中获取关联数组并返回：

echo json_encode($db_query);

在Angular中，您可以将范围设置为等于响应ex：

$scope.items = data.items;

然后，您可以在视图中访问此内容（请确保ng-controller="ShoppingListController"）。

迭代数据：

ng-repeat="item in items"

然后，您可以通过item.id或阵列中的任何密钥访问每个部分。

此外，无需在顶部设置$scope.items。

编辑：

.controller('ShoppingListController', function($scope, $http, $log) {
    function select() {
        $http.get('mod/select.php')
            .success(function(data) {
                $scope.items = data.items;

                if(data.types) {
                    $scope.types = data.types;
                    $scope.type = $scope.types[0].id;
                }
            });
    }

    select();

});

Answer 2

删除var_dump();并且您不需要与var_dump()回应，它已经附加到输出缓冲区。 var_dump()使您的回复无效JSON。

如何使用简单的php api通过$ http.get使用数据

2 个答案: