PHP接收json 并将接收数据插入数据库的实现代码
author:一佰互联 2019-04-27   click:320

最近有一个需求,前端向后台提交json,后台解析并且将提交的值插入数据库中,难点1、php解析json(这个不算难点了,网上实例一抓一大把)2、解析json后,php怎样拿到该拿的值

<?phprequire ("connect.php");/*本例用到的数据:post_array={"order_id":"0022015112305010013","buyer_id":"2","seller_id":"1","all_price":"100.00","json_list":[{"product_id":"3","product_number":"3"},{"product_id":"8","product_number":"2"},{"product_id":"10","product_number":"4"}]} */$post_array=$_POST["post_array"];//--解析Json,获取对应的变量值$obj=json_decode($post_array,TRUE);$order_id = $obj["order_id"];$buyer_id = $obj["buyer_id"];$seller_id = $obj["seller_id"];$all_price = $obj["all_price"];$i=0;//循环变量//--得到Json_list数组长度$num=count($obj["json_list"]);//--遍历数组,将对应信息添加入数据库for ($i;$i<$num;$i++){	$list_product_id[]=$obj["json_list"][$i]["product_id"];	$list_product_number[]=$obj["json_list"][$i]["product_number"];	$insert_order_product_sql="INSERT INTO tbl_order_product (order_id,product_id,product_number) VALUES (?,?,?)";	$result = $sqlconn -> prepare($insert_order_product_sql);	$result -> bind_param("sss", $order_id,$list_product_id[$i],$list_product_number[$i]);	$result->execute();}//--添加订单信息$insert_order_sql="INSERT INTO tbl_order (order_id,buyer_id,seller_id,all_price) VALUES (?,?,?,?)";$result=$sqlconn->prepare($insert_order_sql);$result->bind_param("ssss",$order_id,$buyer_id,$seller_id,$all_price);$result->execute();$result -> close();$sqlconn -> close();?>

投稿者信息昵称: HolaEmail: jamcistos@outlook.com