Kevin
Senior Member
So I have an ajax script that extracts the 6 most recent urls from a database and puts them in image tags, in the sidebar. It connects to the db-server by calling it 'localhost' and works fine when it runs on a page on my server.
Now, I want to have that script run in the sidebar of a tumblelog. But it won't run because it can't connect to the database because of that 'localhost'. I tried Google and turns out I had to replace it by the ip of my server. I tried it, but it won't let me because my hosting is on a share IP.
Soooo... I need a way of accessing those 6 urls from the tumblr-server (by modifying my current script). Or a way of saving the most recent urls to a text file, that I can then process from the tumblr-server (probably through javascript because PHP doesn't work).
Javascript
The important part of the PHP
Now, I want to have that script run in the sidebar of a tumblelog. But it won't run because it can't connect to the database because of that 'localhost'. I tried Google and turns out I had to replace it by the ip of my server. I tried it, but it won't let me because my hosting is on a share IP.
Soooo... I need a way of accessing those 6 urls from the tumblr-server (by modifying my current script). Or a way of saving the most recent urls to a text file, that I can then process from the tumblr-server (probably through javascript because PHP doesn't work).
Javascript
Code:
function freshpics(){ var ajax = createREQ(); var rand = parseInt(Math.random()*99999999); var url = "http://kevindendievel.com/weeds/freshpics.php?rand=" + rand; ajax.onreadystatechange = function(){ var status = ajax.responseText.substr(0,2); var tekst = ajax.responseText.substr(3); if(ajax.readyState == 4 && status == "OK"){ document.getElementById("pictures").innerHTML = tekst; } else if(ajax.readyState == 4 && status != "OK"){ document.getElementById("pictures").innerHTML = "<p>Foto's konden niet ingeladen worden</p>"; } } ajax.open("GET", url); ajax.send(null); }The important part of the PHP
Code:
$dbserver = "localhost"; $user = "deb36814_kevin"; $passwd = "***";$dbnaam = "deb36814_weeds";$query = "SELECT id,src from pictures ORDER BY id DESC LIMIT 6";$link = new mysqli($dbserver, $user, $passwd, $dbnaam);
