PHP搜索生成错误

Question

I have a search field feature on my website, with below PHP code:

if(isset($_GET['search_ti']) && $_GET['search_ti'] != "")
{
    $search_ti = preg_replace('#[^a-z 0-9?!-]#i', '', $_GET['search_ti']);

    $sqlCommand = "
    SELECT * FROM page WHERE title LIKE '%$search_ti%' OR body LIKE '%$search_ti%'
    ";

    $result = $mysqli->query($sqlCommand) or die(mysql_error());
    ...

The URL variable search_ti is the keyword for the search, which I got from the URL. For example:

searchResult.php?search_ti=home

If the keyword is found in my MySQL table, the result shows and no errors.

But if the keyword is not found, the web browser generates below error:

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '-4,4' at line 1

What is wrong with my code?

ADDITIONAL INFO:

The complete PHP code: http://jsfiddle.net/eQSZc/

The keyword: home. The error only shows if the keyword is not found in my table.

Answer 1

Change your pagination_mechanism_and_buttons_variables_for_Search.php to the one in http://jsfiddle.net/GsNmw/1/

I have added the following block just before you construct the $limit and change the $limit as well.

$start = ($pn - 1) * $itemsPerPage;

if($start<0){
    $start = 0;
}

$limit = 'LIMIT ' .$start .',' .$itemsPerPage;

This avoids negative limit and hence avoids your error.