How to get print on browser console using PHP - php

Question

I am testing out my php code and want to make see the files that I am getting from json and check how it is displaying so I know how to send the code to mysql
php file
<?php
$json = json_decode(file_get_contents("php://input"));
$obj = json_decode($json,true);
print_r($json);
print_r($obj);
function debug_to_console( $data ) {
$output = $data;
if ( is_array( $output ) )
$output = implode( ',', $output);
echo "<script>console.log( 'Debug Objects: " . $output . "' );</script>";
}
debug_to_console( "Test" );
debug_to_console($obj);
print_r("hello");
echo "<script>console.log('" . json_encode($json) . "');</script>";
echo "<script>console.log('" . $obj . "');</script>";
echo "<script>console.log('" . '$obj' . "');</script>";
var_dump('Hello');
var_dump($obj);
var_dump($json);
header("Content-Type: application/json; charset=UTF-8")
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
?>
None of this prints to the console
Want code to print to the console

Answer 1

You can try the following function
function consoleLogs($data) {
$html = "";
$coll;
if (is_array($data) || is_object($data)) {
$coll = json_encode($data);
} else {
$coll = $data;
}
$html = "<script>console.log('PHP: ".$coll."');</script>";
echo($html);
}
You can use this as :-
consoleLogs(array("test1", "test2"));