php'ISSET'功能不起作用。 或者代码跳过我的if语句

The not working and i get the same error output as defined in the last else statement.

I am new to Php, please help me out. What am I doing wrong?

HERE IS THE CODE

<form action="imagespace.php?groupid=<?php echo $_GET['groupid'] ;?>" method = "POST">
<input type="button" name="Up-Vote" value ="Up-Vote" class="up-votes" />
<input type="button" name="Down-Vote" value="Down-Vote" class="down-votes" />
</form>
<?php

$emailid=$_SESSION['emailid'];
$groupid = $_GET['groupid'];
$emailid = mysqli_real_escape_string($con,$emailid);
$groupid = mysqli_real_escape_string($con,$emailid);

if(isset($_POST['Up-Vote']))
    { echo ' trial';
    $query = "INSERT into VOTES (emailid, groupid, vote) values ('$emailid','$groupid', '1') ";
    $query_run = mysqli_query($con,$query);
    if($query_run)
    {echo 'upvoted';}
    }

else if(isset($_POST['Down-Vote']))
    {
        $query = "INSERT into VOTES (emailid, groupid, vote) values ('$emailid','$groupid', '0') ";
        $query_run = mysqli_query($con,$query);
        if($query_run)
        {echo 'downvoted';}
    }
else{
echo 'error' ;
}
}
?>

</body>
</html>

In addition to the supplied answer(s), I would like to suggest doin the following:

// Before anything else, start your session
if (!isset($_SESSION)) {
    session_start();
}

// Next, check if your form is actually submitted
if ($_SERVER['REQUEST_METHOD'] == 'POST') {
    if (isset($_POST['...'])) {

    } elseif (isset($_POST['...'])) {

    }
}
?>

<!-- Keep PHP and html seperated -->
<form>
...
</form>

By looking at your form, i notice that you are using type='button'. I suggest one of the following:

<input type="submit" ... />

<button type="submit">...</button>

I can't seem to find a submit button in there. Try adding

<input type="submit" />

right before the closing </form>-tag. Then try if the error still occurs if you click on it.

Also, try using checkboxes/radio buttons instead of buttons. I don't know whether it is possible to do that with buttons.

The extra submit button isn't needed, the <button>s act same in default type.