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PHP - Php Code To Detect External Outgoing Link Click
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 Similar Tutorialshey there peoples i have a bit of a problem with my php code. when i click on the name it wont play the video. but if i click on the thumbnail it plays here is the code Code: [Select] echo "<div class='Video_thumbnail'>"; echo "<a href='/Video.php?id={$row['id']}'>"; echo "<img src='http://i.ytimg.com/vi/{$row['link']}/hqdefault.jpg' width='220' height='170' alt='click' /></a>"; echo "<br>"; echo "</br>"; echo "<a href='/Video.php?={$row['id']}'>"; echo "{$row['navn']}</a>"; echo "</div>"; 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"; ?> i am facing a problem regarding the insertion of of link in a text area is not detected but the output is a normal text "unlike facebook comments" is there any php functions ??? thanks in advance Code: [Select] if ($showigiteidimage) { $igitwid_divlnk = "onclick=location.href='" . get_permalink($igitwid_result->ID) . "'; style=cursor:pointer;"; $igitwid_output .= '<li id="igit_wid_rpwt_li" style="height:' . $igitwid_height . 'px; padding-bottom: 10px;" ' . $igitwid_divlnk . '>'; $igitwid_output .= '<div id="igit_wid_rpwt_main_image" style="float:left;"> <a href="' . get_permalink($igitwid_result->ID) . '" target="_top"><img src="<?php $values = get_post_custom_values("tj_video_img_url"); echo $values[0]; ?>" id="igit_rpwt_thumb" ></a></div>'; the prb is image src cant read the php code , what to do ? please helpppppppppp i'm trying to use this code to detect if the browser of the user is IE (any version) and id true (if it is IE), then it's redirected to an error page; otherwise, it's ok and sent ot the index. I get the following error... Warning: Cannot modify header information - headers already sent by (output started at /index.php:1) in /xie.php on line 13 any ideas.. ??? <?php function ae_detect_ie() { if (isset($_SERVER['HTTP_USER_AGENT']) && (strpos($_SERVER['HTTP_USER_AGENT'], 'MSIE') !== false)) return true; else return false; } if (ae_detect_ie()) { header("Location: error.php"); } else { header("Location: index.php"); } ?> Hi there, I've looked into the PHP superglobal array $_SERVER, but could not find a variable that stores the plugins a client has on his/her browser such as firefox and google chrome. The point here is to detect if a user has hacking addons such as firebug and inspect element installed, and displays an error message telling the user to disable such plugins in order to access site content. Is it possible to accomplish such tasks? Please help. 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. 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 I've just setup an Apache server with a basic LAMP stack. We've decided to use Google Apps (gmail) as our mail server / host. So our ubuntu server is only serving up HTTP files. Because the server is not configured as a mail server, and we're using Gmail for our MX and mail servers, how do I go about using sending out emails via PHP on the Apache Box? I've tried installing Postfix, and configuring using this tutorial, and still a no go. http://www.blogternals.com/2009/04/30/postfix-google-apps-gmail-smtp-relay/ Any Ideas? This topic has been moved to PHP Freelancing. http://www.phpfreaks.com/forums/index.php?topic=358545.0 i have this simple table browse get me the filter let say 15 items of a table of 74 but after i click in the link forward the filter erase and disply me the all records... waht im doing wrong.... TNX in advance <img src="images/misc/search.JPG" width="367" height="100" /> <br /> <?php $strng = $_POST['search:strng']; $separa = "%"; $all = $separa.$strng.$separa; $mensaj="U Look for : "; $mensaj2=$mensaj.$stng; $regis=" u search get :"; $regis2=" Items"; $conn = mysql_connect('localhost','zerocctv_Admin','password') or trigger_error("almacen", E_USER_ERROR); $db = mysql_select_db('zerocctv_almacen',$conn) or trigger_error("almacenz", E_USER_ERROR); $sql = "SELECT COUNT(*) FROM modelos where descrip like '$all'"; $result = mysql_query($sql, $conn) or trigger_error("SQL", E_USER_ERROR); $r = mysql_fetch_row($result); $numrows = $r[0]; $rowsperpage = 10; $totalpages = ceil($numrows / $rowsperpage); // get the current page or set a default if (isset($_GET['currentpage']) && is_numeric($_GET['currentpage'])) { $currentpage = (int) $_GET['currentpage']; } else { $currentpage = 1; } if ($currentpage > $totalpages) { $currentpage = $totalpages; } if ($currentpage < 1) { $currentpage = 1; } $offset = ($currentpage - 1) * $rowsperpage; $sql = "SELECT modelo, paginal,precio,foto FROM modelos where descrip like '$all' LIMIT $offset, $rowsperpage"; $result = mysql_query($sql, $conn) or trigger_error("SQLkk2", E_USER_ERROR); echo $Mensaj2 ; echo $regis. $numrows . $regis2 ; echo "<table width=\50%\ border=\1\ bordercolor=\#0033FF\ cellspacing=\1\">" ; while ($list = mysql_fetch_assoc($result)) { echo "<tr>"; echo "<td>><font color='red'>" . $list['model'] . "</font></td>"; echo "<td><font color='black'>" . $list['pagl'] . "</font></td>"; echo "<td><font color='black'>" . $list['prize'] . "</font></td>"; echo "<td><font color='black'>" . $list['pict'] . "</font></td>"; echo "<tr></tr>"; echo "</tr>"; } echo "</table>"; //****** build the pagination links ******/ $range = 3; if ($currentpage > 1) { echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=1'><<</a> "; $prevpage = $currentpage - 1; echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$prevpage'><</a> "; } for ($x = ($currentpage - $range); $x < (($currentpage + $range) + 1); $x++) { if (($x > 0) && ($x <= $totalpages)) { if ($x == $currentpage) { echo " [<b>$x</b>] "; } else { echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$x'>$x</a> "; } } } if ($currentpage != $totalpages) { $nextpage = $currentpage + 1; echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$nextpage'>></a> "; echo " <a href='{$_SERVER['PHP_SELF']}?currentpage=$totalpages'>>></a> "; } ?> Every other time a user/users click on a link I want the link to change to a different url the next time a user clicks onto that link.
For example two links
url1.html
url2.html
url1.html shows up when a user clicks that link the link auto changes to then when a new user clicks the linke url2.html then back to url1.html and so on?
Don't know where to start on how I would do this let alone have the server remember to keep which ever link to be held for the next user?
Hello,
I am using this collapse menu from bootstrap:
http://getbootstrap....cript/#collapse
What I did is instead of putting text in the content section I added many links. The links only show up when one of the menu link is clicked.
Now my problem is that in one of the menu sub, there are about 30 links, therefore the user has to go down the page to see the content.
Is there a way to have the menu collapsing on itself when a link is clicked please? Like a reset to reset it like on page load
Thanks all,
Ben
how can i do this pls? i just want to display a link "Free Chat" when i click on it, it must display a dropdown with options "yes" and "no" the link must not go anywhere yet, only when yes is selected it must go to a page i have the code: Code: [Select] <a href="" id="link" name="link" onClick="yesnolist(1)">Free Chat</a> then i have javascript: Code: [Select] <script type="text/JavaScript"> function yesnolist() { var e = document.getElementById("link"); var strUser1 = e.options[e.selectedIndex].value; if (strUser1 == "1") { window.location.href = "http://........."; } return strUser1; } </script> the above is not correct, can some-one help me please? thanks 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. <html> <?php $id = $_GET['id']; $dbusername="web148-matt"; $dbpassword="matt"; $dbdatabase="web148-matt"; mysql_connect(localhost,$dbusername,$dbpassword); @mysql_select_db($dbdatabase) or die( "Unable to select database"); mysql_query("UPDATE count SET clicks=clicks+1 WHERE id='$id'"); $sql = mysql_query("SELECT link FROM count WHERE id='$id'"); $fetch = mysql_fetch_row($sql); $result = mysql_query("SELECT * FROM count"); while($row = mysql_fetch_array($result)) { echo "<a href=" .$row['link']. ">Link</a>"; } ?> <a href='http://www.google.com'>Google</a> <a href='/index.php?id=2'>link2</a> </html> I'm using a templated PHP script and have successfully added a link (to an image) I am having trouble figuring this out myself, so hopefully someone can help me out. What I am trying to do is write a code to link an image to a logged in users "edit profile" page. Right now, the page address the image needs to be linked to would be: "MyCollegeFleaMarket.com/users/Username" Currently, the code I have in place is this: "users/<?php global $user; if ($user->uid) { print $user->name; } ?>"> but this is creating a link to, "MyCollegeFleaMarket.com/content/users/Username. Thanks for your help! |
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