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Well, when registering the user has its ACTIVE set to 0, how to send an email with a link so that when clicking this link the user updates the asset to 1.
1 answer
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I Gero a token and saved it on my user, then I search in the database the user that contains this token, so you recover the user without much work. To generate the token I use the following methods
public function onRegistrationSuccess($user)
{
$user->setEnabled(false);
if (null === $user->getConfirmationToken()) {
$user->setConfirmationToken($this->generateToken());
}
$this->sendConfirmationEmailMessage($user);
}
public function generateToken()
{
return rtrim(strtr(base64_encode($this->getRandomNumber()), '+/', '-_'), '=');
}
private function getRandomNumber()
{
return hash('sha256', uniqid(mt_rand(), true), true);
}
You can use openssl to change the randomNumber method to this one
private function getRandomNumber()
{
$nbBytes = 32;
$bytes = openssl_random_pseudo_bytes($nbBytes, $strong);
if (false !== $bytes && true === $strong) {
return $bytes;
}
throw new \Exception('OpenSSL não produziu um número aleatório seguro.');
}
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I do it on the page? $token = generateToken(); and $Numero_aleatorio =getRandomNumber();
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But then I do the same process of my previous code to send it by email as link?
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perform the record you generate the token saved it to the user (in a column in the user table) and then send this token to the user’s email as a link (example.com?token=d8s9dad8as). When the user clicks the link he will be taken to your site, recover it from the bank, searching for the user who has the same token and activate it
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@Iwannaknow I added another method to explain better
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Before Exception is a "same"?
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I got it!!! But with the first getRamdomNumber function (do you have a problem using it?). Does this second one have PDO? You’re making a mistake because of public/private and this.
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1
Sorry, I didn’t take the scope of the method, is that I use this in an OO approach, can remove them. About "" is to pick up the object in a global namespace
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What’s the mistake? What’s the problem? Try to describe the situation better.
– rray
Cara Voce didn’t post many details of the problems but I think I know what’s going on. In mysql functions, like mysql_query, it was optional to send or not your connection variable. Already in the mysqli functions, you have to send the connection variable if not from Warning, try to do like this: mysqi_query($sql, $con);
– Euler01
Check whether the variable
linkhas the connection, tbm put the part of the code where you callmysqli_insert_id– rray
The link variable has the connection.
– I Wanna Know
And the code of the Insert?
– rray
That must be the error then because I have nothing from Insert other than that line, but that’s how it worked before with mysql.
– I Wanna Know
If you were able to solve the problem create an answer and explain what you had to do, when you have a problem other than the original question you better add a new one as a reference.
– rray
All right, I’ll fix it.
– I Wanna Know