在 php mysql 数据库中插入嵌套 JSON 数组作为参数答案

【问题标题】：Insert nested JSON array as parameter in php mysql database在 php mysql 数据库中插入嵌套 JSON 数组作为参数
【发布时间】：2016-01-18 06:35:55
【问题描述】：

我想在我的 phpmyadmin 数据库中插入 JSON 参数。我尝试使用下面的代码，但没有奏效。请帮助我。

我的 JSON 参数是：

 `$ "address" = "bharuch"
 "customer_id" = "108"
 "products" = "[{"product_id":"1","product_name":"Potato","category_id":"1","subcategory_id":"1","product_memberprice":"11","product_nonmemberprice":"14","product_minquantity":"500 gms","product_image":"http:\/\/gaubharat.in\/gaubharat\/images\/potato.png","product_brand":"","sub_total":"28","user_qty":"2"},{"product_id":"2","product_name":"Tomato","category_id":"1","subcategory_id":"1","product_memberprice":"15","product_nonmemberprice":"18","product_minquantity":"500 gms","product_image":"http:\/\/gaubharat.in\/gaubharat\/images\/tomato.png","product_brand":"","sub_total":"18","user_qty":"1"}]"
"pincode" = "392025"
"order_totalamount" = "46"`

我的 PHP 代码是：

 <?php
    require("config.inc.php");
        $address = $_POST['address'];
        $customerid = $_POST['customer_id'];
        $amount = $_POST['order_totalamount'];
        $pincode = $_POST['pincode'];
        $product = json_decode($_POST['products']);

        foreach( $product as $key => $val)
        {
        $product_id = $val['product_id'];
        $product_name = $val['product_name'];
        $category_id = $val['category_id'];
        $subcategory_id = $val['subcategory_id'];
        $product_memberprice = $val['product_memberprice'];
        $product_nonmemberprice = $val['product_nonmemberprice'];
        $product_minquantity = $val['product_minquantity'];
        $product_image = $val['product_image'];
        $product_brand = $val['product_brand'];
        $sub_total = $val['sub_total'];
        $user_qty = $val['user_qty'];

        $query = "INSERT INTO `order`(cm_id,product_id,product_quantity,sub_total,order_totalamount,order_id,address,pincode,order_date) VALUES ('$customerid','$product_id','$user_qty','$sub_total','$amount','1','$address','$pincode',CURDATE())";



        if(!mysqli_query($db,$query))
        {
            die('Error : ' . mysql_error());
        }
        else{
           $response["success"] = 1;
           $response["message"] = "You order placed successfully!";
           echo json_encode($response);
        }
     }
   ?>

请帮帮我。

【问题讨论】：

你用什么代码把json转成php数组？
这不是一个有效的 json
@MahaDev : 我使用简单的 post 方法将 json 参数转换为 php。
@devpro : 它们只是 JSON 的参数。
你得到，$address = $_POST['address']; $customerid = $_POST['customer_id']; $amount = $_POST['order_totalamount']; $pincode = $_POST['pincode'];这些来自帖子和产品也？？？对

标签： php mysql json sql-insert

【解决方案1】：

你需要为$_POST['products']添加json_decode函数为：

$address = $_POST['address'];
$customerid = $_POST['customer_id'];
$amount = $_POST['order_totalamount'];
$pincode = $_POST['pincode'];
$products = json_decode($_POST['products']);
foreach($products as $key => $val){
...

更新 1：

您的查询有问题。您正在使用 9 列并插入 10 个值。

$query = "INSERT INTO `order`(cm_id,product_id,product_quantity,sub_total,order_totalamount,order_id,address,pincode,order_date) VALUES ('$customerid','$product_id','$user_qty','$sub_total','$amount','1','$address',,'$pincode',NOW())";

$address 后面的 ,, 是什么？

更新 2：

修改后的代码：

<?php
    require("config.inc.php");
    $address = $_POST['address'];
    $customerid = $_POST['customer_id'];
    $amount = $_POST['order_totalamount'];
    $pincode = $_POST['pincode'];
    $product = json_decode($_POST['products']);

    $values = array();
    foreach($product as $key => $val)
    {
        $product_id = $val->product_id;
        $product_name = $val->product_name;
        $category_id = $val->category_id;
        $subcategory_id = $val->subcategory_id;
        $product_memberprice = $val->product_memberprice;
        $product_nonmemberprice = $val->product_nonmemberprice;
        $product_minquantity = $val->product_minquantity;
        $product_image = $val->product_image;
        $product_brand = $val->product_brand;
        $sub_total = $val->sub_total;
        $user_qty = $val->user_qty;

        $values[] = "('$customerid','$product_id','$user_qty','$sub_total','$amount','1','$address','$pincode',CURDATE())";
    }

    if(count($values) > 0){
        $query = "INSERT INTO `order` (cm_id,product_id,product_quantity,sub_total,order_totalamount,order_id,address,pincode,order_date) VALUES ";

        $query .= implode(",",$values);
        if(!mysqli_query($db,$query))
        {
            echo "Error";
        }
        else
        {
           $response["success"] = 1;
           $response["message"] = "You order placed successfully!";
           echo json_encode($response);
        }
    }
?>

【讨论】：

@user3424093 是的，它不会工作你在你的查询中有一个错误 chk my update 1
@user3424093: $result = $stmt->execute($query_params);,,, ?? 中的 $query_params 在哪里？
请在错误报告上，在页面顶部，您不需要在循环内使用连接，第三，您要插入多条记录还是单条记录？ @user3424093
是的，我想按照我在格式中提到的插入多个记录。我还更正了有关连接的代码。
我已经这样做了。 n 我的查询运行成功，数据已经插入。