CODEWITHSUNDEEP
php
user5248362
user5248362

Reputation: 1

why in this isset and !empty if block else statement executing by default?

<?php
if( isset($_POST['pass']) && !empty($_POST['pass']) )
{
    $pass = $_POST['pass'];
    echo 'ok';
} else {
    echo 'Enter Password';
}
?>
<form action="md.php" method="POST">
Password: <input type="password" name="pass"><br><br>
<input type="submit" value="Submit">
</form>

The problem is that when I run this code then by default I see the message Enter Password displayed above the password field somehow the else condition is getting executed when I open this page on xampp server that else block is supposed to execute when the user submits empty form and not when I open the page on the server.

Upvotes: 0

Views: 118

Answers (4)

RiggsFolly
RiggsFolly

Reputation: 94642

The first time you run this form, from entering the url in a browser address bar of clicking on a link or menu item on another page, the $_POST array will be empty.

So your code will cause the else to be run and echo 'Enter Password'; will be executed

When you press the Submit button, assuming you enter a password, then the $_POST will exist and $_POST['pass'] will contain a value and you should see the ok message. If you do not enter anything in the pass password field you will of course see the echo 'Enter Password'; again

If you only want to see the message when you have forgotten to enter password, but pressed the Submot button, which is what I assume you are saying then try this change

<?php
if ( $_SERVER["REQUEST_METHOD"] == 'POST' ) {
    // only set when the form was actually posted
    // not set when you run form from a link or the address bar

    if( !empty($_POST['pass']) ) {
        $pass = $_POST['pass'];
        echo 'ok';
    } else {
        // pressed submit without entering a password
        echo 'Enter Password';
    }

}
?>
<form action="md.php" method="POST">
Password: <input type="password" name="pass"><br><br>
<input type="submit" value="Submit">
</form>

Upvotes: 3

DirtyBit
DirtyBit

Reputation: 16772

The first time the page is opened, it gives you the else block because the if statement is false. So edit the else-if statement as describe below, Furthermore, leave the action = "" So it doesn't direct you to another page when you press the submit button. Try this:

<?php
if( isset($_POST['pass']) && !empty($_POST['pass']) )
{
       $pass = $_POST['pass'];
    echo 'ok';
} else if( isset($_POST['pass']) && empty($_POST['pass']) ) {
    echo 'Enter Password';
}
?>
<form action="" method="POST">
Password: <input type="password" name="pass"><br><br>
<input type="submit" value="Submit">
</form>

The above code should display the form, the very first time you open it and accordingly as you input/leave the password field afterwards.

Upvotes: 0

phpchap
phpchap

Reputation: 382

Try changing your PHP code to:

<?php
if( isset($_POST['submit'])
{
    if( isset($_POST['pass']) && !empty($_POST['pass']) )
    {
        $pass = $_POST['pass'];
        echo 'ok';
    } else {
        echo 'Enter Password';
    }
}
?>

This way the code will only run once you click the Submit button and not when you load the page.

Upvotes: 0

Halcyon
Halcyon

Reputation: 57709

Change the else to:

} else if (isset($_POST["pass"]) && empty($_POST["pass"])) {

You code should have 3 paths:

  • There is a password and it's not empty (show ok + form)
  • There is a password and it's empty (show entry pw + form)
  • There is no password (show form)

Right now you only have 2 paths.

Upvotes: 0

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