Getting Proper JSON response in PHP - php

Question

I am pretty new to PHP, I have referred some examples and made a code for getting data from database. but if the database is not found I am getting a text response , Can anyone suggest how to get a proper JSON response back if no database found or misconfigured
This is my $http.get method
$http.get('client/php/popData.php')
.success(function(data) {
$scope.blogs = data;
})
.error(function(err) {
$log.error(err);
})
popdata.php for getting data from database
<?php
$data = json_decode(file_get_contents("php://input"));
include('config.php');
$db = new DB();
$data = $db->qryFire();
echo json_encode($data);
?>
This is my config.php
<?php
define("__HOST__", "localhost");
define("__USER__", "username");
define("__PASS__", "password");
define("__BASE__", "databasename");
class DB {
private $con = false;
private $data = array();
public function __construct() {
$this->con = new mysqli(__HOST__, __USER__, __PASS__, __BASE__);
if(mysqli_connect_errno()) {
die("DB connection failed:" . mysqli_connect_error());
}
}
public function qryPop() {
$sql = "SELECT * FROM `base` ORDER BY `id` DESC";
$qry = $this->con->query($sql);
if($qry->num_rows > 0) {
while($row = $qry->fetch_object()) {
$this->data[] = $row;
}
} else {
$this->data[] = null;
}
$this->con->close();
}
public function qryFire($sql=null) {
if($sql == null) {
$this->qryPop();
} else {
$this->con->query($sql);
$this->qryPop();
}
// $this->con->close();
return $this->data;
}
}
?>

Answer 1

Use Exceptions:
Change your class DB like this:
public function __construct() {
$this->con = new mysqli(__HOST__, __USER__, __PASS__, __BASE__);
if(mysqli_connect_errno()) {
throw new Exception("DB connection failed:" . mysqli_connect_error());
}
}
Then change your popdata.php like
<?php
$data = json_decode(file_get_contents("php://input"));
include('config.php');
try {
$db = new DB();
$data = $db->qryFire();
} catch (Exception $e) {
echo json_encode(['error' => $e->getMessage()]);
exit();
}
echo json_encode($data);
This way you will get error response in JSON for any Exception thrown while constructing the DB class and while executing DB::qryFire
if you want to catch your warnings you can try modifying your DB class like the below:
public function __construct() {
ob_start();
$this->con = new mysqli(__HOST__, __USER__, __PASS__, __BASE__);
$warning = ob_clean();
if ($warning) {
throw new Exception($warning);
}
if(mysqli_connect_errno()) {
throw new Exception("DB connection failed:" . mysqli_connect_error());
}
}
You can also turn off your warnings and notices by adding
error_reporting(E_ERROR);
on the top of your file

Answer 2

replace the line die("DB connection failed:" . mysqli_connect_error()); in DB class __construct function with
die( json_encode( array('status' => 'error') ) );
when ever the app will fail to connect to the data base it will give
{"status": "error"}
i did not checked this. but i hope it will work
btw it's my 1st answer on stackoverflow. i'm sorry for my mistakes. correct them