my php code shows as text

Question

I'm trying to create a search on a website I'm making for homework with a database called animal that is saved on an easy php server. However, when I run the search, the php code shows up as text. Can someone please help me with this?

    <?php
    $criteria = $_GET["criteria"];
    $Field = $_GET["animal"];
    $link = mysql_connect('127.0.0.1', 'criteria', 'animal');
    if (!$link) {
        die('Could not connect: ' . mysql_error());
    }
    echo 'Connected successfully';
    $result = array(connection -> query("SELECT * FROM animal WHERE $Field Like %criteria%"));
    mysql_close($link);
    echo $result;
?>

You are also setting the $result = the query then echo $result right after. If you are planning on executing that query you need to call a method to do so first. — Jacob H, May 23 '17 at 17:39
Possible duplicate of [PHP code is not being executed, instead code shows on the page](https://stackoverflow.com/questions/5121495/php-code-is-not-being-executed-instead-code-shows-on-the-page) — Tom Udding, May 23 '17 at 17:42
You also have a space in your opening php tag ` php` should just be ` — Jonathan Kuhn, May 23 '17 at 17:46

score 2 · Answer 1 · edited May 23 '17 at 17:46

2

I see that you have a space between the <? and php in the opening php tag.

Remove the space to use php tags. It should be <?php.

edited May 23 '17 at 17:46

Jonathan Kuhn

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answered May 23 '17 at 17:45

Erik

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score 1 · Answer 2 · answered May 23 '17 at 17:40

1

    $result = array(connection -> query("SELECT * animal WHERE $Field Like %criteria%"));

should be

    $result = array(connection -> query("SELECT * FROM animal WHERE $Field Like %$criteria%"));

answered May 23 '17 at 17:40

Omis Brown

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my php code shows as text

2 Answers2