Send JSON data to PHP and use it to operate on MySQL - php

Question

What i want to do: copy all selected rows from FIRST table and insert them into SECOND table. In json im sending event type ( insert or delete) and array of IDs rows to copy.
How i can acces to this JSON data in PHP file?
JS:
var data_to_send = {
event : "accept",
ids : ids_to_accept,
}
console.log(ids_to_accept);
$.ajax({
type: "POST",
url: "request.php",
contentType: "application/json; charset=utf-8",
data: data_to_send,
success: function(response){
console.log("Dodałem rekord!");
},
});
PHP:
<?php
//include db configuration file
include_once("connect.php");
if(isset($_POST["data"])){
$data = $_POST["data"];
$ids_to_insert = array();
$ids_to_insert = $data["ids"];
$row_to_insert = $mysqli->query("SELECT * FROM oczekujace WHERE ID='".$ids_to_insert[0]."'");
$inser_row = $mysqli->query("INSERT INTO zaakceptowane(nazwa) VALUES('".$row_to_insert['nazwa']."')");
}
?>
EDIT
Ok, i fixed this. Just deleted contentType...
$.ajax({
type: "POST",
url: "request.php",
//dataType : "json",
//contentType: "application/json; charset=utf-8",
data: {data : data_to_send},
success: function(response){
console.log(response);
console.log("Dodałem rekord!");
},
});

Answer 1

You can use json_decode() function. Its return php array which you can use to insert data in mysql. To read more about json_decode() visit this link.
http://php.net/manual/en/function.json-decode.php

Answer 2

In AJAX use dataType: 'json',
var data_to_send = {
event : "accept",
ids : ids_to_accept,
}
console.log(ids_to_accept);
$.ajax({
type: "POST",
url: "request.php",
dataType: 'json',
contentType: "application/json; charset=utf-8",
data: data_to_send,
success: function(response){
console.log("Dodałem rekord!");
},
});
And in your PHP File use json_decode()
include_once("connect.php");
if($_POST){
$data = json_decode($_POST);
//manipulate $data
$ids_to_insert = array();
$ids_to_insert = $data["ids"];
$row_to_insert = $mysqli->query("SELECT * FROM oczekujace WHERE ID='".$ids_to_insert[0]."'");
$inser_row = $mysqli->query("INSERT INTO zaakceptowane(nazwa) VALUES('".$row_to_insert['nazwa']."')");
}
It will work as you desire.

Answer 3

$_POST['data'] is undefined, because its not present in data_to_send.
You have to either access the posted elements directly in $_POST:
$event = $_POST["event"]);
$ids_to_insert = $_POST["ids"];
or you can change the JS part and provide the data element:
$.ajax({
type: "POST",
url: "request.php",
data: {
"data": data_to_send,
},
success: function(response){
console.log("Dodałem rekord!");
}
});
no need for json_decode, jQuery already converts the data object into a urlencoded parameter string, which PHP in turn converts into an array.