Insert nested JSON array as parameter in php mysql database

Question

I want to insert JSON parameters in my phpmyadmin database. I tried with below code but it didn't work. Kindly help me in this.

my JSON parameters are:

 `$ "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"`

and my PHP code is:

 <?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);
        }
     }
   ?>

Kindly help me out in this.

@MahaDev : I used simple post method to convert json parameters into php. — user3424093
– user3424093, Commented Jan 18, 2016 at 6:50
you are getting, $address = $_POST['address']; $customerid = $_POST['customer_id']; $amount = $_POST['order_totalamount']; $pincode = $_POST['pincode']; these from post and products also??? right — devpro
– devpro, Commented Jan 18, 2016 at 6:55

devpro · Accepted Answer · 2016-01-19 11:13:07Z

0

You need to add json_decode function for $_POST['products'] as:

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

UPDATE 1:

You have an issue in your query. You are using 9 columns and insert 10 values.

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

What is this ,, after $address?

UPDATE 2:

Modified code:

<?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);
        }
    }
?>

edited Jan 19, 2016 at 11:13

answered Jan 18, 2016 at 7:15

devpro

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18 Comments

devpro Over a year ago

@user3424093 yes it will not work u hav an error in yur query chk my update 1

devpro Over a year ago

@user3424093: where is $query_params in $result = $stmt->execute($query_params);,,, ??

devpro Over a year ago

please on error reporting , on top of the page, u dont need to use connection inside the loop, third, do u want to insert multiple records or a single record?? @user3424093

user3424093 Over a year ago

Ya i want to insert multiple recorded as per I mentioned in my format. I also correct my code regarding connection.

user3424093 Over a year ago

I already done this. n my query is run successfully and data are already inserted.

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