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我正在用 Json 传递公司信息，我也想用相同的方法传递这些区域，但是有些地方不太好，有人可以帮我吗？

    if($_GET['method']=="get_all_companies"){
    $query = "SELECT company.id as c_id, company.c_name as c_name, company.c_desc as c_desc,company.c_address as c_address FROM company";
    $result = mysql_query($query);
    if($result){
        $query_areas = "SELECT company.id as c_id, bussiness_area.* FROM company
                        LEFT JOIN company_area
                        ON company_area.id_company = company.id
                        LEFT JOIN bussiness_area
                        ON bussiness_area.id = company_area.id_area
                        WHERE company_id = c_id";
        $result_areas = mysql_query($query_areas);


        $return_arr = array();
        while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {

        $areas_arr =  array();
        while ($row2 = mysql_fetch_array($result_areas, MYSQL_ASSOC)) {
            $row_array['c_id'] = $row['c_id'];
            $row_array['c_name'] = $row['c_name'];
            $row_array['c_desc'] = $row['c_desc'];
            $row_array['c_address'] = $row['c_address'];
            $row_array['bu_title'] = $row2['bu_title'];

            array_push($return_arr,$areas_arr,$row_array);
            }
        }

        echo json_encode($return_arr, $areas_arr);
    }
    else
    {
        echo json_encode(array('error' => 'true'));
    }
}
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1 回答 1

1

从一开始json_encode()，$areas_arr变量就不是必需的。

json_encode($return_arr);

于 2013-05-29T11:13:48.443 回答