Random number generator game in PHP using a form

Question

I am trying to create a webpage that prompts the user to guess a number from 1 to 5 which is randomly generated and stored in the form.

Then, when the form is submitted, the guess is checked against the random number and if the guess is incorrect, the user is prompted to guess again but the random value stays the same with each guess.

My issue is that once the user submits the form, the random number is not showing up when the user is prompted to try again.

else if (isset($_POST["firstname"])) {
    echo $_POST['firstname'] . $_POST['lastname'];
    echo "Hi " . $_POST['firstname'] . " " . $_POST['lastname'] . "!";
    ?>
    <html>
        <form name="number" action="random.php" method="post">
            <p>Enter a guess: <input type="text" name="guess" /></p>
            <input type="hidden" name="numtobeguessed" value="<?php echo $_POST["numtobeguessed"]; ?>" ></p>
            <input type="submit" value="Guess!" />
        </form>
    </html>
    <?php
    $_POST["numtobeguessed"] = rand(1, 5);
    $guessed = $_POST["numtobeguessed"];
    echo "Number to be guessed " . $guessed;

} else if (isset($_POST["guess"])) {
    if ($_POST["guess"] != $_POST["guessed"]) {
        echo $_POST["guess"] . " is not correct";
        ?>
        <form name="number1" action="random.php" method="post">
            <p>Enter a guess: <input type="text" name="guess" /></p>
            <input type="hidden" name="numtobeguessed" value="<?php echo $_POST["guessed"]; ?>" ></p>
            <input type="submit" value="Guess!" />
            <?php
        }
    }
    ?>

On a side note, I'd recommend using $_POST instead of $_REQUEST. I don't know why you use both. Also, you do not need the == true in the if statements. — Jonathan Lam
– Jonathan Lam, Commented Dec 3, 2015 at 2:37
Why do you start your code with an else if? Is there code before this that may be interfering? — Jonathan Lam
– Jonathan Lam, Commented Dec 3, 2015 at 2:39
@Jonathan There is code before this that does work. I didn't include it to save space. The code I've posted is where I'm having the issue. — ilikecake1
– ilikecake1, Commented Dec 3, 2015 at 2:54
Should the logic always check the $_POST["guess"] if it has a value? i.e when $_POST['numtobeguessed'] has a value then the guess can have a value entered by the user? — Ryan Vincent
– Ryan Vincent, Commented Dec 3, 2015 at 3:23

Guffa · Accepted Answer · 2016-02-18 11:12:05Z

1

You are sending the value in a field with the name numtobeguessed but you try to read it with the name guessed.

This:

if ($_POST["guess"] != $_POST["guessed"]) {

should be:

if ($_POST["guess"] != $_POST["numtobeguessed"]) {

and this:

<input type="hidden" name="numtobeguessed" value="<?php echo $_POST["guessed"]; ?>" ></p>

should be:

<input type="hidden" name="numtobeguessed" value="<?php echo $_POST["numtobeguessed"]; ?>" ></p>

answered Feb 18, 2016 at 11:12

Guffa

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