Question

我试图将这些结果放到PHP中的数组中，以便我可以将它们编码为json对象。

我希望json对象看起来像这样：

{"ShopApp":[{"cid":"31","category_name":"Category 1","category_image":"99520_category1.png","subcat":[{"sub_cat_id":"243","sub_cat_name":"Sub Category1"},{"sub_cat_id":"244","sub_cat_name":"Sub Category2"},{"sub_cat_id":"245","sub_cat_name":"Sub Category3"},{"sub_cat_id":"245","sub_cat_name":"Sub Category4"}]},{"cid":"31","category_name":"Category 2","category_image":"99520_category1.png","subcat":[{"sub_cat_id":"1243","sub_cat_name":"Sub Category5"},{"sub_cat_id":"1244","sub_cat_name":"Sub Category6"},{"sub_cat_id":"1245","sub_cat_name":"Sub Category7"}]}]}

这是我的代码。这将返回一个json对象数组

$SQL = "SELECT cid,category_name,category_image FROM tbl_category";

$result = mysql_query($SQL);

$arr = array();
while ($row = mysql_fetch_assoc($result)) {

    $arr[$row['id']]['cid'] = $row['cid'];
    $arr[$row['id']]['category_name'] = $row['category_name'];
    $arr[$row['id']]['category_image'] = $row['category_image'];

    $SQL1 = "SELECT * FROM tbl_sub_category WHERE cat_id = '".$arr[$row['id']]['cid']."'";

    $result1 = mysql_query($SQL1);

    while ($row1 = mysql_fetch_assoc($result1)) {   

        $temp = array('sub_cat_id' => $row1['sub_cat_id'], 'sub_cat_name' => $row1['sub_cat_name']);
        $arr[$row['id']]['cats'][] = $temp;
    }

}

$base_out = array();

foreach ($arr as $key => $record) {

    $base_out['ShopApp'][] = $record;
}

$json = json_encode($base_out);

echo $json;

Answer 1

在我看来，你只需要这样做：

$base_out['ShopApp'] = $arr;

Answer 2

终于解决了自己

$SQL1 = "SELECT cid,category_name,category_image FROM tbl_category";

                $result1 = mysql_query($SQL1);

                        $arr = array();
                        while ($row1 = mysql_fetch_assoc($result1)) 
                        {

                                $catArr=array();
                                $catArr['cid'] = $row1['cid'];
                                $catArr['category_name'] = $row1['category_name'];
                                $catArr['category_image'] = $row1['category_image'];

                                $SQL2 = "SELECT * FROM tbl_sub_category WHERE cat_id = '".$row1['cid']."'";
                        $result2 = mysql_query($SQL2);

                                $subcatArr=array();
                                while ($row2 = mysql_fetch_assoc($result2)) 
                                {   

                                    $temp = array('sub_cat_id' => $row2['sub_cat_id'], 'sub_cat_name' => $row2['sub_cat_name']);
                                    $subcatArr[]=$temp;
                                }
                                $catArr['subcat']=$subcatArr;
                                $arr[$row1['cid']]=$catArr;

                        }


                $shop_out = array();
                foreach ($arr as $key => $record) {

                        $shop_out['ShopApp'][] = $record;

                }

                header( 'Content-Type: application/json; charset=utf-8' );
                $json = json_encode($shop_out);

                echo $json;

嵌套对象json使用php

2 个答案: