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PHP - Showing Rss Feed With Photos On Another Page
If you goto http://www.actionfx.net/bfd/photos.php you will see that im trying to show an rss feed from http://picasaweb.google.com/data/feed/base/user/114565750484201639035?alt=rss&kind=album&hl=en_US&access=public
My code works well if its just text on the rss but it does not work on this particular rss. I want the photos to show up on my page also. Im hoping someone here can be so kind to help me out with this coding. Below is my code. Code: [Select] <?php $feed_url = "http://picasaweb.google.com/data/feed/base/user/114565750484201639035?alt=rss&kind=album&hl=en_US&access=public"; // INITIATE CURL. $curl = curl_init(); // CURL SETTINGS. curl_setopt($curl, CURLOPT_URL,"$feed_url"); curl_setopt($curl, CURLOPT_RETURNTRANSFER, 1); curl_setopt($curl, CURLOPT_CONNECTTIMEOUT, 0); // GRAB THE XML FILE. $xmlTwitter = curl_exec($curl); curl_close($curl); // SET UP XML OBJECT. // Use either one of these, depending on revision of PHP. // Comment-out the line you are not using. //$xml = new SimpleXMLElement($xmlTwitter); $xml = simplexml_load_string($xmlTwitter); // How many items to display $count = 10; // How many characters from each item // 0 (zero) will show them all. $char = 100; foreach ($xml->channel->item as $item) { if($char == 0){ $newstring = $item->description; } else{ $newstring = substr($item->description, 0, $char); } if($count > 0){ //in case they have non-closed italics or bold, etc ... echo"</i></b></u></a>\n"; echo" <div style='font-family:verdana; font-size:.12;'> <b>{$item->title}</b><br /> $newstring ... <span class='redlink' id='redlink' style='redlink'> <a href='{$item->guid}'>read more</a> </span> <br /><br /> </div> "; } $count--; } ?> Similar TutorialsThis topic has been moved to Third Party PHP Scripts. http://www.phpfreaks.com/forums/index.php?topic=354562.0 Hey all, I'm currently making myself a blog website from scratch. It's sort of my personal project; I'm using it to get a feel for PHP (and it'd look great on my resume!). Anyway, I'm using the following PHP code to get the items from my RSS feed, and display them. Problem is, my first item shows twice! I'm thinking it's a problem with my loop, but I'm not sure. Any ideas: My PHP code to render my RSS is as follows: <?php //Purpose: This loads the XML document using the DOMParser, and mixes it w/ HTML to style and format it. try { $xmldoc = new DOMDocument(); $xmldoc->load('test_RSS_feed.xml'); $content = null; //HTML mixed with XML results go here $allitems = $xmldoc->getElementsByTagName('item'); foreach ($allitems as $item) { //get ALL instances of wanted element(s) $titles = $item->getElementsByTagName('title'); $dates = $item->getElementsByTagName('pubDate'); $descriptions = $item ->getElementsByTagName('description'); //will include the CSS classes for these eventually! $content.= (' <ul class ="blog_entry"> <li class="title_entry"> '.$titles->item(0)->nodeValue. '</li> <li class="date_entry"> '.$dates->item(0)->nodeValue. '</li> <br> <li class="description_entry"> '.$descriptions->item(0)->nodeValue. '</li> </ul> <br><br><br>'); echo $content; } } catch(Exception $e) { echo $e->getMessage(); } ?> And the XML in my RSS feed is as follows: <?xml version= "1.0" encoding = "UTF-8" ?> <!--Every XML doc starts with this --> <?xml-stylesheet type="text/css" href="sample_xml_css.css" ?> <!--stylesheet for RSS feed--> <rss version = "2.0"> <channel> <!--RSS feed information--> <title>My first RSS feed</title> <description>This is my first foray into RSS. I may use this to create a blog!</description> <lastBuildDate>Mon, Oct. 11, 2010 13:13:26 GMT</lastBuildDate> <!--date format complies to different standards--> <webMaster>mserrano0410@gmail.com</webMaster> <!--has email of webmaster--> <item> <title>My first item; this is the title.</title> <link>http://www.example.com/sample_item.html</link> <!--provides a link to the RSS item; required--> <pubDate>Mon, Oct 11, 2010 13:15:20 GMT</pubDate> <!--Every date complies to formatting standards--> <description>We are now inside of an item; this is a description tag.</description> </item> <item> <title>My second item.</title> <link>http://www.google.com</link> <pubDate>Fri, Oct 15, 2010 23:02:40 GMT</pubDate> <description>This is a second helping of examples biatch!</description> </item> <!--can create as many items as you want, but it MUST have: 1) Title 2) Link to item; when clicked, takes you directly to that item. This is a SEPARATE HTML document! 3) Description --> </channel> </rss> Thanks in advance, Marvin I am pretty new to PHP and am trying to create a simple (so I assumed) page to takes data from one html page(works fine) and updates a MYSQL Database. I am getting no error message, but the connect string down to the end of the body section is showing up as plain text in my browser window. I do not know how to correct this. I have tried using two different types of connect strings and have verified my names from the HTML page are the same as listed within the php page. Suggestions on what I need to look for to correct would be great. I have looked online, but so far all I am getting is how to connect, or how to create a comment, so I thought I would try here. Thank you for any assistance I may get!! - Amy - Code: [Select] <body><font color="006600"> <div style="background-color:#f9f9dd;"> <fieldset> <h1>Asset Entry Results</h1> <?php // create short variable names $tag=$_POST['tag']; $serial=$_POST['serial']; $category=$_POST['category']; $status=$_POST['status']; $branch=$_POST['branch']; $comments=$_POST['comments']; if (!$tag || !$serial || !$category || !$status || !$branch) { echo "You have not entered all the required details.<br />" ."Please go back and try again."; exit; } if (!get_magic_quotes_gpc()) { $tag = addslashes($tag); $serial = addslashes($serial); $category = addslashes($category); $status = addslashes($status); $branch = addslashes($branch); $comments = addslashes($comments); } //@ $db = new mysqli('localhost', 'id', 'pw', 'inventory'); $db = DBI->connect("dbi:mysql:inventory:localhost","id","pw") or die("couldnt connect to database"); $query = "insert into assets values ('".$serial."', '".$tag."', '".$branch."', '".$status."', '".$category."', '".$comments."')"; $result = $db->query($query); if ($result) { echo $db->affected_rows." asset inserted into Inventory."; } else { echo "An error has occurred. The item was not added."; } $db->close(); ?> </fieldset> </div> </body> Hey all, Been coding another script for my website which allows me to update a users Rank, allthough, when I try and view the page its giving me a White Blank Page. I carn't accually see whats wrong in my Script... <?php session_start(); include ("../includes/config.php"); include ("../includes/function.php"); logincheck(); ini_set ('display_errors', 1); error_reporting (E_ALL); $username = $_SESSION['username']; // Owner Script - Change Users Rank $userlevel = mysql_query ("SELECT * FROM users WHERE username = '$username' LIMIT 1") or die ("Error Finding Right Users " . mysql_error()); $levelright = mysql_fetch_object($userlevel); // If userlevel isn't 5, move them on... if ($levelright->userlevel == "1" || $levelright->userlevel == "2" || $levelright->userlevel == "3" || $levelright->userlevel == "4"){ header ("Location: 404.php"); die(); } // Start Change Rank... if (strip_tags($_POST['changerank'])){ $name = $_POST['name']; $userchange = $_POST['userchange']; $real = mysql_query ("SELECT * FROM users WHERE username = '$userchange' LIMIT 1"); $number = mysql_num_rows($real); if ($number != "1"){ echo ("$userchange, is NOT registered! - Simple Words : No Such User!"); } if ($name == 1){ $reprand = rand(100, 50000); mysql_query ("UPDATE users SET rank = 'Learner Driver (1)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 1 if ($name == 2){ $reprand = rand(50000, 200000); mysql_query ("UPDATE users SET rank = 'Passed Driver (2)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 2 if ($name == 3){ $reprand = rand(200000, 500000); mysql_query ("UPDATE users SET rank = 'New Driver (3)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 3 if ($name == 4){ $reprand = rand(500000, 2500000); mysql_query ("UPDATE users SET rank = 'Speeding Driver (4)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 4 if ($name == 5){ $reprand = rand(2500000, 7000000); mysql_query ("UPDATE users SET rank = 'Skilled Driver (5)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 5 if ($name == 6){ $reprand = rand(7000000, 20000000); mysql_query ("UPDATE users SET rank = 'Maniac Driver (6)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 6 if ($name == 7){ $reprand = rand(20000000, 45000000); mysql_query ("UPDATE users SET rank = 'Wanted Racer (7)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 7 if ($name == 8){ $reprand = rand(45000000, 90000000); mysql_query ("UPDATE users SET rank = 'Pro Driver (8)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 8 if ($name == 9){ $reprand = rand(90000000, 115000000); mysql_query ("UPDATE users SET rank = 'Supreme Racer (9)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 9 if ($name == 10){ $reprand = rand(115000000, 150000000); mysql_query ("UPDATE users SET rank = 'Super Supreme Racer (10)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 10 if ($name == 11){ $reprand = rand(150000000, 210000000); mysql_query ("UPDATE users SET rank = 'Show Off Driver (11)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 11 if ($name == 12){ $reprand = rand(210000000, 280000000); mysql_query ("UPDATE users SET rank = 'Extreme Show Off Driver (12)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 12 if ($name == 13){ $reprand = rand(280000000, 350000000); mysql_query ("UPDATE users SET rank = 'Trained Driver (13)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 13 if ($name == 14){ $reprand = rand(350000000, 430000000); mysql_query ("UPDATE users SET rank = 'Super Trained Driver (14)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 13 if ($name == 15){ $reprand = rand(430000000, 520000000); mysql_query ("UPDATE users SET rank = 'Legendry Racer (15)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 15 if ($name == 16){ $reprand = rand(520000000, 600000000); mysql_query ("UPDATE users SET rank = 'Most Legendry Racer (16)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 16 if ($name == 17){ $reprand = rand(600000000, 1000000000); mysql_query ("UPDATE users SET rank = 'Official SD Legend (17)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 17 if ($name == 18){ $reprand = rand(1000000000, 1000000010); mysql_query ("UPDATE users SET rank = 'Hidden Rank! (18)' AND rep = '$reprand' WHERE username = '$userchange'") or die ("Error Updating rank " . mysql_error()); } // Name 18 echo ("Successfully updated $userchange Rank!"); mysql_query("INSERT INTO `inbox` ( `id` , `to` , `from` , `message` , `date` , `read` , `saved` , `event_id` ) VALUES ( '', '$userchange', 'System', 'Your Rank has now been changed!', '$date', '0', '0', '0')") or die ("Error updating there inbox!" . mysql_error()); } // Rank Change if (strip_tags($_POST['selectname'])){ $writenname = $_POST['nameinput']; $rankinput = $_POST['rankinput']; $repinput = $_POST['repinput']; $findme = mysql_query ("SELECT * FROM users WHERE username = '$writenname' LIMIT 1") or die ("Error Searching Names " . mysql_error()); $found = mysql_num_rows($findme); if ($found != "1"){ echo ("$writenname, isn't Registered! - Simple Terms : No Such User"); }else{ mysql_query ("UPDATE users SET rank = '$rankinput' AND rep = '$repinput' WHERE username= '$writenname'") or die ("Error Updating Rank + Rep " . mysql_error()); echo ("$writenname Rank has now been changed!"); mysql_query("INSERT INTO `inbox` ( `id` , `to` , `from` , `message` , `date` , `read` , `saved` , `event_id` ) VALUES ( '', '$writenname', 'System', 'Your Rank has now been changed!', '$date', '0', '0', '0')") or die ("Error updating there inbox!" . mysql_error()); } // Found? } // Post selectname ?> <html> <head> <title>Change Users Rank</title> </head> <body class='body'> <form action='' method='POST' name='Change Rank'> <table width='40%' border='1' cellpadding='0' cellspacing='0' class='table' align='center'> <tr> <td class='header' align='center' colspan='2'>Change Users Rank?</td> </tr> <tr> <td class='omg' align='center'>This will change the users Current Rank!</td> </tr> <tr> <td align='right'>Username:</td><td><input type='text' name='userchange' class='input'></td> </tr> <tr> <td><select name='name' size='1' class='dropbox'> <option value="1">Learner Driver (1)</option> <option value="2">Passed Driver (2)</option> <option value="3">New Driver (3)</option> <option value="4">Speeding Driver (4)</option> <option value="5">Skilled Driver (5)</option> <option value="6">Maniac Driver (6)</option> <option value="7">Wanted Racer (7)</option> <option value="8">Pro Driver (8)</option> <option value="9">Supreme Racer (9)</option> <option value="10">Super Supreme Racer (10)</option> <option value="11">Show Off Driver (11)</option> <option value="12">Extreme Show Off Driver (12)</option> <option value="13">Trained Driver (13)</option> <option value="14">Super Trained Driver (14)</option> <option value="15">Legendry Racer (15)</option> <option value="16">Most Legendry Racer (16)</option> <option value="17">Official SD Legend (17)</option> <option value="18">Hidden Rank! (18)</option></td> </tr> <tr> <td class='omg' align='center' colspan='2'><input type='submit' name='changerank' class='button'></td> </tr> </table> <br /> <table width='40%' border='1' cellpadding='0' cellspacing='0' class='table' align='center'> <tr> <td class='header' align='center' colspan='2'>Change Rank?</td> </tr> <tr> <td class='omg' align='center'>This will change the <strong>posted</strong> users Rank and Rep!</td> </tr> <tr> <td align='right'>Username:</td><td><input type='text' name='nameinput' class='input'></td> </tr> <tr> <td align='right'>Rank:</td><td><input type='text' name='rankinput' class='input'></td> </tr> <tr> <td align='right'>Rep:</td><td><input type='text' name='repinput' class='input'></td> </tr> <tr> <td align='center' colspan='2'><input type='submit' name='selectname' class='button'></td> </tr> </table> </form> </body> </html> Anyone able to see where I'm going wrong? Thanks Hi. im baffled on this. boss wants a traffic feed for the office so im using SimplePie to grab rss feeds and the google maps traffic overlay as a view of the roads. The trouble is when i do it via xampp locally it looks fine, but when i load into onto our local webserver (ubuntu 18.04 with apache) it only shows a small line and cant figure out why? any help is appreciated. The attached images are how my local xampp sees it and the other is via the ubuntu 18.04 with apache, php & mysql installed My Code is below
<?php
// Include the SimplePie library
require_once('php/autoloader.php');
// Because we're using multiple feeds, let's just set the headers here.
header('Content-type:text/html; charset=utf-8');
// These are the feeds we want to use
$feeds = array(
'http://m.highwaysengland.co.uk/feeds/rss/AllEvents.xml',
'https://traffic.wales/feeds/incidents-events/rss.xml',
'https://traffic.wales/feeds/roadworks/rss.xml',
'https://trafficscotland.org/rss/feeds/currentincidents.aspx',
'https://trafficscotland.org/rss/feeds/roadworks.aspx'
);
// This array will hold the items we'll be grabbing.
$first_items = array();
// Let's go through the array, feed by feed, and store the items we want.
foreach ($feeds as $url)
{
// Use the long syntax
$feed = new SimplePie();
$feed->set_feed_url($url);
$feed->init();
// How many items per feed should we try to grab?
$items_per_feed = 5;
// As long as we're not trying to grab more items than the feed has, go through them one by one and add them to the array.
for ($x = 0; $x < $feed->get_item_quantity($items_per_feed); $x++)
{
$first_items[] = $feed->get_item($x);
}
// We're done with this feed, so let's release some memory.
unset($feed);
}
// We need to sort the items by date with a user-defined sorting function. Since usort() won't accept "SimplePie::sort_items", we need to wrap it in a new function.
function sort_items($a, $b)
{
return SimplePie::sort_items($a, $b);
}
// Now we can sort $first_items with our custom sorting function.
usort($first_items, "sort_items");
// Begin the (X)HTML page.
?>
<!DOCTYPE html>
<html>
<head>
<meta name="viewport" content="initial-scale=1.0, user-scalable=no">
<meta charset="utf-8">
<title>Traffic Layer</title>
<link rel="stylesheet" href="https://stackpath.bootstrapcdn.com/bootstrap/4.2.1/css/bootstrap.min.css" integrity="sha384-GJzZqFGwb1QTTN6wy59ffF1BuGJpLSa9DkKMp0DgiMDm4iYMj70gZWKYbI706tWS" crossorigin="anonymous">
<style type="text/css">
body {
height: 100%;
font:12px/1.4em Verdana, sans-serif;
color:#333;
background-color:#fff;
}
a {
color:#326EA1;
text-decoration:underline;
padding:0 1px;
}
a:hover {
background-color:#333;
color:#fff;
text-decoration:none;
}
div.header {
border-bottom:1px solid #999;
}
div.item {
padding-left:5px;
border-bottom:1px solid #999;
}
div#site {
padding:5px 0;
width:300px;
}
/* Always set the map height explicitly to define the size of the div
* element that contains the map. */
#map {
height: 100%;
}
/* Optional: Makes the sample page fill the window. */
html, body {
height: 100%;
margin: 0;
padding: 0;
}
footnote, p{
color:red;
font:9px/1.4em Verdana, sans-serif;
font-weight:bold;
}
</style>
</head>
<body>
<!-- NEW -->
<div class="row">
<div class="col-md-2">
<!-- BOF Simple Pie News Feed Info -->
<div id="site">
<?php
foreach($first_items as $item):
$feed = $item->get_feed();
?>
<div class="chunk">
<h6><b><?php echo html_entity_decode($item->get_title(), ENT_QUOTES, 'UTF-8'); ?></a></b></h6>
<?php echo $item->get_content(); ?>
<?php if ($enclosure = $item->get_enclosure()): ?>
<div>
<?php echo $enclosure->native_embed(array(
)); ?>
</div>
<?php endif; ?>
<p class="footnote">Source: <?php echo $feed->get_title(); ?></a> | <?php echo $item->get_date('j M Y | g:i a'); ?></p>
</div>
<?php endforeach; ?>
</div>
<!-- EOF Simple Pie News Feed Info --> </div>
<div class="col-md-10">
<!-- BOF Map Info -->
<div id="map"></div>
<script>
function initMap() {
var map = new google.maps.Map(document.getElementById('map'), {
zoom: 7.4,
center: {lat: 53.8059821, lng: -1.6057714}
});
var trafficLayer = new google.maps.TrafficLayer();
trafficLayer.setMap(map);
}
</script>
<script async defer
src="https://maps.googleapis.com/maps/api/js?key=<My Google API Key>&callback=initMap">
</script>
<!-- EOF Map Info --></div>
</div>
<!--OEF Test -->
</script>
</body>
</html>
I've got another funny one... I've noticed that sometimes on some of my pages (all of which are generated by PHP), when i click view source, the source it shows me is not from the page i'm looking at. This isn't a problem that affects me in any way (not yet anyway), it's just a bit odd. I've noticed it happens a little more often in chrome than any other browser. Has anyone else noticed anything like this before? I'm intrigued to know the cause. The only real google response was this: http://www.mail-archive.com/debian-bugs-dist@lists.debian.org/msg144609.html and i haven't seen another issue similar on here. The php web video script that I'm using displays user/member account pages in google searches. For example, the url displaying the page is the https://webvideosite.com/@chrisj I'm looking for guidance/suggestions on how to have the user/member account pages not appear in searches, either by blocking that somehow, or only having access to it if you are logged into the site. I look forward to any replies i have built pages that paginate with 10 rows per page (some pages show more but for the moment i want to focus on this particular page)
//Define Some Options for Pagination
$num_rec_per_page=10;
if (isset($_GET["page"])) { $page = $_GET["page"]; } else { $page=1; };
$start_from = ($page-1) * $num_rec_per_page;
$results = mysql_query("SELECT * FROM `ecmt_memberlist` WHERE toonCategory='Capital' AND oldMember = 0 ORDER BY CONCAT(MainToon, Name) LIMIT $start_from, $num_rec_per_page") or die(mysql_error());
$results_array = array();
while ($row = mysql_fetch_array($results)) {
$results_array[$row['characterID']] = $row;
}
The above sets the variables for the pagination and below the results are echo with 10 rows per page
then i show the pagination links:
<?php
$sql = "SELECT * FROM `ecmt_memberlist` WHERE toonCategory='Capital' AND oldMember = 0 ORDER BY CONCAT(MainToon, Name)";
$rs_result = mysql_query($sql); //run the query
$total_records = mysql_num_rows($rs_result); //count number of records
$total_pages = ceil($total_records / $num_rec_per_page);
?>
<table width="100%" border="0" cellspacing="0" cellpadding="0">
<tr>
<td align="center"><div style="width:100%; text-align:center;">
<ul class="pagination">
<li class="selected">
<?php
echo "<a href='capitalmember.php?page=1'>".'«'."</a> ";?>
</li>
<?
for ($i=1; $i<=$total_pages; $i++) {
echo "<li><a href='capitalmember.php?page=".$i."'>".$i."</a></li> ";
};
?>
<li class="selected">
<?
echo "<a href='capitalmember.php?page=$total_pages'>".'»'."</a> "; // Goto last page
?></li>
</ul></div>
<?php
$pageNr = $page; // Get Current Page Number
$from = $pageNr * $rowsPerPage; // 3 * 10 = 30 // 3 * 10 = 30
$to = $from + $rowsPerPage; // 30 + 10 = 40
echo $pageNr;
/* Result: From page 30 to 40 */
?></td>
</tr>
</table>
this works great and shows me 10 results per page,
what i need to work out and work on next is:
echo the number of results above the records (for example: "showing records 1 to 10" and then on page 2 "showing records 11 to 21" and page 3 "showing records 22 to 32"
how can i work out the maths for this echo?
i was thinking along the lines of;
<?php
$pageNr = $page; // Gets Current Page Number
$from = $pageNr * $rowsPerPage; // 3 * 10 = 30
$to = $from + $rowsPerPage; // 30 + 10 = 40
// Now Show Results
echo $from; // Echo from
echo $to // Echo to
?>
but i'm still working on this..
then i need to look at shortening the amount of page links on the page if for example i have 500 records it shows me links 1 to 50 and eats up the page....
Appreciate any help and light onto my problems.
Many thanks
Edited by jacko_162, 11 January 2015 - 05:43 PM. I had a wordpress site setup for my maintenance company. I am not trying to setup a blog on the site. I have the blogroll showing up on this page: http://handymore.com/blog/ but when you click the individual articles the single post page shows a Error 404 message. Could anyone point me in the right direction as to how to fix this so that the individual post pages are found? Any guidance would be appreciated. I have some thousands of photos about nature I ll let visitors/members to see them one by one, but I dont want to show them the same photo again after they visit 1 week later How can I do this ? What I think as a solution is; For members; I can store the ids (like "everest01") of the photos that member has visited , and show user the most visited photos that he/she has not see for next visit. But what I m wondering is, how will I take the photos from DB ? select * from photos WHERE id not in ( $thousandsofvisitedphotoids ) ?? I m stuck here ? For visitors ( not members ) ; I can set a cookie that keeps the ids of visited photos.. when visitor visits the website again, I take the cookie and sent to $thousandsofvisitedphotoids and make a query again ? I m stuck here, How you guys do this ? what's the logic of this ? How should I go about keeping track of photos that a user has rated, since I only want the user to rate the photo once? Can I store arrays in a mysql database? thanks, George Hi! I am creating a small project somewhat like a photo gallery as my first practice page. I just would like to know how can I alter photos or file in php. Or should I just do a delete and then upload a new one? Thanks in advance... I have a page for image uploads and I just realized it will only work if a user already has on picture uploaded. If they don't it won't work. The ones that do fail some photos anyway which is probably that they don't pass the image check but when I put echoes in there to trace what happens any user with an empty gallery can't upload a photo because the page says there is no file in the $_FILES['image']['name'] variable. Here are the initial conditions and the form (leaving out the image processing etc since that works): Code: [Select] if (!isset($_SESSION['user'])) die("<br /><br /> You need to log in to view this page"); $user = sanitizeString($_SESSION['user']); $view = sanitizeString($_GET['view']); $dir = './grafik/users/'.$user.'/big/'; $files = scandir($dir); $len = count($files); $nr= $len-1; $maxPhotos = 8; if ($view == $user) { echo "view is user"; if(!file_exists("grafik/users/$user")) { mkdir("grafik/users/$user"); mkdir("grafik/users/$user/big/");} } if (!isset($_FILES['image']['name'])) echo "There is no file <br />$dir - $user - $nr"; if (isset($_FILES['image']['name'])) { echo "<br />...is a file <br />$dir - $user - $nr"; $photoName="$dir$user$nr.jpg"; move_uploaded_file($_FILES['image']['tmp_name'], $photoName); $typeok = TRUE; .... <form method='post' action='gallery.php?view=$user' enctype='multipart/form-data'> Upload another photo: <br /> Max $maxPhotos allowed, max filesize 2Mb <br /> <input type='file' name='image' size='10' /><br /> <input type='submit' value='Upload' /> </form> I can't see why it wouldn't let me but I have a feeling someone here knows why. Does anyone know of a good tutorial on uploading a picture file to a folder using php and copying the name to the database in mysql? Resizing photos on upload is helpful also...If you know it works...some I have tried do not work. Not asking for someone to write code for me but info or tutorial would be nice. Or maybe a code that has worked for you that is similar that I could learn from and edit... I can upload the actual photo into the database but it slows it way down. I have heard of loading the name only and resizing the photo and sending the actual photo to a file folder on the server. The few codes I have tried were not successful. Thanks for any guidance. I appreciate it. Hi friends, I have two mysql db tables, photos and album, I would like to list photos by album how do i do that ? CREATE TABLE `album` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `album_name` varchar(95) NOT NULL, `album_desc` text NOT NULL, PRIMARY KEY (`id`) ); CREATE TABLE `photos` ( `id` int(11) unsigned NOT NULL AUTO_INCREMENT, `album_id` int(11) NOT NULL, `thumb_name` varchar(255) NOT NULL, `photo_name` varchar(250) NOT NULL, PRIMARY KEY (`id`) ) Hello. My website has a photo gallery of thumbnails that is created by reading all photo files in a specified directory. Here is my function that builds the array which is ultimately displayed in the gallery...
<?php
function getPhotoFilesArray($photoPath){
/**
* Takes path to photo-directory, and returns an array containing photo-filenames.
*/
// Initialize Array.
$photoFiles = array();
// Check for Photo-Directory.
if (is_dir($photoPath)){
// Photo-Directory Found.
// Open Directory-Handle.
$handle = opendir($photoPath);
// Open Photo-Directory.
if($handle){
// Initialize Key.
$i = 1001;
// Iterate through Photo-Directory items.
while(($file = readdir($handle)) !== FALSE){
// Return next Filename in Directory.
// Define fullpath to file/folder.
$fullPath = $photoPath . $file;
// Populate Array.
if(!is_dir($fullPath)
&& preg_match("#^[^\.].*$#", $file)){
// Not Directory.
// Not Hidden File.
// Add to array.
$photoFiles[$i] = $file;
$i++;
}//End of POPULATE ARRAY.
}//End of ITERATE THROUGH PHOTO-DIRECTORY ITEMS
closedir($handle);
}//End of OPEN PHOTO-DIRECTORY
}else{
// Photo-Directory Not Found.
// Redirect to Page-Not-Found.
header("Location: " . BASE_URL . "/utilities/page-not-found");
// End script.
exit();
}//End of CHECK FOR PHOTO-DIRECTORY
return $photoFiles;
}//End of getPhotoFilesArray
?>
Everything works fine locally in DEV, but when I uploaded my website (and photos) onto a webserver, the photos are appearing in a backwards order in PROD. This is annoying, because I want the photos displayed chronologically from oldest (first) to newest (last). I'm not sure where the problem is happening, because each photo was taken with my camera and by nature of the camera, photo names are incremented by one, so IMG_001.jpg would have been taken FIRST, followed by IMG_002.jpg, IMG_003.jpg, and so on. How can I fix things so the photos are displayed in the order they were physically taken AND match how things display locally in DEV? Thanks!
My photo files are not being displayed in my table? They get sent to the mySQL database, then the server and it does grab all the other variables in the table and displays them, but the .jpg's are not shown, instead theres just the file name?? Code: [Select] <?php error_reporting(E_ALL); ini_set("display_errors", 1); echo '<pre>' . print_r($_FILES, true) . '</pre>'; //This is the directory where images will be saved $target = "/home/users/web/b109/ipg.removalspacecom/images/COMPANIES"; $target = $target . basename( $_FILES['upload']['name']); //This gets all the other information from the form $company_name=$_POST['company_name']; $basicpackage_description=$_POST['basicpackage_description']; $location=$_POST['location']; $postcode=$_POST['postcode']; $upload=($_FILES['upload']['name']); // Connects to your Database mysql_connect("server****", "username***", "password****") or die(mysql_error()) ; mysql_select_db("DB") or die(mysql_error()) ; //Writes the information to the database mysql_query("INSERT INTO `Companies` (company_name, basicpackage_description, location, postcode, upload) VALUES ('$company_name', '$basicpackage_description', '$location', '$postcode', '$upload')") ; echo mysql_error(); //Writes the photo to the server if(move_uploaded_file($_FILES['upload']['tmp_name'], $target)) { //Tells you if its all ok echo "The file ". basename( $_FILES['upload']['name']). " has been uploaded, and your information has been added to the directory"; } else { //Gives and error if its not echo "Sorry, there was a problem uploading your file."; } ?> "upload" is the variable that isnt displaying in my table how i want it to? Have you guys any ideas how to get it displayed correctly? |
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