CODEWITHSUNDEEP
phpmysqljson
Akashii
Akashii

Reputation: 2281

Convert MySQL database to JSON

I'm using this code below to convert database to JSON.

<?php
header('Content-type:application/json');

mysql_connect('localhost','root','')or die('cannot connect');
mysql_select_db('sanpham');
$sql = "SELECT * FROM cpu";
$result = mysql_query($sql);
if(!$result)
{
    echo 'Lỗi SQL: '.mysql_error();
    echo "\r\n<br />";
    echo 'SQL: '.$sql;
    exit;
}
$rows = array();
while($row=mysql_fetch_array($result)){
    $rows[]=$row;
}
echo json_encode($rows);


?>

In my PHP file no error, no database, nothing display.

Upvotes: 1

Views: 13529

Answers (3)

Masivuye Cokile
Masivuye Cokile

Reputation: 4772

Deprecated : mysql_connect(): The mysql extension is deprecated and will be removed in the future: use mysqli or PDO instead in

As the warning suggest use either pdo or mysqli.

mysqli : 

<?php

$servername = "localhost";
$username   = "root";
$password   = "";
$dbname     = "sanpham";

// Create connection
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = new mysqli($servername, $username, $password, $dbname);
$conn->set_charset('utf8');

$sql = "SELECT * FROM cpu";
$result = $conn->query($sql);

if ($result->num_rows > 0) {
    $rows = $result->fetch_all(MYSQLI_ASSOC);

    echo json_encode($rows);
} else {
    echo "no results found";
}

PDO : 

<?php

$host    = 'localhost';
$db      = 'sanpham';
$user    = 'root';
$pass    = '';
$charset = 'utf8';

$dsn = "mysql:host=$host;dbname=$db;charset=$charset";
$opt = [
    PDO::ATTR_ERRMODE            => PDO::ERRMODE_EXCEPTION,
    PDO::ATTR_DEFAULT_FETCH_MODE => PDO::FETCH_ASSOC,
    PDO::ATTR_EMULATE_PREPARES   => false,
];

$dbh = new PDO($dsn, $user, $pass, $opt);

$result = $dbh->query("SELECT * FROM cpu");
$rows = $result->fetchAll();
echo json_encode($rows);

Upvotes: 8

Kevin Kyei
Kevin Kyei

Reputation: 585

Use this,

<?php
    //open connection to mysql db
    $connection = mysqli_connect("hostname","username","password","dbname") or die("Error " . mysqli_error($connection));

    //fetch table rows from mysql db
    $sql = "select * from tablename";
    $result = mysqli_query($connection, $sql) or die("Error in Selecting " . mysqli_error($connection));

    //create an array
    $emparray = array();
    while($row =mysqli_fetch_assoc($result))
    {
        $emparray[] = $row;
    }
    echo json_encode($emparray);

    //close the db connection
    mysqli_close($connection);
?>

Upvotes: -1

Mohd Shibli
Mohd Shibli

Reputation: 988

You can simply use the json_encode method on an array storing the data of your mysql database ex:

<?php
$host = "localhost";
$db = "psychocodes";
$user = "your username";  //enter your database username
$pass = "your password";  //enter your database password
$conn = new mysqli($host,$user,$pass,$db); 
$rows = array();

$sql = "SELECT * FROM data";
$result = $conn->query($sql) or die("cannot write");
while($row = $result->fetch_assoc()){
    $rows[] = $row;
}

echo "<pre>";
print json_encode(array('serverres'=>$rows));
echo "</pre>";

?>

Reference : How to Convert MySQL database table into JSON using PHP

Upvotes: 0

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