Hi,
i’m new to php,html and mysql. i have been assign by my teacher to do the below website
i have a html page that call for php function and retrieve data from mysql. i did manage to do it somehow. But it bothers me when the result of the search is display on another blank page using my php.
So, now can i display the result on my search.html?
Below is my code
in my html
[code]
In my php
[php]<?php
// Get the search variable from URL
$var = @$_GET[‘term’] ;
$trimmed = trim($var); //trim whitespace from the stored variable
// rows to return
$limit=10;
// check for an empty string and display a message.
if ($trimmed == “”)
{
echo “
Please enter a search…
”;exit;
}
// check for a search parameter
if (!isset($var))
{
echo “
We dont seem to have a search parameter!
”;exit;
}
//connect to your database ** EDIT REQUIRED HERE **
mysql_connect(“localhost”,“root”,“shaun830511”); //(host, username, password)
//specify database ** EDIT REQUIRED HERE **
mysql_select_db(“mysql”) or die(“Unable to select database”); //select which database we’re using
// Build SQL Query
$query = “select * from car where class like “%$trimmed%”
order by type”; // 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 “
Results
”;echo “
Sorry, your search: “” . $trimmed . “” returned zero results
”;// google
echo “
<a href=“http://www.google.com/search?q=”
. $trimmed . “” target=”_blank" title=“Look up
" . $trimmed . " on Google”>Click here to try the
search on google
}
// next determine if s has been passed to script, if not use 0
if (empty($s)) {
$s=0;
}
// get results
$query .= " limit $s,$limit";
$result = mysql_query($query) or die(“Couldn’t execute query”);
// display what the person searched for
echo “
You searched for: “” . $var . “”
”;// begin to show results set
echo “Results”;
$count = 1 + $s ;
// now you can display the results returned
while ($row= mysql_fetch_array($result)) {
echo 'No: '.$row[‘id’];
echo '
type: '.$row[‘class’];
echo ’
category: '.$row[‘cat’];
echo ’
brand: '.$row[‘brand’];
echo ’
year: '.$row[‘year’];
echo ’
colour: '.$row[‘colour’];
echo ’
mileage: '.$row[‘mileage’];
echo ’
note: '.$row[‘note’];
echo ’
description: '.$row[‘descp’];
echo ’
path: ‘.$row[‘name’];
echo ‘
’;
echo “
| <img src='images/” .$row[‘name’]."’> | ";
echo “$count.) $title” ;
$count++ ;
}
$currPage = (($s/$limit) + 1);
//break before paging
echo “
”;
// next we need to do the links to other results
if ($s>=1) { // bypass PREV link if s is 0
$prevs=($s-$limit);
print " <a href="$PHP_SELF?s=$prevs&q=$var"><<
Prev 10  ";
}
// 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 = $s + ($limit) ;
if ($a > $numrows) { $a = $numrows ; }
$b = $s + 1 ;
echo “
Showing results $b to $a of $numrows
”;?>[/php]
May i know what is the option and how i going to make the result display on the same page as on my html. i have tough luck on getting it for hours. been look through many tutorial but still cant get it.
Please advice.
TQ