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PHP - Can Not Get Rss Feed To Generate
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 ? Similar TutorialsCan this be done with php? aaa aab aac ... zzz 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 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 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>"; hi, im trying to create a website and only now started thinking about the security part(noob mistake). say for example i have home.php page and an index.php page. index.php is where users would sign up/log in. the login and sign up processes are all done but i was thinking of creating a unique id of some sort for when the user logs in. or something like this site (forum.phpfreaks) when we sign in, you are signed but the url stays the same = forums.phpfreaks.com. like if we were signed out we will be permanantly signed out and typing in forums.phpfreaks.com would just land us at the main page where we need to sign in.
right now ,my home.php can be accessed with or without logging in even with sessions.
hope im making sense, thanks in advanced!
**haha that rhymed.
i tried adding:
<?php echo $_SERVER[PHP_SELF] . '?name=' . $userData['name'];?> in the index.php:
<?php
ob_start();
session_start();
if(isset($_POST['login']))
{
$email = $_POST['email'];
$password = $_POST['pass'];
require "connection.php";
$emails = mysqli_real_escape_string($con, $email);
$query = "SELECT id, name, email, password, salt FROM users WHERE email = '$emails';";
$result = mysqli_query($con, $query);
if(mysqli_num_rows($result) == 0) // User not found. So, redirect to login_form again.
{
echo "<script>alert(\"User does not exist!\")</script>";
}
$userData = mysqli_fetch_array($result, MYSQLI_ASSOC);
$hash = hash('sha256', $userData['salt'] . hash('sha256', $password) );
if($hash != $userData['password'])
{
echo "<script>alert(\"Incorrect Password!\")</script>";
}else{
session_regenerate_id();
$_SESSION['sess_user_id'] = $userData['id'];
$_SESSION['sess_name'] = $userData['name'];
session_write_close();
header('Location: home.php?user=');
}
}
ob_flush();
?>
<!DOCTYPE html>
<form name="login" method="post" action="<?php echo $_SERVER[PHP_SELF] . '?name=' . $userData['name'];?>">
but i got access forbidden!
Edited by noobdood, 19 May 2014 - 10:05 PM. Hi guys, so i have this file upload script. When i upload a file it gets stored in /uploads and keeps the same file name. So if i upload a file "test.exe" the file will be available at uploads/test.exe
What i want is that it generates a new file name like: "9daln292os.exe" so upload/9daln292os.exe
This is my code:
<?php
// Where the file is going to be placed
$target_path = "uploads/";
/* Add the original filename to our target path.
Result is "uploads/filename.extension" */
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
$_FILES['uploadedfile']['tmp_name'];
?>
<?php
$file_type = $_FILES['userfile']['type'];
$file_name = $_FILES['userfile']['name'];
$file_ext = strtolower(substr($file_name,strrpos($file_name,".")));
if (!in_array($file_type, $FILE_MIMES) && !in_array($file_ext, $FILE_EXTS) )
$message = "Sorry, $file_name($file_type) is not allowed to be uploaded.";
else
$message = do_upload_function_here($upload_path_here, $upload_ur_upload_url_herel);
?>
<?php
$target_path = "uploads/";
$target_path = $target_path . basename( $_FILES['uploadedfile']['name']);
if(move_uploaded_file($_FILES['uploadedfile']['tmp_name'], $target_path)) {
echo "The file ". basename( $_FILES['uploadedfile']['name']).
" has been uploaded. Here is the link to your file: <a href=uploads/". basename( $_FILES['uploadedfile']['name']). ">". basename( $_FILES['uploadedfile']['name'])."</a>";
} else{
echo "There was an error uploading the file, please try again!";
}
?>
I don't know any basic php i really need someone to give me the code ready please. Thanks much appreciated.
Edited by darox, 21 July 2014 - 01:07 PM.
<?php
if(isset($_POST['submit'])){
//collect form data
$location = $_POST['location'];
$ID = $_POST['ID'];
$section = $_POST['section'];
//check name is set
if($location ==''){
$error[] = 'Name is required';
}
//if no errors carry on
if(!isset($error)){
# Title of the CSV
$Content = "location, ID, section\n";
//set the data of the CSV
$Content .= "$location, $ID, $section\n";
# set the file name and create CSV file
$my_file = ("$location$ID$section.cvs");
$handle = fopen("$my_file", "w") or die('Cannot open file: '.$my_file);
header('Content-Type: application/csv');
header('Content-Disposition: attachment; filename="' . $FileName . '"');
echo $Content;
exit();
}
}
//if their are errors display them
if(isset($error)){
foreach($error as $error){
echo "<p style='color:#ff0000'>$error</p>";
}
}
?>
<form action='' method='post'>
<p><label>Location:</label><br><input type='text' name='location' value=''></p>
<p><label>ID:</label><br><input type='text' name='ID' value=''></p>
<p><label>Section:</label><br><input type='text' name='section' value=''></p>
<p><input type='submit' name='submit' value='Submit'></p>
</form>
I have wrote a .php file which is a form with 3 different fields, in these fields I want to write entries which lateron will be submitted into a .csv generating a csv file on my server with the name of the entries.
The issue is that the file is not being generated on my server. With the current code the entries are being recognized and being placed in the filename of the csv.
The point is not to make it downloadable sinds I want to have it on my webserver and keep editing information in it.
My code is top of this post.
My .php file replies that:
Cannot open file: BOD10Buffer.csv
Pardon my sloppy code everyone, I am not really a person that codes but this "form" may save me a lot of time in the longrun, just difficult and I am asking for help on this. Is there anyone that may be kind enough to help me? and as to what I might be doing wrong?
All help is much appreciated!
Hi, can someone help me to understand this? What i want to do is to write some information in a form and after i submit the form that data will be in a new php page. Thanks in advance 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'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>' . $item_desc . '</description>'; print '</item>'; } print '</channel>'; print '</rss>'; ?> This seems to work fine as i get what i expect and i'm assuming i can do the same to output .xml but is there a way to have it in a proper .rss / .xml file so that an aggregator or someone could read this properly. Cheers, Reece Hi, I have a small piece of code that creates an RSS feed using a mysql database. The issue is the page itself is blank however if I right click and view source I can see all the feed there. I convert the dat time field into a standard RSS date field. The web address is http://vinovote.com/news/feed.php My code is as follows Code: [Select] <?php echo '<?xml version="1.0" encoding="UTF-8"?>'; ?> <rss version="2.0"> <channel> <title>Vinovote.com</title> <description>The Latest News And Views From Around The Web</description> <link>http://www.vinovote.com/</link> <copyright>Your copyright information</copyright> <?php require_once('../Connections/connection.php'); mysql_select_db($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 $result['pubDate']; ?></pubDate> </item> <?php } ?> </channel> </rss> Does anybody have any recommendations for PHP scripts to pull in a Twitter feed? I've tried several but each seem to just give me error messages! I basically just need a PHP alternative so that the tweets can be read by search engines. Thanks. Hi, I am reading a feed with this code foreach ($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 I'm not sure you can do this but i would like to grab the image gif from the rss feed of yahoo weather. For instance http://weather.yahooapis.com/forecastrss?p=77056 I pulled this script from a tutorial online and it works great for displaying the temperature and condition and i realize those values are located within the xml. But i would like to grab the image located between the <description> tags (if you look at the source of the rss feed). I don't know if this is possible or not. Thanks. Code: [Select] <!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>Yahoo! Weather API RSS</title> <?php function retrieveYahooWeather($zipCode="92832") { $yahooUrl = "http://weather.yahooapis.com/forecastrss"; $yahooZip = "?p=$zipCode"; $yahooFullUrl = $yahooUrl . $yahooZip; $curlObject = curl_init(); curl_setopt($curlObject,CURLOPT_URL,$yahooFullUrl); curl_setopt($curlObject,CURLOPT_HEADER,false); curl_setopt($curlObject,CURLOPT_RETURNTRANSFER,true); $returnYahooWeather = curl_exec($curlObject); curl_close($curlObject); return $returnYahooWeather; } $localZipCode = "77056"; // Houston, Tx $weatherXmlString = retrieveYahooWeather($localZipCode); $weatherXmlObject = new SimpleXMLElement($weatherXmlString); $currentCondition = $weatherXmlObject->xpath("//yweather:condition"); $currentTemperature = $currentCondition[0]["temp"]; $currentDescription = $currentCondition[0]["text"]; ?> </head> <body> <h1>Houston, TX</h1> <ul> <li>Current Temperatu <?=$currentTemperature;?>°F</li> <li>Current Description: <?=$currentDescription;?></li> </ul> </body> </html> 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 want to add some content to my site using xml rss i have a feed i want to show, and the only thing i can find to show it on my page or widgets but they dont look like i want it to look, so i want to make a php script that reads the xml file and then parses it into my website dont know if this is even possible ... but if its possible, is it also possible to filter the feed on certain keywords? 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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