get result in same page and form - php

Question

hello i have simple problem with my functions there is 2 codes :
i need when i put the right input answer echo $row['answer']; and sucess in next code.
it means i need when $_POST['answer'] = $row['answer']; echo my password
can someone help me with if cond.
this code is working perfect :
<?php
$username = $_POST['username'];
include('config.php');
$result = mysqli_query($con,"SELECT * FROM persons WHERE username='$username'");
while($row = mysqli_fetch_array($result)){
echo $row['username'];
echo "</br>";
echo "</br>";
echo "<p><b>Secret Question</b></p>";
echo $row['secret'];
$freeanswer = $row['answer'];
}
?>
code problem in same page :
</br>
</br>
<form action="" method="POST">
<p><b>Answer is :</b><p>
<input type="text" name="answer">
</br>
</br>
<input type="Submit" value="Submit">
</form>
</div>
<?php
$anme = $_POST['answer'];
if ($anme==$freeanswer) {
echo "sucess";
} else {
echo "wrong";
}
?>

Answer 1

Do you have problem sending the data?
If you want to post the form data to the same page you could use this:
<form action="<?php echo $_SERVER['REQUEST_URI'];?>" method="post">
And this code will catch whatever you posted. (Put it in the same page as the form)
<?php
if(isset($_POST['answer'])) {;
if($name==$freeanswer) {
echo "success";
} else {
echo "wrong";
}
} else {
// Just for testing
var_dump("No data sent yet");
}
?>

Answer 2

Where is the $freeanswer variable declared in the second code?
This if statement is failing cause there is no $freeanswer declared.
$anme = $_POST['answer'];
if ($anme==$freeanswer) { ..
I see you declared it in first code but I dont see it in the second code..
EDIT :
Actually I didn't see It's all the same code. But still the problem is $freeanswer cause It's not declared in public scope It's only in the scope of while loop
Try declaring / writing this line of code here :
$username = $_POST['username']; // Near this line
$freeanswer = ""; // Declare this here

Answer 3

You should put a name to submit button, and the check the existence of $_POST['name_of_button'];
This is a dirty example I wrote quickly so you can get the idea, I didn't test it
<?php
// if the button has not been sent show the form
if(!isset($_POST['submit'])) {
?>
<form action="/test.php" method="POST">
<p><b>Answer is :</b><p>
<input type="text" name="answer">
</br>
</br>
<input type="Submit" value="Submit" name="submit">
</form>
</div>
<?php
// otherwise means the form has been sent, show the password or wrong
} else {
include('config.php');
$result = mysqli_query($con,"SELECT * FROM persons WHERE username='$username'");
$row = mysqli_fetch_array($result);
if($row) {
if($row['answer'] == $_POST['answer']) {
echo 'your password is ' . $row['password'];
} else {
echo 'wrong!';
}
}
}