PHP - Why Is It Not Recognising This Variable And It Is Not Outputting All The Data

    //multi query
    $sql= "SELECT option_value FROM settings WHERE option_name = 'email_temp_issue';";  //1
    $sql .="SELECT title FROM books WHERE id = '$bookid';";  //2
    $sql .="SELECT fullName from students WHERE email = '$username'";  //3
    $sql_run = mysqli_multi_query($connection,$sql);
    if(mysqli_num_rows($sql_run)>0)
    {
        foreach ($sql_run as $row)
        {
            $email_template = $row['option_value']; //retrieve from query 1
            $title = $row['title']; //retrieve from query 2
            $name = $row['fullName']; //retrieve from query 3
        }
    }

I am trying to store the data from the multi query into each variable, which I am not sure how to do..Anyone can help? :)

I am working with an API provided here http://ipndoc.paydotcom.com/ to get IPN call back for payment or refund

With the code he
 

$input = file_get_contents('php://input');
$json = @json_decode($input, true);

$secret = 'SECRETCODE';
$secret = substr(sha1($secret), 0, 32);

$notification = base64_decode($json['notification']);

$iv = base64_decode($json['iv']);

$decoded = openssl_decrypt(
$notification,
'AES-256-CBC',
$secret,
OPENSSL_RAW_DATA,
$iv
);

$data = json_decode($decoded, true);

$trxntype = $data['transactionInfo']['transactionType'];
$paymethod = $data['transactionInfo']['paymentService'];
$f_name = $data['customerInfo']['contactInfo']['firstName'];
$l_name = $data['customerInfo']['contactInfo']['lastName'];

The $f_name and $l_name are correct!
But now, the $ trxntype and $paymethod returns NULL instead of the payment details.

How can I fix this?

I am trying to create a code where the user enters the zip code and if found in the database, a location will be retrieved. I am getting two undefined variables as errors. I am still learning the POST and GET method and I fear that is where I am getting something mixed up. I have also attached an image of my DB.

Thank you so much any input is greatly appreciated.

Notice: Undefined variable: address in C:\xampp\htdocs\index.php on line 53

Notice: Undefined variable: zip in C:\xampp\htdocs\index.php on line 54

index.php file code:

<?php

     include ('header.php');
     include ('function.php');

     ?>


        <div class="wrapper">
           <div>
               <?php
               echo'<form method="POST" action"'.getLocations($conn).'">
                   <input type="text" name="zip" class="search" placeholder="Zip code"><br>
                   <button type="submit" value="submit" id="submit">Submit</button>
                   </form>';
               getLocations($conn);
              ?>         
           </div>
        </div>
        
        <div>
             <!--foreach($zip as $zip) : -->
            <?php 
                  echo $address['address'];
                  echo $zip['zip']; 
            ?>
        </div>
    
    
 function.php file code:
    
    
    <?php
    
    $dBServername = "localhost";
    $dBUsername = "root";
    $dBPassword = "";
    $dBName = "addresses";
    
    // Create connection
    $conn = mysqli_connect($dBServername, $dBUsername, $dBPassword, $dBName);
    
    // Check connection
    if (!$conn) {
        die("Connection failed: " . mysqli_connect_error());
    }
    function getLocations($conn) {
        if (isset($_POST['submit'])){
            
             // Validate Zip code field
            if (!empty ($_POST['zip']) && is_numeric ($_POST['zip'])) {
                
            $zip = (int)$_POST['zip'];
        
            
        $query = "SELECT * FROM locations WHERE zip = '$zip'";
        
        //get the results
        $result = mysqli_query($conn, $query);
        
        //fetch the data
        $zip = mysqli_fetch_all($result, MYSQLI_ASSOC);
        var_dump($zip);
        
        
        mysqli_free_result($result);
        
        //close connection
        mysqli_close($conn);
    
        
       }
    } 
      
    }

Hi, I have a connection set up to an API using a PHP script - the API sends back data in JSON format, and if I capture it and echo it, it displays in extremely unfriendly format on the screen.

I've tried to find a way to convert this data into readable format, ideally in a HTML table, but although there are articles on converting manually inputted JSON data into a HTML table using Javascript, I can't find a way to do it from a PHP variable.

This is what I have so far:

if ($_POST['getcompany']) {
$companyname = $_POST['_Name'];
	
$ch = curl_init();

$data_array2 =  array(
      'token' => $token
);
$make_call2 = json_encode($data_array2);
	//echo 'Token is '.$token;
	$token2 = substr($token, 14);
	$token3 = substr($token2, 0, -3);
	
	//echo 'Token 2 is '.$token3;
	//echo 'Company Name is '.$companyname;

//curl_setopt($ch, CURLOPT_GET, 1);
//curl_setopt($ch, CURLOPT_POSTFIELDS, $make_call2);
curl_setopt($ch, CURLOPT_HTTPHEADER, array('Content-Type: application/json','Authorization: '.$token3.'')); 
curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
curl_setopt($ch, CURLOPT_URL, 'https://connectionurl?countries=GB&name='.$companyname);

//Execute the request
$result = curl_exec($ch);

echo 'Companies: '.$result; //This displays the data provided by the search, but is in JSON format

So I need a way to get the data held in $result, into a HTML table. Any idea how I can do this?