如何解析嵌套的 json 并存储到 mysql 数据库中？答案

【问题标题】：How can I parse nested json and store into mysql database?如何解析嵌套的 json 并存储到 mysql 数据库中？
【发布时间】：2020-01-28 13:17:01
【问题描述】：

我对 PHP 一无所知，谁能帮助我如何将嵌套的 JSON 数据存储到数据库中？我编写代码，但一次只写一行，但如果我想解析嵌套的 JSON，我该怎么办？

[
        {
        "product_name": "Product3",
        "product_price": "500",
        "product_quantity": "5",
        "userid": "1",
        "orderid": "1"
    },
        {
        "product_name": "Product4",
        "product_price": "500",
        "product_quantity": "5",
        "userid": "1",
        "orderid": "1"
    },
        {
        "product_name": "Product5",
        "product_price": "500",
        "product_quantity": "5",
        "userid": "1",
        "orderid": "1"
    }

]

// Creating MySQL connection.
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);

// Storing the received JSON in $json.
$json = file_get_contents('php://input');

// Decode the received JSON and store into $obj
$obj = json_decode($json,true);



$productid = $obj['product_id'];

$productname = $obj['product_name'];

$productprice = $obj['product_price'];

$productquantity = $obj['product_quantity'];

$userid = $obj['userid'];

$orderid = $obj['orderid'];



$query = "INSERT INTO order_product (product_id, product_name, product_price, product_quantity, userid, orderid) VALUES ('$productid','$productname','$productprice','$productquantity','$userid','$orderid')";

if(mysqli_query($con,$query)){

     // On query success it will print below message.
    $MSG = 'Data Successfully Submitted.' ;

    // Converting the message into JSON format.
    $json = json_encode($MSG);

    // Echo the message.
     echo $json ;

 }
 else{

    echo 'Try Again';

 }

我正在尝试使用 Postman 插入，但只插入了一个空白行。请帮助我在命中API后如何存储到数据库中？

【问题讨论】：

添加一个循环，并为该循环内的单个对象执行您当前正在执行的操作……

标签： php mysql json

【解决方案1】：

使用循环插入您的 json 值。我还没有测试，但假设它会为你工作

// Creating MySQL connection.
$con = mysqli_connect($HostName,$HostUser,$HostPass,$DatabaseName);

// Storing the received JSON in $json.
$json = file_get_contents('php://input');

// Decode the received JSON and store into $obj
$obj = json_decode($json,true);

foreach($obj as $product) : 

  $productid = $product['product_id'];
  $productname = $product['product_name'];
  $productprice = $product['product_price'];
  $productquantity = $product['product_quantity'];
  $userid = $product['userid'];
  $orderid = $product['orderid'];

  if(!empty($productid)) :
    $query = "INSERT INTO order_product (product_id, product_name, product_price, 
   product_quantity, userid, orderid) VALUES 
('$productid','$productname','$productprice','$productquantity','$userid','$orderid')";

    if(mysqli_query($con,$query)){

       // On query success it will print below message.
      $MSG = 'Data Successfully Submitted.' ;

      // Converting the message into JSON format.
      $json = json_encode($MSG);

      // Echo the message.
     echo $json ;

    }
    else{

     echo 'Try Again';

    }
  endif;
endforeach;

【讨论】：