Presenting MYSQL nested JSON object with php - php

Question

I have a MYSQL Table stored using 000webhostapp cpanel, here is what it look like.
id | name | J_Object
---------------------------------------------------------
1 | deckA | {"type":"A", "ports": {"hdmi": 1, "usb":2}}
2 | deckB | {"type":"B", "ports": {"hdmi": 3, "usb":2}}
3 | deckC | {"type":"A", "ports": {"hdmi": 1, "usb":2}}
4 | deckD | {"type":"B", "ports": {"hdmi": 3, "usb":2}}
Using PHP, I would like to present it like the below to act as a HTTP to pass the data to other platforms.
[
{
"id": "1",
"name": "deckA",
"J_Object": [{"type":"A", "ports": {"hdmi": 1, "usb":2}}]
},
{
"id": "2",
"name": "deckB",
"J_Object": [{"type":"B", "ports": {"hdmi": 3, "usb":2}}]
},
{
"id": "1",
"name": "deckC",
"J_Object": [{"type":"A", "ports": {"hdmi": 1, "usb":2}}]
},
{
"id": "1",
"name": "deckD",
"J_Object": [{"type":"B", "ports": {"hdmi": 3, "usb":2}}]
}
]
However, with my PHP I written below, what follows after is what I got. And it is definitely not what I wanted.
The PHP code:
$search = "SELECT * FROM `database`.`products`
WHERE JSON_EXTRACT(`J_Object` , '$.ports.usb') > 2
AND JSON_EXTRACT(`J_Object` , '$.ports.hdmi') > 1;";
$result = mysqli_query($link, $search);
if ($result) {
//Printing out the details of the rows
while($array = mysqli_fetch_assoc($result)) {
$jsonData[]=$array;
}
$nonArr = json_encode($jsonData, JSON_PRETTY_PRINT);
$stringArr = (string)$nonArr;
$nonSlash = str_replace("\\","", $stringArr);
echo $nonSlash;
}
mysqli_close($link)
And the below is the unwanted results:
[
{
"id": "1",
"name": "deckA",
"J_Object": "{"type":"A", "ports": {"hdmi": 1, "usb":2}}"
},
{
"id": "2",
"name": "deckB",
"J_Object": "{"type":"B", "ports": {"hdmi": 3, "usb":2}}'
},
{
"id": "1",
"name": "deckC",
"J_Object": "{"type":"A", "ports": {"hdmi": 1, "usb":2}}"
},
{
"id": "1",
"name": "deckD",
"J_Object": "{"type":"B", "ports": {"hdmi": 3, "usb":2}}"
}
]
It looks like its the same but if you notice the [ ] is changed with " " which also means that the nested object is not quite right.
Can any pro out there show me how do I get the correct output?

Answer 1

You could try a much simpler way.
$result = mysqli_query($link, $search);
if ($result) {
//Printing out the details of the rows
while($array = mysqli_fetch_assoc($result)) {
$array['J_Object'] = [json_decode($array['J_Object'], true)];
$jsonData[]=$array;
}
echo json_encode($jsonData, JSON_PRETTY_PRINT);
}

Answer 2

This would work for you. It's just a quick fix with preg_replace.
$result = mysqli_query($link, $search);
if ($result) {
//Printing out the details of the rows
while($array = mysqli_fetch_assoc($result)) {
$jsonData[]=$array;
}
$nonArr = json_encode($jsonData, JSON_PRETTY_PRINT);
$stringArr = (string)$nonArr;
$nonSlash = str_replace("\\","", $stringArr);
// quick fix
$nonSlash = preg_replace("/[\"]{1}[{]{1}/", '[{', $nonSlash);
$nonSlash = preg_replace("/[}]{1}[\"]{1}/", '}]', $nonSlash);
echo $nonSlash;
}
Output:
[
{
"id": "1",
"name": "deckA",
"J_Object": [{"type":"A", "ports": {"hdmi": 1, "usb":2}}]
},
{
"id": "2",
"name": "deckB",
"J_Object": [{"type":"B", "ports": {"hdmi": 3, "usb":2}}]
},
{
"id": "3",
"name": "deckC",
"J_Object": [{"type":"A", "ports": {"hdmi": 1, "usb":2}}]
},
{
"id": "4",
"name": "deckD",
"J_Object": [{"type":"B", "ports": {"hdmi": 3, "usb":2}}]
}
]
Hope you find this helpful. Happy coding..