PHP接收json并將接收數據插入數據庫
最近有一個需求,前端向后臺提交json,后臺解析并且將提交的值插入數據庫中,
難點1、php解析json(這個不算難點了,網上實例一抓一大把)2、解析json后,php怎樣拿到該拿的值<?phprequire (’connect.php’);/*本例用到的數據: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,獲取對應的變量值$obj=json_decode($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;//循環變量//--得到Json_list數組長度$num=count($obj['json_list']);//--遍歷數組,將對應信息添加入數據庫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();}//--添加訂單信息$insert_order_sql='INSERT INTO tbl_order (order_id,buyer_id,seller_id,all_price) VALUES (?,?,?,?)';$result=$sqlconn->prepare($insert_order_sql);$result->bind_param('ssss',$order_id,$buyer_id,$seller_id,$all_price);$result->execute();$result -> close();$sqlconn -> close();?>
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昵稱: Hola
Email: jamcistos@outlook.com
blog: holajelly.cc
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