so I started developing a blog for myself, and I came to the first setback. I want to replace the posts that this function returns with ajax with the next 10. I know javascript quite well but I don't know so much about ajax (all I know that it sends a request to the server where the results are posted back, and then you can replace the data with the new data on your site)
function get_posts($i = 0,$max = 10) {
global $con;
$query = 'SELECT * FROM post';
$result = mysqli_query($con, $query);
$rows = mysqli_num_rows($result);
for (;$i < $rows&&$i < $max;$i++) {
$row_res=mysqli_fetch_array($result);
printf("<h1>%s</h1><p class='user'>Posted By: %s</p>
<p class='date'>%s</p>
<p>%s</p>",
$row_res["title"],$row_res["user"],$row_res["date"],$row_res["content"]);
}
}
this returns well. I made links that execute an AJAX function that should return the next ten. this is the function:
function func1() {
var xmlhttp;
if (window.XMLHttpRequest) {
xmlhttp=new XMLHttpRequest();
}
xmlhttp.onreadystatechange=function() {
if (xmlhttp.readyState==4 && xmlhttp.status==200) {
document.getElementById("test").innerHTML=xmlhttp.responseText;
}
}
xmlhttp.open("GET","connect.php?func=1",true);
xmlhttp.send();
} The connect.php?func=1 points to an if that executes the get_posts() function with new args. but it doesn't show anything or it shows the first post at the top of the page and another first post in the content section.