如何检查 WordPress 中是否存在 user_login 并通过附加到现有 user_login 以编程方式创建新用户

Question

In my WP v5.8.1, I am creating new user programmatically from public forms, using wp_create_user . New user_login will be created by combining first_name and last_name

$new_username = strtolower(str_replace(' ', '', $_POST['first_name']) . '-' . str_replace(' ', '', $_POST['last_name']));

However, chances are there might be an existing user with same user_login. Hence I want to append number (1,2,3,4,etc) to the new user_login with below function:

if (username_exists($new_username)) {
    create_new_username();
} else {
    $new_username = $new_username;
}

function create_new_username() {
    $count = 0;
    while (username_exists($new_username)) :
        $new_username = $new_username . '-' . $count + 1;
        $count++;

        if (!username_exists($new_username)) {
            $create_username = $new_username;
        }
    endwhile;

    return $new_username;
}

echo create_username();

If user_login exists, the above function returns existing user_login user ID; it is not creating new user_login with appending with number.

Edit 1:

I have modified the code as below without writing a separate function:

$new_username = strtolower(str_replace(' ', '', $_POST['first_name']) . '-' . str_replace(' ', '', $_POST['last_name']));

if (username_exists($new_username)) {
    $count = 0;
    while (username_exists($new_username)) :
        $new_username = $new_username . '-' . ($count + 1);
        $count++;

    endwhile;
}
echo $new_username;

This is returning a new user_login if user exists; however instead of appending a new series number from ($count + 1) , it is adding another new number as below:

• first-last
• first-last-1
• first-last-1-2
• first-last-1-2-3

Whereas I need the usernames like this:

• first-last
• first-last-1
• first-last-2
• first-last-3

Answer 1

Your function create_new_username() needs the variable $new_username it's not automatically passed to this function, and therefore is undefined.

function create_new_username($new_username) {
    $count = 0;
    while (username_exists($new_username)) :
        $new_username = $new_username . '-' . $count + 1;
        $count++;

        if (!username_exists($new_username)) {
            $create_username = $new_username;
        }
    endwhile;

    return $new_username;
}

Then however you're getting to your if(username_exists()) call... I'd have to assume through some other hook, you would have to pass the variable $new_username to your username_exists() function.

if (username_exists($new_username)) {
    create_new_username($new_username);
} 

Since your else doesn't do anything, it's not necessary. $new_username already is equal to $new_username

Answer 2

Answering my own question, comments from @Howard E has helped me to achieve what I am looking for:

Below is the new user name created from the public form submission:

$new_username = strtolower(str_replace(' ', '', $_POST['first_name']) . '-' . str_replace(' ', '', $_POST['last_name']));

With below code, I am able to check if the user_login is exists, if exists appending a serial number to create new user_login :

    if (username_exists($new_username)) { // check if user_login exists
    while (username_exists($new_username)) :
        $count = preg_match_all("/.*?(\d+)$/", $new_username, $matches); // check if the existing user_login has serialised number at the end
        if ($count) { // if user_login has number at the end
            $new_username = rtrim($new_username, '-' . $count); // remove the number and '-' also before the number 
        } else {
            $count = 0; // else keep the count as 0 to add to the submitted user_login from the public form
        }
        $new_username = $new_username . '-' . ($count + 1); // this will create a new serialised user_login with numbering continuation from previous, else create new user_login as per form submission
        $count++;

    endwhile;
}
echo $new_username;

Now this is creating user_logins as below:

• first-last
• first-last-1
• first-last-2
• first-last-3

Please comment if the code can be still be more simpler and efficient?