Suggestions de recherche PHP

Question

J'ai un script PHP qui fonctionne avec jQuery pour fournir des suggestions de recherche. Il tire les résultats d'une base de données MySQL. Cependant, je veux seulement 5 résultats à afficher à la fois pour la lettre que l'utilisateur a tapé mais il semble que tous les résultats apparaissent. Pourquoi cela pourrait-il être?

Mon code est:

<p id="searchresults"><?php 

$db=new mysqli('localhost','username','password','database'); 

if(isset($_POST['queryString'])){ 
$queryString=$db->real_escape_string($_POST['queryString']); 
      if(strlen($queryString)>0){ 
       $query = $db->query("SELECT * FROM search s WHERE name LIKE '%" . $queryString . "%'"); 
       if($query){ 
        while ($result = $query ->fetch_object()){ 
         echo '<a href="/search/'.$result->name.'/1/">';      
         $name=$result->name;    
         echo ''.$name.''; 
        } 
       } 
      } 
     } 
?></p> 

J'espère que vous pouvez comprendre ce que je suis en train de décrire.

Answer 1

changement "SELECT * FROM search s WHERE name LIKE '%" . $queryString . "%'"

-"SELECT * FROM search s WHERE name LIKE '%" . $queryString . "%' LIMIT 5"

si vous voulez limiter à 5 résultats.

Answer 2

Vous devez ajouter le code de pagination à votre page:

Il y a l'exemple de code:

<?php 
// Connects to your Database 
mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()); 
mysql_select_db("address") or die(mysql_error()); 
//This checks to see if there is a page number. If not, it will set it to page 1 
if (!(isset($pagenum))) 
{ 
$pagenum = 1; 
} 
//Here we count the number of results 
//Edit $data to be your query 
$data = mysql_query("SELECT * FROM topsites") or die(mysql_error()); 
$rows = mysql_num_rows($data); 
//This is the number of results displayed per page 
$page_rows = 4; 
//This tells us the page number of our last page 
$last = ceil($rows/$page_rows); 
//this makes sure the page number isn't below one, or more than our maximum pages 
if ($pagenum < 1) 
{ 
$pagenum = 1; 
} 
elseif ($pagenum > $last) 
{ 
$pagenum = $last; 
} 
//This sets the range to display in our query 
$max = 'limit ' .($pagenum - 1) * $page_rows .',' .$page_rows; 
//This is your query again, the same one... the only difference is we add $max into it 
$data_p = mysql_query("SELECT * FROM topsites $max") or die(mysql_error()); 
//This is where you display your query results 
while($info = mysql_fetch_array($data_p)) 
{ 
Print $info['Name']; 
echo "<br>"; 
} 
echo "<p>"; 
// This shows the user what page they are on, and the total number of pages 
echo " --Page $pagenum of $last-- <p>"; 
// First we check if we are on page one. If we are then we don't need a link to the previous page or the first page so we do nothing. If we aren't then we generate links to the first page, and to the previous page. 
if ($pagenum == 1) 
{ 
} 
else 
{ 
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=1'> <<-First</a> "; 
echo " "; 
$previous = $pagenum-1; 
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$previous'> <-Previous</a> "; 
} 
//just a space 
echo " -- "; 
//This does the same as above, only checking if we are on the last page, and then generating the Next and Last links  
if ($pagenum == $last) 
{ 
} 
else { 
$next = $pagenum+1; 
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$next'>Next -></a> "; 
echo " "; 
echo " <a href='{$_SERVER['PHP_SELF']}?pagenum=$last'>Last ->></a> "; 
} 
?> 

Source: www.twitter.com/ZishanAdThandar