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PHP - Link Script
I would like it to list a bunch of music in a folder but the user dosent have accesses to except me.
So the page he can view will list all the files in the other folder and has a downloadable link to it so he can download the files Thanks Dan Similar TutorialsSeems to be a basic question, but I couldn't find an answer nor figure it out on my own. Basically I have a script that takes out specific data out of the database, the script works on its own, now I just need a way to make the user execute it with a link or a button. Example: Category: [Smileys] - [Category2] - [Category3] - etc. As soon as the user clicks on [Smileys] all data in the database which contains the word smileys in the category field gets selected and outputted as a list. Again the script works, I just need to be able to execute it with a button. If I understood it correctly I have to run the script by adding a if (isset($_POST['Smileys'])) { in front of the script. But how do I build the connection with the text link? in need of an external link counter script! please help cheers matt Hello, I had a web page which had sharing (share issue on facebook, twitter....) and I want to record what each user had shared. So, they will click on alink to share it. I want to call a php function when the user click on a link, whats the best method to do this? Thanks in advance This is my script I want to display but it does not show in the page just blank. <td><object width="220" height="148"><param name="movie" value="http://www.youtube.com/user/THEWORLDOFTRAVEL#p/search/2/uX9nd3xM_MY=1&h1=enUS&color1=0x234900&color3=34900&"></param><param name="allowFullScreen" value="true"></param><param name="allowscriptaccess" value="always"></param><embed src="http://www.youtube.com/user/THEWORLDOFTRAVEL#p/search/2/uX9nd3xM_MY=1&h1=enUS&color1=0x234900&color3=34900&" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="220" height="148"></embed></object></td> Hello there Really new by in PHP, I find out your forums and look really professional, so I'm dropping in some questions. Maybe can have some help. I'm setting an old plugin of WP and got some issues. The plugin is really nice and dedicated to create a directory inside a website in WP but regretfully no one take care of that code, so no one to ask for. There are some errors. Example, lets go with the first one. When I (or someone) add a new website in the directory the script does not manage correctly a duplicate submissions. I can submit more times the same url. It is supposed to manage that problem, here is the code that take care of all the errors. Code: [Select] if($title == "" || $url == " " || $url == "http://" || $description == "" || $description == " " || $keyword == "" || $contact_name == "" || $contact_email == "" || !$cat_id) { if($title == "") $error_msg .= "- " . 'Please insert the title of the website!' . " - <br />"; if($url == " " || $url == "http://" ) $error_msg .= "- " . 'Please insert the URL of the website!' . " - <br />"; if(!$cat_id) $error_msg .= "- " . 'Please select a category for the website!' . " - <br />"; if($description == " ") $error_msg .= "- " . 'Please write the description of the website!' . " - <br />"; if($description == "") $error_msg .= "- " . 'Please write the description of the website!' . " - <br />"; if($keyword == "") $error_msg .= "- " . 'Please write the keywords of the website!' . " - <br />"; if($contact_name == "") $error_msg .= "- " . 'Please insert your name!' . " - <br />"; if($contact_email == "") $error_msg .= "- " . 'Please insert your email address!' . " - <br />"; } else { $link_id = $class_link->GetLinkId($cat_id, $url, $title); if ($link_id != "") { $error_msg .= "- " . 'This link currently exist!' . " - <br /> "; } } It is working fine for all the errors but NOT for the last one... is not working, no message ""This link currently exist"" and no error come out after the submission that go in smoothly. So here is the problem.... but can't solve it. Code: [Select] } else { $link_id = $class_link->GetLinkId($cat_id, $url, $title); if ($link_id != "") { $error_msg .= "- " . 'This link currently exist!' . " - <br /> "; } } Any idea? Hi all, I'm not great with PHP and I have this one problem that I hope you can help with. I have a script that generates an URL. The output string is $shorturl and displays properly when I echo it using Code: [Select] <?php echo '$shorturl' ?>. However, when I try the following script for a Facebook share button Code: [Select] <fb:share-button> <a name="fb_share" type="button" share_url="<?php echo '$shorturl' ?>" href="http://www.facebook.com/sharer.php" layout="button_count">Share</a><script src="http://static.ak.fbcdn.net/connect.php/js/FB.Share" type="text/javascript"></script> </fb:share-button> The resulting URL it produces is: http://</?php+echo+%27http%3A%2F%2Fshelb... Any idea's? Thank you! Hi All, The Problem: I'm trying to learn how to take records from the NAME row of one table (i.e. Tom, Jim, Chris, Mike) and put them into a SELECT (drop down menu) box, then select a name and save it to a DIFFERENT table where the row is also called NAME, and THEN have that selection I just saved show in the SELECT box as SELECTED when the page refreshes. I have found a hundred websites showing how to bind a SELECT box to a table and put them into a SELECT box via $key => $val and even how to save it to ANOTHER textbox on the page, which is a lot easier, but never a step further as described in the first paragraph above. Seems unbelievable to me since it's so common. Maybe I'm just not using the proper search terms/keywords. Here's an example: Let's say you wanted to notify 3 people of your birthday. So, on the notification page, you would enter three names of people you would want notified all in separate input textboxes with [keys]. So, you enter John, Chris, and Tom into three text boxes and save the page. Done. Now, on another page, I have -- say .... five select boxes. For arguments sake, let's just say I click on all of them one at a time. In each of them, I would only see three names (John, Chris, and Tom). Now, in the first SELECT box I click on John. In the second select box I click on Chris, and on the third I click on Tom. Then save the form. I would like to see John, Chris, and Tom in the first three boxes after refreshing and I would like to see SELECT A NAME on the last two Select boxes that we never did anything with. Remember, there are NO default names in the first table with row called NAMES. They are populated by the user and saved to the row. I'm attaching an image of what the page looks like that I want to see the names and be able to select them. Does anyone have a link to a site that shows how to do this... or maybe have something very basic that would show me the process? Or, if not too difficult, maybe you could show me? Thanks in advance for any help! Attached Files Untitled.jpg 24.97KB 0 downloads Goal: To have a gallery that downloads images from the folder I previously uploaded to in a previous script. Bug: When I load the page the thumbnail comes up as broken and when I click on the thumbnail to get the bigger picture it comes up with the following error message: "Firefox doesn't know how to open this address, because the protocol (c) isn't associated with any program." <?php include 'db.inc.php'; //connect to MySQL $db = mysql_connect(MYSQL_HOST, MYSQL_USER, MYSQL_PASSWORD) or die ('Unable to connect. Check your connection parameters.'); mysql_select_db(MYSQL_DB, $db) or die(mysql_error($db)); //change this path to match your images directory $dir ='C:/x/xampp/htdocs/images'; //change this path to match your thumbnail directory $thumbdir = $dir . '/thumbs'; ?> <html> <head> <title>Welcome to our Photo Gallery</title> <style type="text/css"> th { background-color: #999;} .odd_row { background-color: #EEE; } .even_row { background-color: #FFF; } </style> </head> <body> <p>Click on any image to see it full sized.</p> <table style="width:100%;"> <tr> <th>Image</th> <th>Caption</th> <th>Uploaded By</th> <th>Date Uploaded</th> </tr> <?php //get the thumbs $result = mysql_query('SELECT * FROM images') or die(mysql_error()); $odd = true; while ($rows = mysql_fetch_array($result)) { echo ($odd == true) ? '<tr class="odd_row">' : '<tr class="even_row">'; $odd = !$odd; extract($rows); echo '<td><a href="' . $dir . '/' . $image_id . '.jpg">'; echo '<img src="' . $thumbdir . '/' . $image_id . '.jpg">'; echo '</a></td>'; echo '<td>' . $image_caption . '</td>'; echo '<td>' . $image_username . '</td>'; echo '<td>' . $image_date . '</td>'; echo '</tr>'; } ?> </table> </body> </html> Any help appreciated. Hello. I am working on a php script for searching a database table. I am really new to this, so I used the this tutorial http://www.phpfreaks.com/tutorial/simple-sql-search I managed to get all the things working the way I wanted, except one important and crucial thing. Let me explain. My table consist of three columns, like this: ID(bigint20) title(text) link (varchar255) ============================= ID1 title1 link-1 ID2 title2 link-2 etc... Like I said, I managed to make the script display results for a search query based on the title. Want I want it to do more, but I can't seem to find the right resource to learn how, is to place a "Download" button under each search result with its corresponding link from the table. Here is the code I used. <?php $dbHost = 'localhost'; // localhost will be used in most cases // set these to your mysql database username and password. $dbUser = 'user'; $dbPass = 'pass'; $dbDatabase = 'db'; // the database you put the table into. $con = mysql_connect($dbHost, $dbUser, $dbPass) or trigger_error("Failed to connect to MySQL Server. Error: " . mysql_error()); mysql_select_db($dbDatabase) or trigger_error("Failed to connect to database {$dbDatabase}. Error: " . mysql_error()); // Set up our error check and result check array $error = array(); $results = array(); // First check if a form was submitted. // Since this is a search we will use $_GET if (isset($_GET['search'])) { $searchTerms = trim($_GET['search']); $searchTerms = strip_tags($searchTerms); // remove any html/javascript. if (strlen($searchTerms) < 3) { $error[] = "Search terms must be longer than 3 characters."; }else { $searchTermDB = mysql_real_escape_string($searchTerms); // prevent sql injection. } // If there are no errors, lets get the search going. if (count($error) < 1) { $searchSQL = "SELECT title, link FROM db WHERE title LIKE '%{$searchTermDB}%'"; $searchResult = mysql_query($searchSQL) or trigger_error("There was an error.<br/>" . mysql_error() . "<br />SQL Was: {$searchSQL}"); if (mysql_num_rows($searchResult) < 1) { $error[] = "The search term provided {$searchTerms} yielded no results."; }else { $results = array(); // the result array $i = 1; while ($row = mysql_fetch_assoc($searchResult)) { $results[] = "{$row['title']}<br /> Download - this is the button I want to link to the title results - and maybe other links too - <br /> "; $i++; } } } } function removeEmpty($var) { return (!empty($var)); } ?> <?php echo (count($error) > 0)?"The following had errors:<br /><span id=\"error\">" . implode("<br />", $error) . "</span><br /><br />":""; ?> <form method="GET" action="search?" name="searchForm"> Search for title: <input type="text" name="search" value="<?php echo isset($searchTerms)?htmlspecialchars($searchTerms):''; ?>" /> <input type="submit" name="submit" value="Search" /> </form> <?php echo (count($results) > 0)?"Rezultate lucrari de licenta sau disertatie pentru {$searchTerms} :<br /><br />" . implode("", $results):""; ?> $results = array(); // the result array $i = 1; while ($row = mysql_fetch_assoc($searchResult)) { $results[] = "{$row['title']}<br /> Download - this is the button I want to link to the title results - and maybe other links too - <br /> "; $i++; I would like the results to be displayed like this Results for SearchItem: Result 1 Download | Other link Result 2 Download | Other link etc.... or something like this. So, how do I add the data from the link row into a text(Dowload), within an <a href> tag (well, at least I guess it would go this way) ? My first tries (fueled by my lack of knowledge) where things like $results[] = "{$row['title']}<br /> <a href="{$row['link']}">Download</a> <br /> "; but I keep getting lots of errors, and then I don't know much about arrays and stuff (except basic notions); So there it is. I am really stuck and can't seem to find any workaround for this. Any suggestions? (examples, documentation, anything would do, really) Thanks, Radu Hi, n0obie here. I'm trying to identify where my customers are coming from via emails I've sent/received. However, many email providers (namely Gmail) have circumvented this by disallowing IP address geolocation/image caching. I am having problems understanding the reason for why the user has to click logout twice, here's the bulk of the code: <?php ini_set('display_errors',0); require_once 'header.html'; require_once 'db.functions.php'; require_once 'config.php'; $database = dbConnect($host, $username, $password, $database); // should output 1 or nothing at all! if($database == true) { // now connected? // carry on with logic of outputting the blog contents: $result = entries("SELECT * FROM entries"); printf("<table>"); while($row = mysql_fetch_array($result)) { printf(" <tr> <td>%s</td> <td>%s</td> </tr> <tr> <td colspan=\"2\">%s</td> </tr> ", $row[2], $row[4], $row[3]); } printf("</table>"); printf("\n\n"); session_name("jeremysmith_blog"); session_start(); if(array_key_exists('login',$_SESSION)) { if($_SESSION['login'] == 1) // change this to correspond with session on the login.php script { printf("<p>Welcome %s</p> <p>To logout, click <a href=\"index.php?action=logout\">here</a></p> ",$_SESSION['username']); } } else { printf("<p>You are not logged in, please click <a href=\"login.php\">here</a> to login.</p>"); } } else { printf("\n<p id=\"error\">Could not connect to database, please try again later.</p>"); } // init the logout script? if(array_key_exists('action',$_GET)) { if($_GET['action'] == 'logout') { // log user out of the system: unset($_SESSION['login']); unset($_SESSION['username']); session_destroy(); } } printf("\n"); // just for output format! require_once 'footer.html'; Why does the user have to click logout twice, have I missed anything? Any helps appreciated thanks. Hi I've got this database I created with fields ProductId ProductName Image I've managed to get it to list the ID,productname, and Image urls in a list. My next step is to have the image field actually display an image and make it clickable: heres what I've done so far: Code: [Select] <?php $con = mysql_connect("localhost","root",""); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("productfeed", $con); $result = mysql_query("SELECT * FROM productfeeds"); echo "<table border='0'> <tr> <th>Firstname</th> <th>Lastname</th> <th>Image</th> </tr>"; while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; echo "<td>" . $row['ImageURL'] . "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Heres what I want to do: Code: [Select] while($row = mysql_fetch_array($result)) { echo "<tr>"; echo "<td>" . $row['ProductID'] . "</td>"; echo "<td>" . $row['ProductName'] . "</td>"; // my changes beneath echo "<td>" . <a href="<?php echo $row['ImageURL'];?>"> <img src="<?php echo $row['LinkURL']; ?>"> </a>. "</td>"; echo "</tr>"; } echo "</table>"; mysql_close($con); ?> Can you guys point me in the right direction? Many thanks Hello.
I have a bit of a problem. When I fetch the link field from the database.i don't see an actual link on the page.
One more thing, what type of field should I use to store the link in the database? Probably there is where I went wrong.
All help is
Hi Support, I have a form, where it collects user description input. I can collect the inputs and store it with newline. The issue is - how to collect the http link to actual hyperlink ref during display. The following is my code: <textarea name="description" cols="50" rows="10" id="description"><?php echo str_replace("<br>", "\n", $description);?></textarea></td> For example, User input: Hi, Check it out - http://www.google.com/ I would like to display google link as href so that Viewers can click the link and go to the page. Right now, it is not href and user need to copy the link to new tabs or pages and then it can come. Thanks for your help. Regards, Ahsan hi hope you all are fine. i have been working on a Email Form (like user fills up the form which send the information to our email) but i was having problem with (URL field i created) link of form is (http://services.shadowaura.com/allquotations/static.php) field which is not working is "Inspirational Website:" when i submit the form it says (Forbidden You don't have permission to access /allquotations/staticworking.php on this server. Additionally, a 404 Not Found error was encountered while trying to use an ErrorDocument to handle the request.) Can some one help me out ????????????? code behind this form is: Code: [Select] <?php /* Email Variables */ $emailSubject = 'Shadow Aura Contact Info!'; $webMaster = '*****@shadowaura.com'; /* Data Variables */ $Name = $_POST['Name']; $email = $_POST['email']; $Cell = $_POST['Cell']; $Phone = $_POST['Phone']; $CompanyName = $_POST['CompanyName']; $TypeOfBusiness = $_POST['TypeOfBusiness']; $Address = $_POST['Address']; $YourBudget = $_POST['YourBudget']; $HaveDomain = $_POST['HaveDomain']; $RunningWeb = $_POST['RunningWeb']; $WebLink = $_POST['WebLink']; $Inspiration1 = $_POST['Inspiration1']; $Inspiration2 = $_POST['Inspiration2']; $NumberPages = $_POST['NumberPages']; $UseFlash = $_POST['UseFlash']; $TimeFrame = $_POST['TimeFrame']; $Provided = $_POST['Provided']; $Comments = $_POST['Comments']; $body = <<<EOD <h1> Static Website Quotation </h1> <br> <b>Name of Client:</b>$Name<br> <b>Your Email:</b>$email<br> <b>Cell Number:</b>$Cell<br> <b>Line Phone Number:</b>$Phone<br> <b>Company Name:</b>$CompanyName<br> <b>Type of Business:</b>$TypeOfBusiness<br> <b>Address:</b>$Address<br> <b>Your Budget:</b>$YourBudget <br> <b>Do you have Domain:</b>$HaveDomain<br> <b>Your Site is Running:</b>$RunningWeb <br> <b>Website Link:</b><a href="$WebLink">$WebLink</a><br> <b>Inspiration:</b>$Inspiration1<br> <b>2nd Inspiration:</b>$Inspiration2<br> <b>Number of Pages:</b>$NumberPages<br> <b>Use Flash:</b>$UseFlash <br> <b>Time Frame:</b>$TimeFrame<br> <b>You will provide:</b>$Provided<br> <b>Comments:</b>$Comments<br> EOD; $headers = "From: $email\r\n"; $headers .= "Content-type: text/html\r\n"; $success = mail($webMaster, $emailSubject, $body, $headers); /* Results rendered as HTML */ $theResults = <<<EOD <html> <head> <title>sent message</title> <meta http-equiv="refresh" content="3;URL=http://services.shadowaura.com/"> <style type="text/css"> <!-- body { background-color: #8CC640; font-family: Verdana, Arial, Helvetica, sans-serif; font-size: 20px; font-style: normal; line-height: normal; font-weight: normal; color: #fec001; text-decoration: none; padding-top: 200px; margin-left: 150px; width: 800px; } --> </style> </head> <div align="center">Your email will be answered soon as possible! You will return to <b>Shadow Aura Services</b> in a few seconds !</div> </div> </body> </html> EOD; echo "$theResults"; ?> here's my code that i've used to send an email. Code: [Select] $link = "<a href=\"http://www.example.com/" . $num . "\">" . $num . "</a>"; $query = "SELECT content FROM emails"; $result = mysql_query($query) or die(); $email_content = mysql_result($result, 0); $email = sprintf($email_content, $first, $name, $from, $link, $record, $rec, $inc, $max); $email_body = stripslashes(htmlentities($email, ENT_QUOTES, 'UTF-8')); // this is sent to another php script via post.... $subject = $_POST['subject']; $message = nl2br(html_entity_decode($_POST['email_body'])); $to = "me@whatever.com"; $charset='UTF-8'; $encoded_subject="=?$charset?B?" . base64_encode($subject) . "?=\n"; $headers="From: " . $userEmail . "\n" . "Content-Type: text/html; charset=$charset; format=flowed\n" . "MIME-Version: 1.0\n" . "Content-Transfer-Encoding: 8bit\n" . "X-Mailer: PHP\n"; mail($to,$encoded_subject,$message,$headers); in the db, emails.content is of the text type and contains several lines of text with %4$s which inserts the value of $link into the body. when the email arrives, there is a link and it appears fine, with the value of $num hyperlinked. however when you click on it it doesn't go anywhere. when copying the link location from the email it gives me x-msg://87/%22http://www.example.com/16 what is x-msg? how can i get this to work properly? I have a line like this it prints text link but I prefer image link how should i edit it I would appreciate some feedback Code: [Select] $templates['etiket'] = array('name' => t('ETİKET'), 'module' => 'uc_invoice_pdf', 'path' => $templates_uc_invoice_pdf_path, 'pdf_settings' => $pdf_settings); Hi guys, I'm using a twitter script that grabs the title and publishings it like so: "Title - Read More at..." I was wondering how i would be able to post the direct link into twitter.. like news.php?id=1 for example. Code: [Select] $tweet->post('statuses/update', array('status' => ''.$_POST[title].' - more at MY URL')); This part is in the script to publish automatically when the users adds to the news database. How am i able to get the ID just after the posting of the news? Thanks! Hi everyone! I've been working on a php script to replace links that contain a query with direct links to the files they would redirect to. I'm having trouble echoing $year in my script. Listed below is the script, just below ,$result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error());, in the script I try to echo $year. It doesn't show up in the table on the webpage. Everything else works fine. Any help wold be appreciated greatly. Thanks in advance. <?php include 'config2.php'; $search=$_GET["search"]; // Connect to server and select database. mysql_connect($dbhost, $dbuser, $dbpass)or die("cannot connect"); mysql_select_db("vetman")or die("cannot select DB"); $result = mysql_query("SELECT * FROM $dbname WHERE class LIKE '%$search%'") or die(mysql_error()); // store the record of the "" table into $row //$current = ''; echo "<table align=center border=1>"; echo "<br>"; echo "<tr>"; echo "<td align=center>"; ?> <div style="float: center;"><a><h1><?php echo $year; ?></h1></a></div> <?php echo "</td>"; echo "</tr>"; echo "</table>"; // keeps getting the next row until there are no more to get if($result && mysql_num_rows($result) > 0) { $i = 0; $max_columns = 2; echo "<table align=center>"; echo "<br>"; while($row = mysql_fetch_array($result)) { // make the variables easy to deal with extract($row); // open row if counter is zero if($i == 0) echo "<tr>"; echo "<td align=center>"; ?> <div style="float: left;"> <div><img src="<?php echo $image1; ?>"></div> </div> <?php echo "</td>"; // increment counter - if counter = max columns, reset counter and close row if(++$i == $max_columns) { echo "</tr>"; $i=0; } // end if } // end while } // end if results // clean up table - makes your code valid! if($i > 0) { for($j=$i; $j<$max_columns;$j++) echo "<td> </td>"; echo '</tr>'; } mysql_close(); ?> </table> |
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