php Laravel：生成随机唯一令牌
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Laravel: Generate random unique token
提问by user5486944
I have a table in my database called keysthat has this structure:
我的数据库中有一个名为keys具有以下结构的表：
id | user_id | token_id | token_key
Every time a user logs into my site, I need to generate a new token_idand token_keyset for that user. How can I generate a random token for both the token_idand the token_keywhile keeping the two values unique?
每次用户登录到我的网站，我需要生成一个新的token_id，并token_key为该用户设置。如何为 thetoken_id和 the生成随机令牌，token_key同时保持两个值的唯一性？
For example, if:
例如，如果：
token_idisdfbs98641aretwsg,token_keyissdf389dxbf1sdz51fga65dfg74asdf
token_id是dfbs98641aretwsg，token_key是sdf389dxbf1sdz51fga65dfg74asdf
Meaning:
意义：
id | user_id | token_id | token_key
1 | 1 | dfbs98641aretwsg | sdf389dxbf1sdz51fga65dfg74asdf
There can be no other row in the table with that combination of tokens. How can I do this?
表中不能有具有该标记组合的其他行。我怎样才能做到这一点？
回答by James
In terms of generating the tokens, you could use one of Laravel's Helper Functions; str_random().
在生成令牌方面，您可以使用Laravel 的辅助函数之一；str_random().
This will generate a random string of a specified length, e.g str_random(16)will generate a random string of 16 characters (upper case, lower case, and numbers).
这将生成一个指定长度的随机字符串，例如str_random(16)将生成一个 16 个字符（大写、小写和数字）的随机字符串。
Depending on how you are using the tokens, do they really need to be completely unique? Given that they will match to a user, or I assume you may be using the token_idand then verifying this against the token_keydoes it really matter if there is a double up of one of them? - although the chances of this are extremely small!
根据您使用令牌的方式，它们真的需要完全独一无二吗？鉴于它们将与用户匹配，或者我假设您可能正在使用token_id，然后根据 验证这token_key一点，如果其中一个加倍是否真的很重要？——虽然这种可能性极小！
However, if you do need them to be truly unique you can always use a validator with the uniqueconstraint. Using this packageyou could also test that the two of them are unique too with unique_with. And then if the validator fails then it generates a new token as needed.
但是，如果您确实需要它们真正独一无二，您始终可以使用带有unique约束的验证器。使用此包，您还可以使用unique_with. 然后，如果验证器失败，则它会根据需要生成一个新令牌。
Based off your examples, you would be using str_random(16)for token_idand str_random(30)for the token_key.
根据您的示例，您将使用str_random(16)fortoken_id和str_random(30)for token_key.
回答by Oliver Maksimovic
I'd avoid including an extra package for a case like this one. Something like:
我会避免为这样的案例包含一个额外的包。就像是：
do {
$token_id = makeRandomToken();
$token_key = makeRandomTokenKey();
} while (User::where("token_id", "=", $token_id)->where("token_key", "=", $token_key)->first() instanceof User);
...should do. Replace model name with yours, if different from 'User', and use your or suggested functions for creating random strings.
...应该做。如果与“用户”不同，请将模型名称替换为您的名称，并使用您或建议的函数来创建随机字符串。
回答by Yallison Gabriel Reis
You can use a dependency to do this. Dirape laravel-token
您可以使用依赖项来执行此操作。Dirape laravel-token
Run the command
运行命令
composer require dirape/token
In your controller use
在您的控制器中使用
use Dirape\Token\Token;
You can use it like this:
你可以这样使用它：
User::create([
'name' => $data['name'],
'email' => $data['email'],
'password' => bcrypt($data['password']),
'token_key' => (new Token())->Unique('users', 'api_token', 60),
'active' => 1
])
回答by daisura99
Following @ivanhoe suggestion... this is what I came up with:-
遵循@ivanhoe 的建议……这就是我想出的：-
$token = new Token;
//in case there are duplicate
for ($x = 0; $x < 10; $x ++) {
$token->access_token = str_random(16);;
try {
if ($token->save()) {
break;
}
}catch (QueryException $e) {
}
}
