I am trying to write an if statement for image in my database. Example: if no image is relating to the recipeid do not post a blank image box leave blank as if there is nothing there..only show image for recipeid that has a picture.
This is my code that I have but I can not figure how to write the if code:
<?php
$con = mysql_connect("localhost", "test", "test") or die('Could not connect to server');
mysql_select_db("recipe", $con) or die('Could not connect to database');
$recipeid = $_GET['id'];
$query = "SELECT title,poster,shortdesc,ingredients,directions, calories, picture from recipes where recipeid = $recipeid";
$result = mysql_query($query) or die('Could not find recipe');
$row = mysql_fetch_array($result, MYSQL_ASSOC) or die('No records retrieved');
$title = $row['title'];
$poster = $row['poster'];
$shortdesc = $row['shortdesc'];
$ingredients = $row['ingredients'];
$directions = $row['directions'];
$calories = $row['calories'];
$picture = $row['thumbnail'];
$ingredients = nl2br($ingredients);
$directions = nl2br($directions);
echo "<h2>$title</h2>\n";
echo "by $poster <br><br>\n";
echo $shortdesc . "<br><br>\n";
echo "<h3>Ingredients:</h3>\n";
echo $ingredients . "<br><br>\n";
echo "<h3>Directions:</h3>\n";
echo $directions . "\n";
echo "<h3>Calories:</h3>\n";
echo $calories . "<br><br>\n";
echo "<img src=\"showimage.php?id=$recipeid\" border=\"0\" ></a>\n";
echo "<br><br>\n";
?>