How to store data into json.file with variable name from database using php

Question

I want to store the JSON data store.json file like below json format.

store.json

  { "data" :
        [
            {
                "id": 1,
                "customerid": "balaji",
                "name": "Balaji",
                "email": "karurbalamathi@gmail.com",
                "phone": "9566711194"
            }
        ]
    }

But the data is stored like this

[
  {
    "id": 1,
    "customerid": "balaji",
    "name": "Balaji",
    "email": "karurbalamathi@gmail.com",
    "phone": "9566711894"
  }
]

php.php

<?php
$json = $conn->prepare("SELECT * from users");
        $json->execute();
        $json = $json->get_result();
        $response = array();
        $posts = array();
        while ($row = $json->fetch_assoc()) {
            $rows[] = $row;
        }
        $rows1[] = '{ "data" : '.$rows.'}';
        print_r($rows1);
        $rows = json_encode($rows1);
        if(file_exists('data.json')){
            unlink('data.json');
        }
        $myFile = "data.json";
        $fh = fopen($myFile, 'w') or die("can't open file");
        fwrite($fh, $rows);
        fclose($fh);
?>

So I can't select the data for display. Please help me to solve .

Answer 1

You have added data key yourself in this line $rows1[] = '{ "data" : '.$rows.'}'; .

Replace

$rows1[] = '{ "data" : '.$rows.'}';
print_r($rows1);
$rows = json_encode($rows1);
if(file_exists('data.json')){
    unlink('data.json');
}

$myFile = "data.json";
$fh = fopen($myFile, 'w') or die("can't open file");
fwrite($fh, $rows);
fclose($fh);

with below

$json_data = [];
$json_data['data'] = $rows;
$myFile = "data.json";
$fh = fopen($myFile, 'w') or die("can't open file");
fwrite($fh, json_encode($json_data));
fclose($fh);

• You have to assign it as a key to the array to get the format you desired and not do any string manipulation.

• Also, file_exists() and unlink() is redundant since you are opening the file in w mode.

Answer 2

Change the way you store the data to...

$rows1['data'] = $rows;

When you use...

$rows1[] = '{ "data" : '.$rows.'}';

$rows is an array, so this will not work(array to string conversion).