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PHP - Dynamic Rss Feed Not Working..
Hi,
I have a problem with an RSS Feed. the error im getting on the W3C validator is this: This feed does not validate. * line 1, column 0: XML parsing error: <unknown>:1:0: syntax error [help] Query couldn't be executed In addition, interoperability with the widest range of feed readers could be improved by implementing the following recommendation. * "text/xml" media type is not specific enough [help] .htaccess AddType application/x-httpd-php .xml rss.php <? //Set XML header for browser header('Content-type: text/xml'); //Create the heading/channel info for RSS Feed $output = '<rss version=\"2.0\">'; $output .= '<channel>'; $output .= '<title>Muzo Feed</title>'; $output .= '<description>Your RSS Feed Description</description>'; $output .= '<link>http://mi-linux.wlv.ac.uk/~0703272</link>'; //Individual Items of our RSS Feed //Connect to a database and and display each new item in our feed //Connect to DB $host = "localhost"; //Name of host $user = "0703272"; //User name for Database $pass = "rachel123"; //Password for Database $db = "db0703272"; //Name of Database mysql_connect($host,$user,$pass); //Create SQL Query for our RSS feed $sql = "SELECT `title`, `webaddress`, `message`, `date_added` FROM `messages`ORDER BY `date_added` DESC LIMIT 0 , 15"; $result = mysql_query($sql) or die ("Query couldn't be executed"); //Create Loop for the individual elements in the RSS item section while ($row = mysql_fetch_array($result)) { $output .= '<item>'; $output .= '<title>'.$row['title'].'</title>'; $output .= '<link>http:/mi-linux.wlv.ac.uk/~0703272/blog.php'.$row['link'].'</link>'; $output .= '<message>'.$row['message'].'</description>'; $output .= '<date_added>'.$row['date'].'</pubDate>'; $output .= '</item>'; } //Close RSS channel $output .= '</channel>'; $output .= '</rss>'; //Display output in browser echo $output; ?> any ideas? Similar Tutorialshi i know eval is not best way to run php code via database but its required to me. What is happening i am reading a feed and if i use eval it just repeat same rss links but if i put a direct code it show proper feed. here is my code which works proper $xml = simplexml_load_file($feedurl); foreach ($xml->channel->item as $item) { echo $item->title . "\n"; echo $item->link . "\n"; } results of above code are fine. Code: [Select] Feed one title feed one link feed two title feed two link feed three title feed three link and so on for 10 times as 10 feeds are available...... But if i use eval my database contains <?php echo $item->title . "\n"; echo $item->link . "\n"; ?> and i run it like this $xml = simplexml_load_file($feedurl); foreach ($xml->channel->item as $item) { eval('?>' . $fromadatabase['code'] . '<?php '); } and i get this results Code: [Select] Feed one title feed one link Feed one title feed one link Feed one title feed one link and so on...... means it just repeat first feed for 10 times Thanks for any help. Hello PHP Freaks! I have the following PHP Curl code which is supposed to fetch the twitter timeline feed for a specific username and parse and display it on a page, but for some reason Curl is unable to fetch data from twitter: Code: [Select] <?php function get_data($url) { $ch = curl_init(); $timeout = 5; curl_setopt($ch,CURLOPT_URL,$url); curl_setopt($ch,CURLOPT_RETURNTRANSFER,1); curl_setopt($ch,CURLOPT_CONNECTTIMEOUT,$timeout); $data = curl_exec($ch); curl_close($ch); return $data; } $json = get_data("https://api.twitter.com/1/statuses/user_timeline.json?include_entities=true&include_rts=true&screen_name=digioz&count=2"); if ($json != false) { $obj = json_decode($json); foreach($obj as $var => $value) { echo "Message number: $var <br/>"; echo "Name: " . $obj[$var]->user->name; echo "Handle: " . $obj[$var]->user->screen_name . "<br/>"; echo "Message: " . $obj[$var]->text; echo "Created" . $obj[$var]->created_at . "<br/>"; echo "URL" . $obj[$var]->user->url . "<br/>"; echo "Location" . $obj[$var]->user->location . "<br/>"; echo "<br/>"; } } else { echo "Could not fetch Twitter Data"; } ?> Anyone have any idea why the data is not being fetched? If I copy and paste the URL into my browser window it returns results just fine, so I know the problem is not with the URL. Thanks, Pete Im rebuilding my website and have decided to make the dynamic images for my main photography gallery be layed out using div tags and css. I can get the images to load in and layout correctly but for some reason the links on the images only works on the last loaded in image. I check the code on the page and all the links are there but only the last one is active. The code worked when it was all wrapped in table tags. Is this a PHP or css issue stopping all the links expect for the last to be active. This is the css code I am using. #maincontent{width:1000px;margin:0 auto;padding:0 20px} #maingallery{height:382px;position:relative;width:1000px;text-align:center;margin-top:15px} #maingallery #text{position:absolute;top:50px} .gallery_item {float:left; margin: 20px 0 0 0; width: 200px; height: 67; padding: 0 10px 0 10px} .gallery_item .p {padding: 10px; font-size:12px} .gallery_item a:hover, a:active { display : block; border : none; } Any help would be greatly appreciated. Not sure if there would be any thing conflicting the links being active in either php or css. Thanks Matt I'm trying to create a dynamic option menu with one alert selected based on the first query to the db. Any help would be greatly appreciated. Code: [Select] //function to get alerts and create select menu with current alerts pre-selected function getALERTS1($id){ require('db.php'); $alert = mysqli_query($conn, "SELECT alert1 FROM visit_data WEHERE patientid = $id AND discharged IS NULL"); $row = mysqli_fetch_array($alert); $selects=null; $query = mysqli_query($conn, "SELECT alertid, name FROM alerts"); while($row1 = mysqli_fetch_array($query)) { $selects .= "<option value=\"" . $row1['alertid'] . "\"> if($row1['alertid']==$row['alert1']) { echo ' selected'; } ".$row1['name']."</option>"; } return $selects; } Folks, I need help (Php code ) to generate a Dynamic Text on a Base Image. What i want to do is, to make this Image as header on my Site and to make this Header Specific to a Site, i want to Add the Domain Name on the Lower Left of the Image. Got the Idea? Here is the Image link: Quote http://img27.imageshack.us/i/shoppingheader1.jpg/ PHP Variable that holds the Domain name is: $domain All i need the Dynamic PHP Codes that i can put on all my sites to generate this Text on Image (Header) Dynamically... May Anyone Help me with this Please? Cheers Natasha T. Hi all I need to combine these two scripts: Firstly, the following decides which out of the following list is selected based on its value in the mySQL table: <select name="pack_choice"> <option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option> <option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option> <option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option> </select> Although this works OK, I need it also to show dynamic values like this: select name="category"> <?php $listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC"); while($categoryReturned=mysql_fetch_array($listCategories)) { echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>"; } ?> </select> I'm not sure if this is possible? Many thanks for your help. Pete I'm lost on this one. This code works fine in Firefox, but not my IE8. I think it has something to do with the <?php echo '<?xml version="1.0" encoding="utf-8"?>'; ?> line, but i'm not sure. IE says "Internet Explorer cannot display this feed" and "A name contained an invalid character. Line: 7 Character: 701." Any ideas? <?php header('Content-Type: text/xml'); ?> <?php echo '<?xml version="1.0" encoding="utf-8"?>'; ?> <?php include('../vars.php'); ?> <rss version="2.0"> <channel> <title>Newsfeed</title> <description>News and Updates</description> <link>http://www.example.com</link> <language>en-us</language> <?php require_once("../conn.php"); $query = "SELECT * FROM news"; $data = mysqli_query($db, $query); while ($row = mysqli_fetch_array($data)) { echo '<item>'; echo ' <title>' . $row['title'] . ' - ' . substr($row['content'], 0, 32) . '...</title>'; echo ' <link>http://www.example.com/new.php?newsid=' . $row['newsid'] . '</link>'; //echo ' <pubDate>' . $row['date'] . '</pubDate>'; echo ' <description>' . $row['content'] . '</description>'; echo '</item>'; } ?> </channel> </rss> Hi I get an error when trying to add a link to my rss feed Code: [Select] <link>http://www.jobjar.co.uk/jobdetails.php?keywords=lon0014&page=1</link>Something to do with the last part of the link because when I remove the page=1 it then works. How can I resolve this? Also, will I have to manually update my xml file for rss feeds or can it be updated automatically? If so how? Thanks Hi All, I'm trying to incorporate a BBC Sport RSS Feed into my website. The following code will show me the news feeds as text and I can include the url's as text also but want to turn this into a clickable url. How do I do this, I cant work it? please help. $feed_url = "http://newsrss.bbc.co.uk/rss/sportonline_uk_edition/football/teams/s/shrewsbury/rss.xml"; // Get data from feed file $response = file_get_contents($feed_url); // Insert XML into structure $xml = simplexml_load_string($response); // Browse structure foreach($xml->channel->item as $one_item) echo $one_item->title."<BR>"; Hello all That forum is my last desperate attemp to do what i want to do. Ok here is the story I want to create a simple rss feed in conjuction with php and mysql. I dont want admin areas ect , i just want when i insert a new listing to my database to be able shown up to my (future) rss subscribers. To be more technically specific i want to show to my surfers updates about 2 tables in my database not all the tables. The example i found so far were about only 1 table, plus i was encounting errors to my script. I would like some ideas, directions if someone is kind enough to help a sad developer Thanks in advance! I'm been searching and can't seem to figure this out. I want to take an RSS feed from a news site and display it on my site. any help? Hi, I've written some code to take information from an SQL database and write it out in the RSS format (Although it doesn't validate). The problem is i'd like the page to have the .rss (or .xml) file extension, I'm not sure if there's any advantages in having this but thought i'd ask. I've got the following code: Code: [Select] <?php header('Content-type: text/xml'); print '<?xml version="1.0"?>'; print '<rss version="2.0">'; print '<channel>'; include("phpfunctions.php"); db_connect(); //select all from users table $select="SELECT title, link, description FROM news"; $result = mysql_query($select) or die(mysql_error()); //If nothing is returned display error no records if (mysql_num_rows($result) < 1) { die("No records"); } //loop through the results and write each as a new item while ($row = mysql_fetch_assoc($result)) { $item_title = $row["title"]; $item_link = $row["link"]; $item_desc = $row["description"]; print '<item>'; print '<title>' . $item_title . '</title>'; print '<link>' . $item_link . '</link>'; print '<description>' . $item_desc . '</description>'; print '</item>'; } print '</channel>'; print '</rss>'; ?> This seems to work fine as i get what i expect and i'm assuming i can do the same to output .xml but is there a way to have it in a proper .rss / .xml file so that an aggregator or someone could read this properly. Cheers, Reece hi all, i am trying to just get a couple things from this RSS feed: http://www.sierraavalanchecenter.org/bottomline-rss.php I would like to grab the danger rose image on the left, the text that tells of the Avalanche danger ie... "Avalanche Danger Remains LOW..." and the date. I have tried using a few things like: $html = 'http://www.sierraavalanchecenter.org/bottomline-rss.php';$dom = new DOMDocument;$dom->loadHTML($html);$xpath = new DOMXPath($dom);$res = $xpath->query('//div[@id="dangericon"]');if ($res->item(0) !== NULL) { $test = $res->item(0)->nodeValue;}() ...without any luck. Anyone have some advice for me? thanks! I have a php script that saves some data as a .xml document. When I view this in my browser I can subscribe to the feed (I am using firefox). However I asked someone else to see if they could subscribe (they were using IE) but the couldn't. Therefore how can users subscribe to my feed? Will only some be able to subscribe? Thanks for any help. Hi, I am reading a feed with this code foreach ($xml->channel->item as $item) { echo $item->title; echo '<br>'; } and outputs Code: [Select] title 1 title 2 title 3 .... and so on until 10 I wants in reverse order to output last rss feed first like: Code: [Select] title 10 title 9 title 8 .... and so on until 1, how can i do it? i tried with rsort($xml->channel->item) butt getting this error: Quote Warning: rsort() expects parameter 1 to be array, object given in Thanks for help Hi, I have a small piece of code that creates an RSS feed using a mysql database. The issue is the page itself is blank however if I right click and view source I can see all the feed there. I convert the dat time field into a standard RSS date field. The web address is http://vinovote.com/news/feed.php My code is as follows Code: [Select] <?php echo '<?xml version="1.0" encoding="UTF-8"?>'; ?> <rss version="2.0"> <channel> <title>Vinovote.com</title> <description>The Latest News And Views From Around The Web</description> <link>http://www.vinovote.com/</link> <copyright>Your copyright information</copyright> <?php require_once('../Connections/connection.php'); mysql_select_db($database_vinovotedb, $vinovotedb); $doGet = mysql_query("SELECT feed_content.feed_content_id, feed_content.feed_id, feed_content.url, feed_content.title, feed_content.content, feed_content.item_time, Date_FORMAT(feed_content.item_time,'%a, %d %b %Y %T') AS pubDate FROM feed_content order by item_time desc LIMIT 50 ", $vinovotedb) or die(mysql_error()); while($result = mysql_fetch_array($doGet)){ ?> <item> <title> <?php echo $result['title']; ?></title> <description> <?php echo $result['content'];?></description> <link><?php echo $result['url'];?></link> <pubDate> <?php echo $result['pubDate']; ?></pubDate> </item> <?php } ?> </channel> </rss> I have tried sitemap.org and a few others but have not found anything that will give the results I am after. I have a blog site that has a URL and a Description for the URL and wish to have the description show as the anchor text which is linked to the URL, all data is taken from MySQL. Does anyone have a code that will generate something like this ? Hello! I'm trying to develop some kind of RSS news aggregator and I want to show only feeds younger than 1day. I figured i could transform RSS pubDate to timestamp (strtotime()), but there are some feeds without timestamp (like: http://izklop.com/xmldata/rsslinks.xml). Is there any other way to do it, so I could find timestamp from those feeds without pubdate? If there isn't any other way, do You think it is ok, to just show last 5 feeds? I hope I made my self clear, and please forgive me for my English |
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