PHP - Drop-down Box Link

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?

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 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!

trying to get my drop down box to display more then one result. I have two users with Mod wrote in the column Monitor
but right now it only displays just the one.
So I need to display all the goaulds that have mod in the monitor column

<?php Code: [Select]

                                          $result = mysql_query("SELECT goauld FROM users WHERE monitor ='mod' ORDER BY id DESC") or die(mysql_error());
                                         
while($row = mysql_fetch_array( $result )) {

                           if($row['goauld'] == $_POST['name_list']){

echo '<div class="containerc">';
echo '<div class="search">';
echo '<form method="post" >';

echo '<select name="name_list" class="textbox" id="name_list">';

echo "<option value=\"".$row['goauld']."\">".$row['goauld']."</option>";

echo "<option selected=\"selected\" value=\"".$row['goauld']."\">".$row['goauld']."</option>";

echo '</select>';

echo  '<input type="submit" name="remove" id="remove" value="Remove">';

}
}
}
'</form>';

'</div>';
'</div>';
?>

hi
how do i get the id value from the dropdown selection from the database

Here is my code. i need the id to store it another table to refer it in my application. its like a categories and subcategories table.

<html>
<form id="name" action="<?php $_POST['SERVER_SELF'] ?>" method="POST" >
<input type="text" name="name1">  <br>
  <select name="select">
  <?php mysql_connect('localhost','root','') or                                                              
                                  die('Could not connect to mysql ' . mysql_error()); // error message
                                                                 
                                 mysql_select_db("dbtest") or die(mysql_error()); 
                                 $query = "select name,id from main"; 
                                   $result = mysql_query($query);
                                    if ($query) 
                                    {  
                             while ($row = mysql_fetch_array($result))
                                  { $strB=$row['name'];
                                   ?>
                                    <option value="<?php $stra ?>"><?php echo $strB ?></option> <br>
                                      
                                   <?php
                                  
                                   }
                                   }
       ?>
 </select>
<input type="text" name="name2">     <br>
<input type="submit" name="submit1">
    </form>
     <?php if(isset($_POST['submit']))
     { //$name=$_POST['val'];
        $strB=$stra;
        mysql_connect('localhost','root','') or                                                              
                                  die('Could not connect to mysql ' . mysql_error()); // error message
                                                                 
                                 mysql_select_db("dbtest") or die(mysql_error());
        $query = "insert into sub (sub) values('$strB')";
            $result = mysql_query($query);
             if ($query) 
                                    {                       
                                    
                                    
                            echo "query executed";         
       }
                                     }
       ?>
   
</html>


thanks

The code for the drop down:

Code: [Select]
echo '<select name="select"  ONCHANGE="goto(this.form)">';
while ($dropDown > 0) {
if ($page == $dropDown) {
$selected = 'selected="selected"';
} else {
$selected = "";
}
echo "<option value=\"searchFlights.php?page=" . $dropDown . "\" $selected>" . $dropDown . "</option>";
$dropDown = $dropDown - 1;
}
echo '</select>';
It works fine, but I want 1 to be at the top of the list and go down from there because right now, it starts with the biggest number and goes down. Any suggestions? Thanks!

Hi all, ive just coded a new thing for my website which lets users upgrade there car to a certain speed, its all coded and works fine, but im just not sure how todo one thing on it.


<?php
$getall = mysql_query("SELECT car, id FROM garage WHERE owner='$username'");

while($row = mysql_fetch_array($getall)){

echo $row['car']. " - ". $row['id'];
echo "<br />";
}
?>


That bit of my code, select the car and the car ID from my table garage, but i'm wanting it in a drop down box which shows all the cars in there garage, but not sure how I would do this.

Any advice?

Thanks.

Hi all, ive got a script ive finishing coding from the other day, but ive got a slight problem with it.


<?php
if ($fetch->missions == "4"){
// Mission 4, Toyota Aygo up to 500MPH
    $car = "Toyota Aygo";
if ($garage->car == $car && $garage->mph == "500"){
$rightcar = "Yes";
}else{
$rightcar = "No";
}
if (strip_tags($_POST['submit'])){
if ($rightcar == "Yes"){
    echo ("<table width='30%' cellpadding='0' align='center' cellspacing='0' border='1' class='table'>
           <tr>
                <td class='header' align='center'>Well Done!</td>
           </tr>
           <tr>
                <td>You Successfully Completed Mission 3!</td>
           </tr>
           </table><br />");
$newmoney = $fetch->money + 70000000;
$newrep = $fetch->rep + 750000;

   mysql_query ("UPDATE users SET money='$newmoney' WHERE username='$username'");
   mysql_query ("UPDATE users SET rep='$newrep' WHERE username='$username'");
   mysql_query ("UPDATE users SET missions='5'  WHERE username='$username'") or die ("Error - Query : Line 121 : " . mysql_error());
   mysql_query ("INSERT INTO `inbox` ( `id` , `to` , `from` , `message` , `date` , `read` , `saved` , `event_id` ) 
VALUES (
'', '$username', 'System', 'Well done! <br /> As You completed Mission 4 Successfully we will reward you with<br /> &pound;70,000,000 Money<br /> 750,000 Rep!', '$date', '0', '0', '0'
)");
}
}
} // Mission 4
?>
<?php if ($fetch->missions == "4"){ ?>
<form action="" method="POST" name="mission4">
<table width="50%" cellpadding="0" cellspacing="0" border="1" class="table" align="center">
<tr>
<td class='header' align='center'>Mission 4</td>
</tr>
<tr>
<td>Now SD Stars can now see your progressing, they have asked you if you can get a Toyota Aygo up to 500MPH. You must get it to them without any damage to prove your werthy!</td>
</tr>
<tr>
<td align='center' class='header'>Select Your Car:</td>
</tr>
<tr>
<td><select class="dropbox">
                                        <option selected>Pick Car</option>
<?php $get=mysql_query("SELECT * FROM garage WHERE owner='$username' AND car='Toyota Aygo' AND mph='500'");
while($stats=mysql_fetch_object($get)){
echo "<option value=?carname=$stats->id>$stats->car, $stats->damage%, $stats->mph MPH</option>";
}
 ?></select></td>
</tr>
<tr>
<td class='omg'><input type="submit" name="submit" class="button" value="Send Car!"></td>
</tr>
                        </table>
</form>
<?php } ?>


I blive it works apart from the Drop Down Box. It selects all the cars its suppose to but when I pick the car and press submit it just basicly refreshes the page :S.

I'm not sure why its doing it .

Can any one else see the problem?

Thanks for any help given

Hello,

must admit i'm very new to this and just a hobiest ,

I am creating a page which will be used for logging information.

With a mysql database i have two tables "tbl_rides" & "tbl_courses" rides is the main table for storing information and has a foreign key associating column "course_id" with course_id in the tbl_courses

What i want to do on a HTML/PHP page is display a drop down box that has the contents of "course" from the tbl_courses but when the user selects it and submits the record is written to the main form "tbl_rides"  "course_id"  so display the friendly name via the foreign key but the record is written to "tbl_rides"

Can that be done .. am i going about this wrong ?

 

Thanks for any help Andrew. 

Hello,

I'm trying to use a dropdown to display different divs.


<html>
<head>
<title>DDlist Div Display</title>
<script type="text/javascript">
function ShowDivArea(info) {
  var sel = document.getElementById('divArea').getElementsByTagName('div'); 
  for (var i=0; i<sel.length; i++) { sel.style.display = 'none'; }
  if (info == '0') { return; }
  document.getElementById('divArea'+info).style.display = 'block';
}

</script>

<style type="text/css">
.divArea { display:none; height:100px; width:200px; border:1px solid red; }
</style>

</head>
<body>

<select id="DDDivList" onchange="ShowDivArea(this.selectedIndex)">
  <option value="0" selected> -- Select A Design Service --</option>
  <option value="1"> QR Bookmark </option>
  <option value="2"> Twitter </option>
  <option value="3"> Ning or Tumblr </option>
  <option value="4"> Flyers </option>
  <option value="5"> Business Card or Brochure</option>
  <option value="6"> Album or Mixtape cover</option>
  <option value="7"> Other</option>
</select>

<div id="divArea">
 <div id="divArea1" class="divArea">


<h1>QR Bookmarks</h1>
<form action='qrgen.php' method='POST'>
<input type='text' name='bmarksite'> URL<br>
<input type='text' name='sitedesc'> Description<br>
<input type='submit' value='Convert'><br>

<?php

include("QrCodes.php")
?>

<?php

$bmarksite = $_POST['bmarksite'];
$sitedesc = $_POST['sitedesc'];

if ($sitedesc&&$bmarksite)
{
$qrcode = new QrCodes;
$qrcode -> IsImage =1;
echo $qrcode -> GetBookmark("$bmarksite","$sitedesc");
echo "</br>";
echo $bmarksite;
echo "</br>";
print $sitedesc;

}

else
{
echo "Please fill in all the required fields.";
}
?>


</div>
 <div id="divArea2" class="divArea">Twitter</div>
 <div id="divArea3" class="divArea">Ning or Tumblr</div>
 <div id="divArea4" class="divArea">Flyers</div>
 <div id="divArea5" class="divArea">Business Card or Brochure</div>
 <div id="divArea6" class="divArea">Album or Mixtape Cover</div>
 <div id="divArea7" class="divArea">Other</div>
</div>

</body>
</html>


The problem is with this line: <form action='qrgen.php' method='POST'>. I am using a wordpress page template. What should I change the qrgen.php to? Or how do I get it to work?

Thank you for your help.

Hi,
I've a two select box with same name enabled multiple. like this
Code: [Select]
<select name="saniSelect[]" multiple = "multiple">
    <option value="a">a</option>
    <option value="b">b</option>
    <option value="c">c</option>
</select>

<select name="saniSelect[]" multiple = "multiple">
    <option value="d">d</option>
    <option value="e">e</option>
    <option value="f">f</option>
</select>

Now from select box first i select "a" and "c" and from 2nd select box i just select "f"

when i submit the form. and watch the result like this:
Code: [Select]
print_r($_REQUEST['saniSelect']);
it should return like this:

Array([0] => Array([0]=>a, [3] => c),
          => Array([3]=>f),
        )

which is what i want .but it didn't.. it shows like this

Array([0] => Array([3]=>a, [3] => c))

why i'm not getting my require result.. it should become array onto an array.. means multidimensional array

Please help

helo
i a newbie in php. i need some help regarding my system. i want my data from database to display in drop down menu form.anyone have idea how to do that? i really need some help. i have search all over and couldn't find any solution. any suggestions would be helpful 

Code: [Select]
$q = mysqli_query($mysqli, "DROP TABLE IF EXIST 'airline_survey'");
  if($q){echo "deleted the table airline_survey....<br>";}
  else{echo "damm... ".mysqli_error($mysqli);}
$mysqli is my connection and it works I had no trouble creating the table but I decided to add these few lines in front of the code to create the table and I get this error.

damm...You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'EXIST 'airline_survey'' at line 1
I gotta believe this is really simple but it's got me stumped.