php echoing images based on $answer [closed] - php

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This is a university project and I have a page will a php calculator where users enter '$valuea' which is likely to be between 0-550 then they will hit submit which subtracts '$valueb' which is 250, giving them '$answer'.
What I need help with is I was to have a particular image echo out once the user hit submit, which depends on the figure of $answer.
Here's what I have so far but it is not working;
<?php
if (isset($_POST['valuea'])) $valuea = $_POST['valuea'];
if (isset($_POST['valueb'])) $valueb = '250';
$answer = $valuea - $valueb;
echo <<<_END
<form method='post' action='calculator.php'>
<table border='0' width='500px' cellpadding='3' cellspacing='1' class="table">
<tr class="calcheading"><td colspan="2"><strong>Work out how much you could be saving</strong></td></tr>
<tr class="calcrow"><td>How much do you spend a year?</td><td align="center"><input type='text' name='valuea' value="$valuea"/></td></tr>
<tr class="calcrow"><td>Minus the average price of an n-power student tarrif* Leave Blank:</td><td align="center"><input type='text' name='valueb' value="$valueb"/></td></tr>
<tr class="submit"><td colspan="2"><input type='submit' value='Calculate'/></td></tr>
_END;
?>
<tr class="calcrow">
<td><i>You could be saving:</td>
<td align="center"><input type="text" value="<?php echo round($answer)?>"></td></i>
</tr>
</table>
</form>
<?php
if($db_server){
$image1 = mysqli_query("SELECT URL FROM images WHERE imagename='image1');"
$image2 = mysqli_query("SELECT URL FROM images WHERE imagename='image2');"
$image3 = mysqli_query("SELECT URL FROM images WHERE imagename='image3');"
$image4 = mysqli_query("SELECT URL FROM images WHERE imagename='image4');"
$image5 = mysqli_query("SELECT URL FROM images WHERE imagename='image5');"
if( $answer > 0 and $answer < 150 )?>
<img src="<?php echo $image1;?>"/>
else if( $answer < 250)
<img src="<?php echo $image2;?>"/>
else if( $answer < 350)
<img src="<?php echo $image3;?>"/>
else if( $answer < 450)
<img src="<?php echo $image4;?>"/>
else if( $answer < 550)
<img src="<?php echo $image5;?>"/>
<?php
else
echo 'error';
}
?>

Answer 1

Your code is a real mess. I've cleaned it a bit.
<?php
$valuea = (isset($_POST['valuea']) && is_numeric($_POST['valuea'])) ? $_POST['valuea'] : 0;
$valueb = 250;
$answer = $valuea - $valueb;
?>
<form method='post' action='calculator.php'>
<table border='0' width='500px' cellpadding='3' cellspacing='1' class="table">
<tr class="calcheading">
<td colspan="2"><strong>Work out how much you could be saving</strong></td>
</tr>
<tr class="calcrow">
<td>How much do you spend a year?</td>
<td align="center"><input type='text' name='valuea' value="$valuea"/></td>
</tr>
<tr class="calcrow">
<td>Minus the average price of an n-power student tarrif* Leave Blank:</td>
<td align="center"><input type='text' name='valueb' value="$valueb"/></td>
</tr>
<tr class="submit">
<td colspan="2"><input type='submit' value='Calculate'/></td>
</tr>
<tr class="calcrow">
<td><i>You could be saving:</td>
<td align="center"><input type="text" value="<?php echo round($answer)?>"></td></i>
</tr>
</table>
</form>
<?php
if($db_server){
switch( $answer ){
case $answer > 0 and $answer < 150 : $image = mysqli_query("SELECT URL FROM images WHERE imagename = 'image1'");
break;
case $answer < 250 : $image = mysqli_query("SELECT URL FROM images WHERE imagename = 'image2'");
break;
case $answer < 350 : $image = mysqli_query("SELECT URL FROM images WHERE imagename = 'image3'");
break;
case $answer < 450 : $image = mysqli_query("SELECT URL FROM images WHERE imagename = 'image4'");
break;
case $answer < 550 : $image = mysqli_query("SELECT URL FROM images WHERE imagename = 'image5'");
break;
}
echo "<img src='$image' alt='' />";
}
?>