How to use php functions in IF statement

Question

INTRO

I am trying to better understand my knowledge of Php and using classes to better prganise my code so this is just an exercise for a better understanding rather than a real world solution.

BRIEF

I am calling in a function from a class which I have just learnt to do but I want to know the best way to do something simple tasks like use the object in an IF statement.

SCENARIO

So for instance I am setting my classes like so:

    class user
    {

        // Get users ID
        function get_user_id()
        {

            global $conn;
            $sql = 'SELECT id FROM user';
            $result = $conn->query($sql);

            if ($result->num_rows > 0) {

                while ($row = $result->fetch_assoc() ){
                    echo $row['id'] . ', '; }

            }
        }

        // Get users name
        function get_user_name()
        {
            global $conn;
            $sql = 'SELECT name FROM user';
            $result = $conn->query($sql);

            if ($result->num_rows > 0) {

                while ($row = $result->fetch_assoc() ){
                    echo $row['name'] . ', ';   }

            }
        }

    }

$userId = new user;
$userName = new user;

I am then initializing in my classes like so:

<?php $userId->get_user_id(); ?>
<?php $userName->get_user_name(); ?>

and THEN I am wanting to performa simple task like show a user based on the value of their ID, the above will return 2 sets of results of 4 so id 1, 2, 3, 4 & Dan, Andy, Ryan, Aran

so I am performing a simple IF statement like so:

if($userId > 1){
    echo $userName;
} else {
    echo 'not working';
}

But it returns 'not working' - I am just wanting to better understand how to use the functions in a way that A works and B best practice.

your $userId is an object, how can you echo that ? What you want, please specify — Rahul
– Rahul, Commented Dec 16, 2016 at 12:20
You need to return value from your function instead of echo. However your code doesn't make sense to get user_id and username ? For whom you are getting it ? do you want to get all user details ? — TIGER
– TIGER, Commented Dec 16, 2016 at 12:21
I would suggest you GOOGLE OOP and go through the docs. — Sougata Bose
– Sougata Bose, Commented Dec 16, 2016 at 12:21
the whole contruct of your class and how you use it is... hm.. unconventional. You need to read the docs again! — Jeff
– Jeff, Commented Dec 16, 2016 at 12:22

Steve · Accepted Answer · 2016-12-16 17:24:34Z

It doen't look like you've understood OOP just yet.

These code examples should hopefully give you an introduction but as in other comments, read up on OOP. I struggled with it at first but keep at it!

Create your user class
This class represents a single user and the actions associated with a user, think of it as a blue print. It should only perform functions related to a user, it shouldn't keed to 'know' about anything else. For example, database functions sholud be done elsewhere.

class User {
   private $id;
   private $name;

   function __construct($array)
   {
       $this->id = $array['id'];
       $this->name = $array['name'];
   }

   function getId()
   {
       return $this->id;
   }

   function getName()
   {
       return $this->name;
   }
}

Load all users into an array

$sql = 'SELECT * FROM user';
$result = $conn->query($sql);

$users = [];
while ($row = $result->fetch_assoc() ){
   $users[] = new User($row);
}

// this array now contains all your users as User objects
var_dump($users);

// echo all user's details
foreach($users as $user) {
   echo $user->getId();
   echo ' - ';
   echo $user->getName();
   echo "\r\n";
}

Load a single user

$sql = 'SELECT * FROM user WHERE id = 1';
$result = $conn->query($sql);

if ($row = $result->fetch_assoc()) {
   $user = new User($row);
} else {
   exit('User ID does not exist');
}

// echo the user's ID and name
echo $user->getId();
echo ' - ';
echo $user->getName();

Resourses
Laracasts - https://laracasts.com/series/object-oriented-bootcamp-in-php
Search PHP OOP explained - https://www.google.co.uk/search?q=php+oop+explained

I really appreciate it Steve, and I am glad I am not the only one!

Nitin Vaghani · Accepted Answer · 2016-12-16 13:13:58Z

<?php

class user {

    // Get users ID
    function get_user_id() {

        global $conn;
        $data = array();
        $sql = 'SELECT id FROM user';
        $result = $conn->query($sql);

        if ($result->num_rows > 0) {
            while ($row = $result->fetch_assoc()) {
                $data[] = $row['id'] . ', ';
            }
        }
        return $data;
    }

    // Get users name
    function get_user_name() {
        global $conn;
        $data = array();
        $sql = 'SELECT name FROM user';
        $result = $conn->query($sql);

        if ($result->num_rows > 0) {

            while ($row = $result->fetch_assoc()) {
                $data[] = $row['name'] . ', ';
            }
        }
        return $data;
    }
}

$userId = new user;

$userName = new user;

// all user ids
$all_ids = $userId->get_user_id();

echo '<pre>';
print_r($all_ids);

// all user name 
$all_name = $userId->get_user_name();
echo '<pre>';
print_r($all_name);`enter code here`

Check first response from both function after use if condition

Sunil Chhimpa · Accepted Answer · 2016-12-16 12:35:53Z

2

You are comparing object with 1 not the value returned by function get_user_id().

So instead of

         <?php $userId->get_user_id(); ?>
         <?php $userName->get_user_name(); ?>

Try

         <?php $id=$userId->get_user_id(); ?>
         <?php $name= $userName->get_user_name(); ?>

and then put in your condition

  if($id > 1){
     echo $name;
  } else {
  echo 'not working';
 }

I will suggest you to replace echo with return statement.

answered Dec 16, 2016 at 12:35

Sunil Chhimpa

4041 gold badge4 silver badges12 bronze badges

Comments

smartnet · Accepted Answer · 2016-12-16 13:04:43Z

call your class as an object $userid = user(); $username = user();

you can also try something like this

class user
{

    // Get users ID
    function get_user_id($id = "")
    {
        global $conn;

        // check if id is empty or not
        if(!empty($id)) {
            $sql = 'SELECT id FROM users WHERE id = '.$id;
        }else{
            $sql = 'SELECT id FROM users';
        }
        $result = $conn->query($sql);

        if ($result->num_rows > 0) {

            while ($row = $result->fetch_assoc() ){
                echo $row['id'] . ', '; }

        }
    }

    // Get users name
    function get_user_name($name = "")
    {
        global $conn;

        // check if name is empty or not
        if(!empty($name)) {
            $sql = 'SELECT name FROM user WHERE name = '.$name;
        }else{
            $sql = 'SELECT name FROM user';
        }

        $result = $conn->query($sql);

        if ($result->num_rows > 0) {

            while ($row = $result->fetch_assoc() ){
                echo $row['name'] . ', ';   }

        }
    }

}

    $userId = new user();
    $userName = new user();

    $userId->get_user_id(1);
    $userName->get_user_name();

    echo $userId;
    echo $userName;

please make sure you sanitize the id and name before use

TIGER · Accepted Answer · 2016-12-16 14:01:03Z

0

IN both get_user_id, get_user_name methods please return $row = $result->fetch_assoc(); so, it will value comes in $userId, $userName and you can access it.

right now you return nothing so $user_id has null value so, it always goes in else condition.

Example

function get_user_id()
    {

        global $conn;
        $sql = 'SELECT id FROM user';
        $result = $conn->query($sql);

        if ($result->num_rows > 0) {
            $value = '';

            while ($row = $result->fetch_assoc() ){
                $value .= $row['id'] . ', ';
            }

           return $value;
        }
    }

edited Dec 16, 2016 at 14:01

TIGER

2,9055 gold badges38 silver badges46 bronze badges

answered Dec 16, 2016 at 12:25

Garvit Joshi

791 silver badge10 bronze badges

2 Comments

danjbh Over a year ago

That isn't quire true, I do get my results back but not in the argument which is failing.

Garvit Joshi Over a year ago

i think this one's help you check it again

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