您如何向PHP中的用户建议唯一的用户名? [英] How do you suggest unique usernames to user in PHP?

问题描述

假设某人想要的用户名被使用，那么我希望系统建议可用用户名的列表，例如5条建议.

Let's say the username someone wants is taken, then I want the system to suggest a list of available usernames, say 5 suggestions.

现在，我可以根据我已有的数据的排列和组合生成多个用户名，例如名称，生日等.但是，如何确保这些生成的用户名可用?我的意思是，如果对于每个生成的用户名，我必须查询数据库以检查可用性，在最坏的情况下，这可能会变成无限循环之类的事情.

Now, I can generate several usernames based on permutations and combinations of data I already have, like their name, birthday, e.t.c. But how do I make sure those generated usernames are available? I mean, if for each generated username, I have to query the database to check availability, in the worst case scenario this might become something like an infinite loop.

有什么想法吗?

推荐答案

我怀疑这个问题是基础问题，因此我将提供一个简单的示例:

I suspect the question is a base one so I'll provide a simple example:

<?php

// Untested code and only one of the many possible ideas
$suggestions = array(
    'foobar' => TRUE,
    'foo1974' => TRUE,
    'foo37' => TRUE,
    'drfoo' => TRUE,
    'mrfoo' => TRUE,
);


$params = $placeholders = array();
foreach(array_keys($suggestions) as $position => $username){
    $params['u' . $position] = $username;
    $placeholders[] = ':u' . $position;
}
$sql = 'SELECT username
    FROM user
    WHERE username IN (' . implode(', ', $placeholders) . ')';

$res = $conn->prepare($sql);
$res->execute($params);
while( $row = $res->fetch(PDO::FETCH_ASSOC) ){
    $suggestions[ $row['username'] ] = FALSE;
}


foreach($suggestions as $username => $available){
    if($available){
        // ...
    }
}

提供无限可用名称列表的唯一方法是使用非常简单的规则，例如添加连续数字.在这种情况下，您可以尝试执行以下操作:

The only way to provide an infinite list of available names is to use a very simple rule such as adding a consecutive number. In such case, you can try something like this:

SELECT username
FROM user
WHERE username REGEXP '^foo[0-9]+'

...然后:

$username = 'foo';

$suggestions = array();
$count = 0;
$names_left = 5;
while($names_left>0){
    $count++;

    if( !in_array($username . $count, $names_taken) ){
        $suggestions[] = $username . $count;
        $names_left--;
    }
}

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