{if...else} how to display an image if a mysql field equals ex.1

i have a question: i have a mysql table that has values such as pages_viewed that increments every time a user accesses a page such as 1,2,3 etc. My question is simple how do i display an image when the field equals or is more than 1?

like

select pages_viewed from table;
if field pages_viewed > 1 than echo images/image.png

开发者_开发知识库

pardon my php i'm not too good at it.

i've been reading here and in other forums but all i can find is drupal or wordpress.

---

something like

header('Content-Type: image/png')
echo file_get_contents(images/image.png)

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Here is everything basically short of database connection, table name, field name.

mysql_connect("localhost", "mysql_user", "mysql_password") or die("Could not connect: " . mysql_error());
mysql_select_db("mydb");

$result = mysql_query("select * from table");

while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
    # Any other code to post the page information
    if($row['pages_viewed'] > 0) echo "<img src='images/image.png' />";
    # Any other code to post the page information
}