PHP receives json and inserts the received data into the database implementation code_php skills

WBOY
Release: 2016-05-16 20:03:54
Original
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There is a requirement recently. The front end submits json to the background, and the background parses and inserts the submitted value into the database.
Difficulty
1. PHP parses json (this is not difficult, there are a lot of examples online)
2. After parsing json, how does php get the value it should get

<&#63;php
require ('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 (&#63;,&#63;,&#63;)";
	$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 (&#63;,&#63;,&#63;,&#63;)";
$result=$sqlconn->prepare($insert_order_sql);
$result->bind_param("ssss",$order_id,$buyer_id,$seller_id,$all_price);
$result->execute();

$result -> close();
$sqlconn -> close();
&#63;>
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