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PHP接收json 并将接收数据插入数据库的实现代码_PHP

WBOY
WBOYOriginal
2016-05-29 11:47:57920browse

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

<&#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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