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PHP - Php Echo To Link
How do I make this output a clickable link:
<?php echo $_POST["website"]; ?><br /> Similar TutorialsI have created a database that echos out the result on a webpage, what i want to do is create one of the echo results into a link - the code is as follows: "Web Address: $web_address <br>" . Is there an easy way to make the result a link as the input will always be a web address? Thanks in advance!
Hi guys,
echo " Download <a href=".$file_path." download>Here</a>"; ?>
Didnt work, just took me to the path of the image. What i want is a link and when clicked the download window pops out.
Thanks hello, im trying to add a hyperlink that launches in a new window to the following (in the last column) any ideas? Code: [Select] echo "<tr>"; echo "<td align='center'>" . $row["ID"] . "</td>"; echo "<td align='center'>" . $row["Name"] . "</td>"; echo "<td align='center'>" . $row["jobNO"] . "</td>"; echo "<td align='center'>" . $epn . "</td>"; echo "<td align='center'>" . $cname . "</td>"; echo "<td align='center'>" . $cadd . "</td>"; //want to add a hyperlink here echo "</tr>"; } echo "</table>"; } Hi all, I am having problems in forming this link: { echo "<a title='page $i' class='current' href='javascript:ajaxpage('view.php?catid=" . $catid ."&" ; if($subid > 0) { echo "subid="; echo $subid . "&"; } echo "p=$i', 'content';'>$i</a> "; } Which currently points to javascript:ajaxpage( and then ends! I have to put a link that can be clicked around the first name and last name variables in the echo statement below. I also need to add a variable from the Select query to the link such as: 'www.phpfreaks.com/admin/customers.php?page=1&cID="VARIABLE HERE" &action=edit' I've tried a bunch of things and none seem to work. while($row = mysql_fetch_array($result)){ echo '<tr><td>' . $row['customers_firstname'] . " " . $row['customers_lastname'] . '</td><td>' . date("M. d, Y", strtotime($row['date'])) . '</td><td>' . $row['plan_id'] . '</td></tr>'; David Trying to figure out how to make it so name is a link to the profile when its echo anyone know how to do this? im at a huge stand still Code: [Select] <?php $sql = "SELECT name FROM users WHERE DATE_SUB(NOW(),INTERVAL 5 MINUTE) <= lastactive ORDER BY id ASC"; $query = mysql_query($sql) or die(mysql_error()); $count = mysql_num_rows($query); $i = 1; while($row = mysql_fetch_object($query)) { $online_name = htmlspecialchars($row->name); echo '<a href="Inbox.php">"'[$goauld]'</a>"'; ?> <html> <?php include connect.inc; $id = $_GET['id']; mysql_query("UPDATE count SET clicks=clicks+1 WHERE id='$id'"); $sql = mysql_query("SELECT link FROM count WHERE id='$id'"); $fetch = mysql_fetch_row($sql); $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { echo "Name :{$row['id']} <br>" . "Subject : {$row['link']} <br>" . "Message : {$row['count']} <br><br>"; } mysql_close(); ?> <a href='/index.php?id=1'>link1</a> <a href='/index.php?id=2'>link2</a> </html> OK, have no idea what's going on... I've done this a million times... why wont this output!?? I must have a major brain meltdown and dont know it yet!!! Code: [Select] <?php // this echoes just fine: echo $_POST['testfield']; // but this wont echo: echo if (isset($_POST['testfield'])) { $_POST['testfield'] = $test; } echo $test; /// or even this DOESNT echo either!: $_POST['testfield'] = $test; echo $test; ?> Hi All, I'm trying to echo the response from an SLA query, the query works and returns the data when I test it on an SQL application.. but when I run it on my webpage it won't echo the result. Please help? <?php $mysqli = mysqli_connect("removed", "removed", "removed", "removed"); $sql = "SELECT posts.message FROM posts INNER JOIN threads ON posts.pid=threads.firstpost WHERE threads.firstpost='1'"; $result = mysqli_query($mysqli, $sql); echo {$result['message']}; ?> So I need to echo a row from my database with php, but where i need to echo is already inside an echo. This is my part of my code: $con = mysql_connect("$host","$username","$password"); if (!$con) { die('Could not connect: ' . mysql_error()); } mysql_select_db("main", $con); $result = mysql_query("SELECT * FROM Vendor"); while($row = mysql_fetch_array($result)) { //I need to echo right here .................. but I get a blank page when I try this. Please Help. echo '<option value=$row['vendor_id']>'; echo $row['vendor_id']; echo '</option>'; } mysql_close($con); Result: A Blank page. Thanks in advance! I have a log system that allows 10 logs on each side(Left and right). I am trying to make it so that the left side has the 10 most recent logs, then the right as the next 10. Any ideas? Hi
I try to echo out random lines of a html file and want after submit password to whole content of the same html file. I have two Problems.
1st Problem When I echo out the random lines of the html file I don't get just the text but the code of the html file as well. I don't want that. I just want the text. How to do that?
for($x = 1;$x<=40;$x++) {
$lines = file("$filename.html");
echo $lines[rand(0, count($lines)-1)]."<br>";
}
I tried instead of "file("$filename.html");" "readfile("$filename.html");" But then I get the random lines plus the whole content. Is there anything else I can use instead of file so that I get the random lines of text without the html code?P.S file_get_contents doesn't work either have tried that one.
2nd Problem:
As you could see in my first problem I have a file called $filename.html. After I submit the value of a password I want the whole content. But it is like the program did forget what $filename.html is. How can I make the program remember what $filename.html is? Or with other words how to get the whole content of the html file?
My code:
if($_POST['submitPasswordIT']){
if ($_POST['passIT']== $password ){
$my_file = file_get_contents("$filename.html");
echo $my_file;
}
else{
echo "You entered wrong password";
}
}
If the password isn't correct I get: You entered wrong password. If the password is correct I get nothing. I probably need to create a path to the file "$filename.html", but I don't know exactly how to do that.
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 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! I saw that <?=$var?> instead of <?php echo $var; ?> can be used for echo, does it work on every server? |
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