Hi everyone, I know I am new and I am also quite new to web-design so if you would bare with me... I am trying to run a script on my website (Quite simple) The script must search my database and find all 12 rows of information on the query I enter... The query I enter could be located in any of the columns and there may be more than one result... My columns are The error from what I can understand is located around line 35 ("$numrows=mysql_num_rows($numresults);") and then even if I delete that I get no results... Here is my HTML: <html> <head> </head> <body> <form name="form" action="search.php" method="get"> <input type="text" name="q" /> <input type="submit" name="Submit" value="Search" /> </form> </body> </html> HTML: and here is my PHP: <?php // Get the search variable from URL $var = @$_GET['q'] ; $trimmed = trim($var); //trim whitespace from the stored variable // rows to return $limit=10; // check for an empty string and display a message. if ($trimmed == "") { echo "<p>Please enter a search...</p>"; exit; } // check for a search parameter if (!isset($var)) { echo "<p>We dont seem to have a search parameter!</p>"; exit; } //connect to your database ** EDIT REQUIRED HERE ** mysql_connect("localhost","user1234","password1234"); //(host, username, password) //specify database ** EDIT REQUIRED HERE ** mysql_select_db("database1234") or die("Unable to select database"); //select which database we're using // Build SQL Query $query = "select * from Table1234 where SURNAME like '$trimmed'"; // EDIT HERE and specify your table and field names for the SQL query $numresults=mysql_query($query); $numrows=mysql_num_rows($numresults); // If we have no results, offer a google search as an alternative if ($numrows == 0) { echo "<h4>Results</h4>"; echo "<p>Sorry, your search: "" . $trimmed . "" returned zero results</p>"; // google echo "<p><a href=\"http://www.google.com/search?q=" . $trimmed . "\" target=\"_blank\" title=\"Look up " . $trimmed . " on Google\">Click here</a> to try the search on google</p>"; } // next determine if s has been passed to script, if not use 0 if (empty($s)) { $s=0; } // get results $result = mysql_query($query) or die("Couldn't execute query"); // display what the person searched for echo "<p>You searched for: "" . $var . ""</p>"; // begin to show results set echo "Results"; $count = 1 + $s ; // now you can display the results returned while ($row= mysql_fetch_array($result)) { $name = $row[1]; $staff_no = $row[2]; $Position = $row[3]; $pass_no = $row[4]; $dob = $row[5]; echo "$count.) $name, $staff_no, $Position, $pass_no, $dob" ; $count++ ; } $currPage = (($s/$limit) + 1); //break before paging echo "<br />"; // next we need to do the links to other results if ($s>=1) { // bypass PREV link if s is 0 $prevs=($s-$limit); print " <a href=\"$PHP_SELF?s=$prevs&q=$var\"><< Prev 10</a>  "; } // calculate number of pages needing links $pages=intval($numrows/$limit); // $pages now contains int of pages needed unless there is a remainder from division if ($numrows%$limit) { // has remainder so add one page $pages++; } // check to see if last page if (!((($s+$limit)/$limit)==$pages) && $pages!=1) { // not last page so give NEXT link $news=$s+$limit; echo " <a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 >></a>"; } $a = $s + ($limit) ; if ($a > $numrows) { $a = $numrows ; } $b = $s + 1 ; echo "<p>Showing results $b to $a of $numrows</p>"; ?> PHP: Thank you in advance,
The replies took too long sorry - I have had to use another script in-order to finish the project intime... Thank you for the reply though...