PHP - How To Link External Files With Php
I know basically nothing about PHP but i have a few websites i manage that are built with PHP.
I plan to learn PHP but for the time being learning Javascript has a wider range of usability for me. That being said, Im trying to link to an XML sitemap using php. in html i would use Code: [Select] <link href="sitemap.xml" /> how can i do this in PHP? I have searched google and such for how to do this but was unable to find an explanation that i could understand as relating to my need. Similar Tutorialshi 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"; ?> Hello, I am trying to make the buy product button on all single product pages to open in a popup iframe or simular! I am new to java, php and such and I am looking for how to achieve this?
I have been researching this for over a month and nothing seems to work. I am affiliated with Amazon so I need my woocommerce buy product buttons to open a new smaller popup window in front of my site!
I have already had it set to open as _blank, But I really need all my external product button links to open as a popup! Is this possible? I have tried editing cart.php and I know nothing about java. Please help!!
Thank you in advance,
Dee
in need of an external link counter script! please help cheers matt Folks, When someone click on a link on my webpage and if that Link takes the visitor off my site, then i want to make an entry in my database. But if that link goes back to some other link in my Site then i do not want to Count that link. So, is there any code that lets me detect any outgoing link Click? Cheers So i have a file uploader that allows people to upload files to the server and then download them via a URL. I want to store uploaded files outside of the Root. (One level up from the root in a folder called Uploads) Code snippet Code: [Select] $server = "http://www.mysite.com"; $name = $_FILES['file']['name']; $temp = $_FILES['file']['tmp_name']; $size = $_FILES['file']['size']; $destination = '../uploads/'. $random; mkdir($destination); move_uploaded_file($temp, $destination."/".$name); $final = $server."/".$destination."/".$name; This... Code: [Select] $destination = '../uploads/' . $random ; Didn't work, any tips? So say i have this: Code: [Select] <link rel="stylesheet" media="all" type="text/css" href="_css/styles.css" /> now the file that is linking to that is 3 directories deep, besides using a rel or absolute link, could i define the path by using something like this Code: [Select] <link rel="stylesheet" href="<?php require($_SERVER['DOCUMENT_ROOT'] . "/_css/portfolio-styles.css"); ?>" type="text/css" /> but this is actually opening the CSS in the header thanks for any explanation on how to do this So far I have managed to create an upload process which uploads a picture, updates the database on file location and then tries to upload the db a 2nd time to update the Thumbnails file location (i tried updating the thumbnails location in one go and for some reason this causes failure) But the main problem is that it doesn't upload some files Here is my upload.php <?php include 'dbconnect.php'; $statusMsg = ''; $Title = $conn -> real_escape_string($_POST['Title']) ; $BodyText = $conn -> real_escape_string($_POST['ThreadBody']) ; // File upload path $targetDir = "upload/"; $fileName = basename($_FILES["file"]["name"]); $targetFilePath = $targetDir . $fileName; $fileType = pathinfo($targetFilePath,PATHINFO_EXTENSION); $Thumbnail = "upload/Thumbnails/'$fileName'"; if(isset($_POST["submit"]) && !empty($_FILES["file"]["name"])){ // Allow certain file formats $allowTypes = array('jpg','png','jpeg','gif','pdf', "webm", "mp4"); if(in_array($fileType, $allowTypes)){ // Upload file to server if(move_uploaded_file($_FILES["file"]["tmp_name"], $targetFilePath)){ // Insert image file name into database $insert = $conn->query("INSERT into Threads (Title, ThreadBody, filename) VALUES ('$Title', '$BodyText', '$fileName')"); if($insert){ $statusMsg = "The file ".$fileName. " has been uploaded successfully."; $targetFilePathArg = escapeshellarg($targetFilePath); $output=null; $retval=null; //exec("convert $targetFilePathArg -resize 300x200 ./upload/Thumbnails/'$fileName'", $output, $retval); exec("convert $targetFilePathArg -resize 200x200 $Thumbnail", $output, $retval); echo "REturned with status $retval and output:\n" ; if ($retval == null) { echo "Retval is null\n" ; echo "Thumbnail equals $Thumbnail\n" ; } }else{ $statusMsg = "File upload failed, please try again."; } }else{ $statusMsg = "Sorry, there was an error uploading your file."; } }else{ $statusMsg = 'Sorry, only JPG, JPEG, PNG, GIF, mp4, webm & PDF files are allowed to upload.'; } }else{ $statusMsg = 'Please select a file to upload.'; } //Update SQL db by setting the thumbnail column to equal $Thumbnail $update = $conn->query("update Threads set thumbnail = '$Thumbnail' where filename = '$fileName'"); if($update){ $statusMsg = "Updated the thumbnail to sql correctly."; echo $statusMsg ; } else { echo "\n Failed to update Thumbnail. Thumbnail equals $Thumbnail" ; } // Display status message echo $statusMsg; ?> And this does work on most files however it is not working on a 9.9mb png fileĀ which is named "test.png" I tested on another 3.3 mb gif file and that failed too? For some reason it returns the following Updated the thumbnail to sql correctly.Updated the thumbnail to sql correctly. Whereas on the files it works on it returns REturned with status 0 and output: Retval is null Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif' Failed to update Thumbnail. Thumbnail equals upload/Thumbnails/'rainbow-trh-stache.gif'The file rainbow-trh-stache.gif has been uploaded successfully. Any idea on why this is? Hello I have a simple question about file handling... Is it possible to list all files in directories / subdirectories, and then read ALL files in those dirs, and put the content of their file into an array? Like this: array: [SomePath/test.php] = "All In this php file is being read by a new smart function!"; [SomePath/Weird/hello.txt = "Hello world. This is me and im just trying to get some help!";and so on, until no further files exists in that rootdir. All my attempts went totally crazy and none of them works... therefore i need to ask you for help. Do you have any ideas how to do this? If so, how can I be able to do it? Thanks in Advance, pros I am using WPSQT plugin in my blog site .I code some files in PHP also.how to add that files in plugin files.
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 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 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 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 $num hyperlinked. however when you click on it it doesn't go anywhere. when copying the link location from the email it gives me x-msg://87/%22http://www.example.com/16 what is x-msg? how can i get this to work properly? I would like a part of my script to link to an external sites script It doesn't seem to be doing it though Othersite.com - index.php?name=hello&status=1 I would like my post.php to run that above. Can it be done? I was just wondering if it was possible to have an external php file that can be included in the head of a web page, like a .js file. If this isn't possible maybe have a .js file containing php code that can be executed regarding the JavaScript around it... Hi all,
So im trying to improve on my PHP as my knowledge isn't that great.
lets say for argument sake my ip is:
192.168.0.1
im trying to set a external variable to set this as the url.
so on each page i use, for each link instead of typing out the ip address i can simply type:
<a href="{url}/index.php">Home</a>
and then if it is possible, to do something like this:
<link rel="stylesheet" href="{url}/main.css" type="text/css">
all in one file so in my php pages i can simple include 1 php file and it will have all of the relevant stylesheets ect linked to it.
Edited by srwright, 30 May 2014 - 01:18 PM. Hi, I have a peice of code which publishes an image with a link from my database. However I cant get it to use external links. My code is: echo "<a href=\"" .$link . "\"> <img src=\"" .$image ."\" /> </a><BR />"; I have tried all the options I can think of but I cant get it work. Can anyone advise please? I'm building a program in php that will be able to view YouTube videos from the URL. So if I type: http://www.youtube.com/watch?v=(Video ID) Then my program works and it will return the Video ID! But then some idiot clicks on a related video and this URL is generated: http://www.youtube.com/watch?v=(Video ID)&feature=related And my program strips the first bit by replacing "http://www.youtube.com/watch?v=" with an empty string "" but then I'm still left with the "&feature=related" I thought about just replacing that with an empty string as well but sometimes there can be a URL like this: &feature=g-vrec&context=G28d9eecRVAAAAAAAABA Which has a different unique code each time. So I thought it'd be much simpler if I could use $_GET[] with an external URL, so the user types in: http://www.youtube.com/watch?v=(Video ID)&feature=related It just gets the "v" value rather than my buggy replace thing. Thanks. 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! |