用户单击提交按钮时如何调用两个不同的网页？

Question

My code looks likes this:

<form action="register.php" method="post">
  <input type="text" name="uname">
  <input type="submit" >
</form>

In register.php file I have a database connection codes. I need to load another html page after submitting the form simultaneously if the database is connected through the register.php file.

How to do that?

Answer 1

Simply redirect the user to the second page once the registration is successful, through means of the header() function:

index.php:

<form action="register.php" method="post">
  <input type="text" name="uname">
  <input type="submit" >
</form>

register.php:

// Process registration
if (valid) {
  header('success.php);
}

success.php:

// Thank the user for registering

Alternatively, you could always simply process your success page on register.php itself.

Hope this helps! :)

Answer 2

Welcome to SO. It sounds to me like you don't actually need to load 2 URLs simultaneously and what you are looking for is an if statement to execute some code only if post data is set

if(!empty($_POST)){
  //code to execute when form is submitted
}
else {
  //code to execute otherwise
}

If you do need to make a request to 2 URLs at the same time you are looking for an AJAX request.

More detail in your question about your setup and end goal would be useful to help further.

Answer 3

in your Register.Php add something like this check your Sql result something like this

$sql='select * Tablename where condition'

$result= fetch($sql)
$row=fetch_array($result)asdsad
if($row>0){
header(Location:'connected.php') // whatever your html
}
else {header(Location:"failed.php")//whenever query got no row
}