Use of undefined constant foundnum line 54

[balam]

hi, can someone help me with this error pls. the error is Use of undefined constant in line 54. what is it thats not defined pls


<?php

include ('connect.php');

error_reporting(E_ALL);
ini_set('display_errors', '1');

$submit = $_GET['submit'];
$search = $_GET['search'];
$x=0;
$construct='';
$foundnum="";

if (!'submit')

echo "you didnt submit a keyword.";

else

{

if (strlen($search)<=2)

echo "search term to short.";
else
{
echo " You searched for <b>$search</b><hr size='1'>";

//connect to our database

$search_exploded = explode(" ",$search);


foreach($search_exploded as $search_each)

{

// construct query

$x++;
if ($x==1)
$construct .= " location LIKE '%$search_each%'";
else
$construct .= " OR rent LIKE '%$search_each%'";

}

// echo out construct

$construct = "SELECT * FROM flats WHERE $construct";
$run = mysql_query($construct);
$foundnum = mysql_num_rows($run);

if (foundnum==0)
echo "No results found.";
else
{
echo "$foundnum result found!<p>";

while ($runrows = mysql_fetch_assoc($run))

{

// get data

$select = $runrows['type'];
$title = $runrows['title'];
$location = $runrows['location'];


echo "

echo $title;
echo <br>;
echo $select;
echo <br>;
echo $rent;
echo <br>;
echo $location";

}


}
}
}


?>

[angelwatt]

Please use the PHP code block when posting code to improve readability.

The error concerns this line,

PHP Code:

if (foundnum==0) 


See the problem? Missing $ in front of variable name.

[SrWebDeveloper]

Also...

PHP Code:

if (!'submit') 


Should be $submit