I have script a.php
<html> <head> <link rel="stylesheet" type="text/css" href="f.css"> </head> <body> <ul> <li><a class="active" href="afisare.php">Home</a></li> <li><a href="#news">News</a></li> <li><a href="#contact">Contact</a></li> <li><a href="#about">About</a></li> </ul> </body> </html>
and script afisare.php:
<?php session_start(); // conectare la server $conn = new mysqli('localhost', 'root', '', 'db'); // verifica conexiunea if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); } //require_once('conectare.php'); echo "<div id=\"continut\">"; $sql = "select denumire,pret,categorie from poze"; $result = $conn->query($sql); $nr=$result->num_rows; for ($i=0;$i<$nr;$i++) { $row=$result->fetch_assoc(); $poza=$row['denumire'].'.jpg'; echo "<img src=\"".$poza."\" style=\"width:200px;height:200px\">"; } echo "</div>"; ?>
if click on link Home, i want display result of script afisare.php in page of a.php, in new div element.
How do it?