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But it still doesn't show anything. So... -
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jQuery: Function gives... Nothing?
Code:<!DOCTYPE html> <html> <head> <title>Example</title> <meta http-equiv="Content-type" content="text/html;charset=UTF-8"> <script type="text/javascript" src="http://ajax.googleapis.com/ajax/libs/jquery/1.5.1/jquery.min.js"></script> <script type="text/javascript"> (function($){
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Thew replied to " Notice: Undefined index: prod_rep in F:\xampp\htdocs\isys\products\process.php on "in PHP1: What an ugly code
Code:mysql_query($query) or die (mysql_error());
Code:$query = "UPDATE tbl_products set prod_name = '$prod_name',cat_id= '$cat_id', prod_desc='$prod_desc', prod_price='$prod_price', prod_rep='$prod_rep',prod_brnd='$prod_brnd', prod_unit='$prod_unit' WHERE prod_id='$prod_id'";
Code:echo 'This
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How to get to apache webserver behind router?
So i have my ip adress going to my router page,
192.168.0.1 to my router page, 192.168.0.100 failing,
And i want to get 83.81.212.133 to my localhost page.
I did everything i needed to do that they told me at portforward.com
So i haven now in my router:
[IMGnothumb]http://i.imgur.com/tvM81.png[/IMGnothumb]
On my firewall:
[IMGnothumb]http://i.imgur.com/l3h2f.png[/IMGnothumb]
...Last edited by Niheel; Mar 1 '11, 05:12 PM. -
I searched on PHP.net and after some time i found this:
Code:$text = base64_decode('Image could not be loaded!'); $im = imagecreatefromstring($text); header('Content-Type: image/png'); imagepng($im); imagedestroy($im);
EDIT: I searched through the GD Functions and imagestring() worked. But...Leave a comment:
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Create image from url: multiple problems
Code:<?php if(isset($_GET['img']) && is_numeric($_GET['img'])){ $img = $_GET['img']; $imgarray = array ( '1' => 'http://www.path/to/image1.png', '2' => 'http://www.path/to/image2.png', '3' => 'http://www.path/to/image3.png' ); $src = $imgarray[$img]; header('Content-type: image/png'); echo file_get_contents($src); } else
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I fixed it, the script im using is so big (876 rows and not finished) that i need to use multiple variables... So i got confused what $getsql and $sql was... But still thanks.Leave a comment:
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Try this:
Code:<?php $newname = dirname(__FILE__).'/upload'.$myvar; $newcopy = dirname(__FILE__).'/copy'.$today.$myvar; if (!file_exists($newname)) { if(!file_exists($newcopy)) { if (move_uploaded_file($_FILES['uploaded_file']['tmp_name'],$newname)) { echo 'File 1 has been uploaded...<br/>'; if (move_uploaded_file
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It was much easier if you tried it, and a lot faster too!
here is the example:
page.php?get=va riable#pointLeave a comment:
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How to get the last 5 topics from a miniBB table?
Code:<?php $getsql = mysql_query("SELECT * FROM minibbtable_topics ORDER BY topic_id DESC LIMIT 5"); if($getsql){ if(mysql_num_rows($sql) == 0){ echo 'No recent topics'; } else { while($getdata = mysql_fetch_assoc($sql)){ echo '<a href="forum/index.php?action=vthread&topic='. $getdata['topic_id']
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Css width not working
Code:<!DOCTYPE html> <html> <head> <title>Layout</title> <style type="text/css"> body { background-color: #DDDDDD; text-align:center; } body, td, div, ul, button, textarea, input { font: 12px helvetica,arial,sans-serif; } #header1 { background-color: #FD0000; }
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Well when i go to my ip adress, i come on my router page. I saw something about ip forwarding, but i didn't understand anything of it?Leave a comment:
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How to host from your own IP address?
I see at some websites, that people host stuff at an IP adress. But im wondering, how do they do that? I have Apache installed, but do i need something else? -
Javascript Slide up en down by click doesn't work
Hi
Im working on an HTML5 Valid and Javascripted page of my website. For that im using Motionpack.
This is my webpage: http://html5.alphenwee r.nl/ (dutch)
I want if someone clicks on radar, there appears a little radar underneath. But then, if someone clicks on that little radar, underneath all maps there comes a bigger radar.
So this is the code i build:
Code:<!DOCTYPE
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Change innerhtml with function gives black page?
Code:<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd"> <html xmlns="http://www.w3.org/1999/xhtml"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title></title> <link href="style.css" rel="stylesheet"
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How to grab image from variable in URL
Its hard to explain.
If i wanna know that if you for example type showimage.php?s how=1, it wil show the image that matches the image from the array.
so if i have this array:
Code:$imgarray = array( '1' => 'http://www.link.to/image1.jpg', '2' => 'http://www.link.to/image2.jpg', '3' => 'http://www.link.to/image3.jpg')
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