[AJAX] Affichage des données
Bonjour
je suis debutant , je veux que j'ajoute un article les autres article affiches dans meme page plus le derniere article sans actualiser la page merci
Code:
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| function getAJAX(page){
var xmlhttp;
if (window.XMLHttpRequest)
xmlhttp=new XMLHttpRequest();
else
{
xmlhttp=new ActiveXObject("Microsoft.XMLHTTP");
if (!xmlhttp)
xmlhttp=new ActiveXObject("Msxml2.XMLHTTP.3.0");
if (!xmlhttp)
xmlhttp=new ActiveXObject("Msxml2.XMLHTTP.6.0");
}
xmlhttp.onreadystatechange =function(){
if (xmlhttp.readyState < 4)
document.getElementById("jax").innerHTML='<img src ="1.gif" border=0 />';
if (xmlhttp.readyState==4 && xmlhttp.status==200)
document.getElementById("jax").innerHTML=xmlhttp.responseText;
}
xmlhttp.open("GET",page,true);
xmlhttp.send();
} |
Code:
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| <?php require_once "functions.php";
if(isset($_POST['title']) && isset($_POST['content'])) {
add_post($_POST['title'], $_POST['content']);
//header("Location:index.php");die();
}
?>
<!doctype html>
<html lang="en">
<head>
<meta charset="UTF-8">
<title>Posts</title>
<script type="text/javascript">
...
</script>
</head>
<body>
<h1>Add</h1>
<form action="#" method="post">
Title:
<input type="text" name="title"> <br>
Content:
<input name="content" type="text"> <br>
<!--<input type="submit" value="Add">-->
<button type="submit" value="add" onclick="getAJAX('index.php')">Change Content</button>
</form>
<div id="jax"></div>
</body>
</html> |