use of undefined constant
Posted: Mon Dec 14, 2009 11:53 am
pickle | Please use [ code=php ], [ code=text ], etc tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read:
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hi, can someone help me with this error pls. the error is Use of undefined constant in line 54. Have i not defined its on top of the script foundnum=" ";??????
pickle | Please use [ code=php ], [ code=text ], etc tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read:
Posting Code in the Forums to learn how to do it too.
hi, can someone help me with this error pls. the error is Use of undefined constant in line 54. Have i not defined its on top of the script foundnum=" ";??????
Code: Select all
<?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";
}
}
}
}
?>pickle | Please use [ code=php ], [ code=text ], etc tags where appropriate when posting code. Your post has been edited to reflect how we'd like it posted. Please read: