Do in_array have a like function? - php

Question

I have an array of people that is registered as online in a html file. I am using this so that each can have an image assigned to them. But when checking to see if using name is already in use the in_array function return false and allow the script to continue.
$user = "< img src='default.jpg' />John";
$explode = array("<img src='tress.jpg' />John");
if(in_array($user, $explode))
{
//show login script if user exists
}
else
{
//continue to script
}
Now the reason this is not working is because the john in the array is not identical to the john in $user. Is there anyway of checking that the name exists in the array? When responding please explain.

Answer 1

Instead of asking, "How do I solve this problem?", you need to start with, "Why am I having this problem?"
$user = "< img src='default.jpg' />John";
Is < img src='default.jpg' />John a user name? Why are you using it as one? I'm guessing there's some clever thought behind this like "Well, I always display a user's image with their name, so I'll just make the image part of their name. This is going to cause far more problems than it solves. This comes back to a big concept in computer science called separation of concerns. An image is not logically a part of a user name, so don't store it as one. If you always display them together, you can use functions to display a user's information in a standard way without making the image part of the user name.
So first off, remove the image from the name. There are several ways to store this separately.
I would suggest using a class:
class User {
public $name;
public $imageSource;
// The following functions are optional, but show how a class
// can be useful.
/**
* Create a user with the given name and URL to their image
*/
function __construct($name, $imageSource) {
$this->name = $name;
$this->imageSource = $imageSource;
}
/**
* Gets the HTML to display a user's image
*/
function image() {
return "<img src='". $this->imageSource ."' />";
}
/**
* Gets HTML to display to identify a user (including image)
*/
function display() {
return $this->image() . $this->name;
}
}
$user = new User("john", "default.jpg");
// or without the constructor defined
//$user = new User();
//$user->name = "john";
//$user->imageSource = "default.jpg";
echo $user->display();
You can use an "array" if you want to be a little lazier, but I don't recommend it in the general case, since you lose the cool features of classes (like those functions):
$user = array(
name => "john",
image => "<img src='default.jpg' />";
);
echo $user["image"] . $user["name"];
In your database (if you're using one), make them separate columns and then use one of the above data structures.
Now that you have this, it's easy to see if a user name is in a given list using a foreach loop:
function userNameInList($user, $users) {
for($users as $current) {
if($user->name == $current) {
return true;
}
}
return false;
}
$newUser = new User("John", "john.jpg");
$currentUsers = array("John", "Mary", "Bob");
if(userNameInList($newUser, $currentUsers) {
echo "Sorry, user name " . $newUser->name . " is already in use!";
}
If you're new to PHP, the normal for loop may be easier to understand:
function userNameInList($user, $users) {
for($i = 0; $i < count($users); ++i) {
$current = $users[$i];
if($user->name == $current) {
return true;
}
}
return false;
}
Let me know if any of this doesn't run, I don't write PHP very often anymore..