CODEWITHSUNDEEP
php
cadavid4j
cadavid4j

Reputation: 171

PHP will not execute properly and only displays as text

My entire PHP page only displays as text and no PHP code is executed. It's weird because when I test it using <? phpinfo(); ?> in a test.php file, I get a successful test and it works on my Apache server. However when I attempt to do anything else. It only shows as text.

Edit: Here is the link to the code. I couldn't figure out how to post it here. Pastebin

<?php
  // create short variable names
  $tireqty = $_POST['tireqty'];
  $oilqty = $_POST['oilqty'];
  $sparkqty = $_POST['sparkqty'];
  $find = $_POST['find'];
?>
<html>
<head>
  <title>Bob's Auto Parts - Order Results</title>
</head>
<body>
<h1>Bob's Auto Parts</h1>
<h2>Order Results</h2>
<?php

    echo "<p>Order processed at ".date('H:i, jS F Y')."</p>";

    echo "<p>Your order is as follows: </p>";

    $totalqty = 0;
    $totalqty = $tireqty + $oilqty + $sparkqty;
    echo "Items ordered: ".$totalqty."<br />";


    if ($totalqty == 0) {

      echo "You did not order anything on the previous page!<br />";

    } else {

      if ($tireqty > 0) {
        echo $tireqty." tires<br />";
      }

      if ($oilqty > 0) {
        echo $oilqty." bottles of oil<br />";
      }

      if ($sparkqty > 0) {
        echo $sparkqty." spark plugs<br />";
      }
    }


    $totalamount = 0.00;

    define('TIREPRICE', 100);
    define('OILPRICE', 10);
    define('SPARKPRICE', 4);

    $totalamount = $tireqty * TIREPRICE
                 + $oilqty * OILPRICE
                 + $sparkqty * SPARKPRICE;

    echo "Subtotal: $".number_format($totalamount,2)."<br />";

    $taxrate = 0.10;  // local sales tax is 10%
    $totalamount = $totalamount * (1 + $taxrate);
    echo "Total including tax: $".number_format($totalamount,2)."<br />";

    if($find == "a") {
      echo "<p>Regular customer.</p>";
    } elseif($find == "b") {
      echo "<p>Customer referred by TV advert.</p>";
    } elseif($find == "c") {
      echo "<p>Customer referred by phone directory.</p>";
    } elseif($find == "d") {
      echo "<p>Customer referred by word of mouth.</p>";
    } else {
      echo "<p>We do not know how this customer found us.</p>";
    }

?>
</body>
</html>

Upvotes: 2

Views: 6931

Answers (1)

Alex
Alex

Reputation: 14618

I am willing to bet 10$ that test.php uses <?php, and the changed code uses <?, while the server does not understand it as an opening tag since short_open_tags is off in php.ini.

A lot of books use <? for open tags, while most servers only support the long version (<?php). If that's the case, then changing all the simple <? to <?php will do the trick.

PHP code is only exposed in 1 case: When PHP interpreter does not identify it as PHP code. That can be caused by only 2 problems:

  • Wrong configuration of Apache (or other http server) which doesn't handle php files at all.
  • Wrong open tags in PHP files, so PHP doesn't know when code begins.

If the file is a *.php, if Apache is turned on serving *.php files through PHP interpreter, if standard open tags are used or PHP is configured to use other types of used tags, and if you're accessing this PHP file through the browser, in no circumstances would PHP expose this code.

Upvotes: 5

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