qmai Posted July 25, 2012 Share Posted July 25, 2012 <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title>testing</title> </head> <body class="holder"> <center> <form action='results.php' method='get'> <input type='text' name='input' size='50' value='<?php echo $_GET['input']; ?>' class="search-field" /> <input type='submit' value='Search' class="seach-button"> </form> </center> <br/> <?php $input = $_GET['input'];//Note to self $input in the name of the search feild $terms = explode(" ", $input); $query = "SELECT * FROM table1 WHERE "; foreach ($terms as $each){ $i++; if ($i == 1) $query .= "Firstname LIKE '%$each%' "; else $query .= "OR Firstname LIKE '%$each%' "; } // connecting to our mysql database mysql_connect('localhost', 'root', ''); mysql_select_db("database1"); $query = mysql_query($query); $numrows = mysql_num_rows($query); if ($numrows > 0){ while ($row = mysql_fetch_assoc($query)){ $id = $row['ID']; $Firstname = $row['Firstname']; $Lastname = $row['Lastname']; $Year = $row['Year']; $Course = $row['Course']; echo "<h2>$Firstname $Lastname $Year $Course</h2> <br /><br />"; } } else echo "No results found for \"<b>$input</b>\""; // disconnect mysql_close(); ?> </body> </html> im getting ' Undefined variable: i in line 23 ' ( $i++; ) can anyone help me solve this? Thanks! Quote Link to comment Share on other sites More sharing options...
Barand Posted July 25, 2012 Share Posted July 25, 2012 put $i = 0; before the foreach loop Quote Link to comment Share on other sites More sharing options...
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