PHP - Quick One: How To Submit Form Results In <strong> Tags

Hi,

having problems getting checkboxes to display all reuslts when a user selects more than one check box say in the category section and one in the location section - see page http://www.partyco.co.uk/event-and-party-venues/ - submit to see reults page:

I managed to get it to display reults if the user only seletc either a right or left column option OR one of each - BUT not when thet select multiple categories and one location - and ideas how to do this ? putting it into an array perhaps - but how - new to some of this....

here is the code for the reults page:
Code: [Select]
<?php
$location = $_POST[location];
$category = $_POST[category];
?>

<?php

$Link = mysql_connect("xxxxxxxxx", "xxxxxxxxx", "xxxxxxxx") or die(mysql_error());
mysql_select_db("xxxxxxxx") or die(mysql_error());

// selects db listings when location not given
if (empty($location)) {
$query = "SELECT * FROM venues WHERE category = '$category' order by title";
$result = mysql_query($query) or die(mysql_error());

// selects db listings when category not given
} elseif(empty($category)) {
$query = "SELECT * FROM venues WHERE location = '$location' order by title";
$result = mysql_query($query) or die(mysql_error());

// selects db listings when both given
}else {
$query = "SELECT * FROM venues WHERE category = '$category' and location = '$location' order by title";
$result = mysql_query($query) or die(mysql_error());
}

while($row = mysql_fetch_array($result)){
echo "<div class=\"resultsShort\" style=\"margin-bottom:10px;\">";
echo "<h2 id=\"resultsHeading\">";
echo $row['title'];
echo "</h2>";
echo "<p class\"resultspara\">". nl2br($row['description']). "</p>";
echo "<h4 style=\"margin:5px 0 0 0; padding:0;\">Contact details</h4>";
echo "<p class\"resultspara\">". nl2br($row['contact']). "</p>";
echo "<div style=\"float:left; width:124px; height:40px; margin:10px 15px 0 0;\">";
echo "<a href=\"/party-supplier-resources/email-supplier.php?title=". $row['title']. "&email=" . $row['email']. "&location=" . $row['location']. "&category=" . $row['category']. "\" title=\"contact this venue here\">";
echo "<img src=\"/images/email-supplier.jpg\" width=\"124\" align=\"right\" height=\"35\" alt=\"contact this supplier button\" border=\"0\" /></a>";
echo "</div>";
echo "</div>";
}


?>


<?php include("../include/shareLinks.php"); ?>


<div id="popupContact">
<a id="popupContactClose" title="close this window">close x</a>
<h1>Supplier Directory Enquiry Form</h1>
<?php include("../include/enquiryform.php"); ?>
</div>
<div id="backgroundPopup"></div>


<?php
mysql_close ($Link);
?>
Any help appreciated!

Aaron

MOD EDIT: [code] . . . [/code] BBCode tags added.

hi. i want to echo part of my code in bold or strong. can anyone help please?

echo "<table border='0'>
<tr>
<th></th>
</tr>";

$row = mysql_fetch_array($result) or die(mysql_error());
echo "<tr>";
echo "<td>"  . $row['question']. "  ". $row['answer']; "</td>";
echo "<tr>";

echo "</table>";


I want to bold the $row question part. Hope someone knows. thanks

Can I replace the set message text for $body and $subject with $_POST['message']; $_POST['subject']; so I can get a users input from a html form via php as the action.  Instead of the set text in the script.
As seen http://email.about.com/od/emailprogrammingtips/qt/PHP_Email_SMTP_Authentication.htm

Hello, first time poster..  I've looked the web over for a long time and can't figure this one out.

- Below is basic code that successfully checks MySQL for a match and displays result.  I was debugging and forced the "height" and "width" to be 24 and 36 to make sure that wasn't the problem.  That's good.. 
- I'd like to give the user ability to select width and height from a form.. and have it do an onchange this.form.submit so the form can be changing as fields are altered (thus the onchange interaction)
- In a normal coding environment I've done this numerous times with no "Page cannot be displayed" problems.  It would simply change one select-option value at a time til they get down the form and click submit...  but in WordPress I'm having trouble making even ONE single onchange work!
- I've implemented the plugins they offer which allows you to "copy+paste" your php code directly into their wysiwyg editor.  That works with basic tests like my first bullet point above.
- I've copied and pasted the wordpress url (including the little ?page_id=123) into the form "action" url... that didn't work...  tried forcing it into an <option value=""> tag.. didn't work.  I'm just not sure. 

I've obviously put xx's in place of private info..

Why does this form give me Page Cannot Be Displayed in WordPress every time?  It won't do anything no matter how simple.. using onchange..

Code..

$con = mysql_connect("xxxx.xxxxxxx.com","xxxxxx","xxxxx");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
mysql_select_db("xxxxxx", $con);
$myprodwidth=24;
$myprodheight=36;
$result = mysql_query("SELECT * FROM product_sizes WHERE prodwidth='$myprodwidth' and prodheight='$myprodheight'");
while($row = mysql_fetch_array($result))
  {
  echo $row['prodprice'];
  }
mysql_close($con);


<form method="post" action="">
<select name="myheight" onchange="this.form.submit();">
<option selected="selected" value="">select height</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">36</option>
<option value="xxxxxxxxx.com/wordpress/?page_id=199&height=36">48</option>
</select>

I know this problem comes up a lot but it can be for various reasons so after reading up I can't decide what might be wrong. I'm still learning!
I have a simple php registration form (first name, second name, email address) and every time a user submits an entry a second blank record is created in the MYSQL database after it.

Any help would be great. (php code in red)


Code: [Select]
[color=red]<?
$firstname=$_POST['firstname'];
$surname=$_POST['surname'];
$email=$_POST['email'];

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query = "INSERT INTO register VALUES ('','$firstname','$surname','$email')";
mysql_query($query);

mysql_close();
?>[/color]



Here is the section on the html HTML page that submits the form

Code: [Select]
<form action="register.php" method="post" class="BagTitle">
        <table width="700" border="0" cellpadding="0" cellspacing="2">
  <tr>
    <td width="100">First Name: </td>
    <td width="594"><input name="firstname" type="text" size="50" /></td>
      </tr>
  <tr>
    <td>Last Name: </td>
    <td><input name="surname" type="text" size="50" /></td>
      </tr>
  <tr>
    <td>E-mail:</td>
    <td><input name="email" type="text" size="50" /></td>
      </tr>
  <tr>
    <td colspan="2">&nbsp;</td>
      </tr>
  <tr>
    <td colspan="2"><input type="Submit" value="Send" />
        <input name="Reset" type="reset" value="Reset Form" /></td>
      </tr>
    </table>
      </form></p>
Also I have no idea about securing this information. Are there any basic steps I can take?

Hi all,

What I am trying to achieve is, I thought quite simple!

Basically, a user signs up and chooses a package, form is submitted, details added to the database, email sent to customer, then I want to direct them to a paypal payment screen, this is where I am having issues!

Is their any way in php to submit a form without user interaction? Here is my code for the form process page
Code: [Select]
<?php
include('config.php');
require('scripts/class.phpmailer.php');

$package = $_POST['select1'];
$name = $_POST['name'];
$email = $_POST['email'];
$password = md5($_POST['password']);

$domain = $_POST['domain'];
$a_username = $_POST['a_username'];
$a_password = $_POST['a_password'];


$query=mysql_query("INSERT INTO orders (package, name, email, password, domain, a_username, a_password)
VALUES
('$package', '$name', '$email', '$password', '$domain', '$a_username', '$a_password')");

if (!$query)
{
echo "fail<br>";
echo mysql_error();
}

else
{
$id = mysql_insert_id();
$query1=mysql_query("INSERT INTO customers (id, name, email, password)
values
('$id', '$name', '$email', '$password')");

if (!$query1)
{
echo "fail<br>";
echo mysql_error();
}


if($package=="Reseller Hosting")
{

//email stuff here - all works - just cutting it to keep the code short


if(!$mail->Send())
{
   echo "Message could not be sent. <p>";
   echo "Mailer Error: " . $mail->ErrorInfo;
   exit;
}
?>


<form name="_xclick" action="https://www.paypal.com/cgi-bin/webscr" method="post">
<input type="hidden" name="cmd" value="_xclick-subscriptions">
<input type="hidden" name="business" value="subscription@jollyhosting.com">
<input type="hidden" name="currency_code" value="USD">
<input type="hidden" name="item_name" value="Jolly Hosting Reseller Packages">
<input type="hidden" name="no_shipping" value="1">

<!--1st month -->
<input type="hidden" name="currency_code" value="USD"> 
<input type="hidden" name="a3" value="3.00">
<input type="hidden" name="p3" value="1">
<input type="hidden" name="t3" value="M">
<input type="hidden" name="src" value="1">
<input type="hidden" name="sra" value="1">
</form>';


<?php
}

//last
}
//end
?>

Hi-
the code below lets me upload a CSV file to my database if I have 1 field in my database and 1 column in my CSV.
I need to add to my db "player_id" from the CVS file and "event_name" and "event_type" from the form... any ideas???

here's the code:
Code: [Select]
<?php
$hoststring =""; 
$database = "";
$username = "";
$password = "";
$makeconnection = mysql_pconnect($hoststring, $username, $password);
?>
<?php
ob_start();
mysql_select_db($database, $makeconnection);
$sql_get_players="
SELECT * 
FROM tabel 
ORDER BY player_id ASC";
//
$get_players = mysql_query($sql_get_players, $makeconnection) or die(mysql_error());
$row_get_players = mysql_fetch_assoc($get_players);
//
$message = null;
$allowed_extensions = array('csv');
$upload_path = '.'; //same directory

if (!empty($_FILES['file'])) {

if ($_FILES['file']['error'] == 0) {
// check extension
$file = explode(".", $_FILES['file']['name']);
$extension = array_pop($file);

if (in_array($extension, $allowed_extensions)) {

if (move_uploaded_file($_FILES['file']['tmp_name'], $upload_path.'/'.$_FILES['file']['name'])) {

if (($handle = fopen($upload_path.'/'.$_FILES['file']['name'], "r")) !== false) {
$keys = array();
$out = array();
$insert = array();
$line = 1;

while (($row = fgetcsv($handle, 0, ',', '"')) !== FALSE) {

foreach($row as $key => $value) {    

if ($line === 1) {
$keys[$key] = $value;       
} else {       
$out[$line][$key] = $value;              
}       
}                
$line++;          
}        

fclose($handle);            

if (!empty($keys) && !empty($out)) {        
$db = new PDO(
'mysql:host=host;dbname=db', 
'user', 
'pw');   

$db->exec("SET CHARACTER SET utf8");        

foreach($out as $key => $value) {        
$sql  = "INSERT INTO `table` (`";
$sql .= implode("`player_id`", $keys);    
$sql .= "`) VALUES (";    
$sql .= implode(", ", array_fill(0, count($keys), "?"));    
$sql .= ")";    
$statement = $db->prepare($sql);    
$statement->execute($value);       
}      

$message = '<span>File has been uploaded successfully</span>';      
}
}
}
} else {
$message = '<span>Only .csv file format is allowed</span>';
}
} else {
$message = '<span>There was a problem with your file</span>';
}
}
ob_flush();?>
<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=UTF-8" />
<title>CSV File Upload</title>
</head>
<body>

<form class="form" action="" method="post" enctype="multipart/form-data">
<h3>Select Your File</h3>
<p><?php echo $message; ?></p>
<input type="file" name="file" id="file" size="30" />
<br/>
<label>Event Name:</label><input name="event_name" type="text" value="" />
<br/>
<label>Event Type:</label><input name="event_type" type="text" value="" />
<br/>
<input type="submit" id="btn" class="button" value="Submit" />
</form>

<br/>
<h3>Results:</h3>
<?php do { ?>
<p><?php echo $row_get_players['player_id'];?></p>
<?php } while ($row_get_players = mysql_fetch_assoc($get_players)); ?>
</body>
</html>

Hello firstly i would like to say im very new to this forum (Signed up 5 minutes ago). I have decided to join as im in my final year at university and have been asked to create a website, so im guessing ill be coming back here a lot for help! .

Anyway my most current problem is i have an if statement that hides the "Register" button if the user is logged in and Shows a welcome message and the users name if the user is logged in. However i would like to make the log in form disappear if the user is logged in. I know that by putting the form code in the else part of my if statement it will disappear if the user is logged in however when i do this it stops my whole website working.

Any help would be very grateful.

Thanks Glen.

Hi, I have search form which works apart from 1 part, I have set the results limit to 10 per page with a next link if there are more than 10 results, however when the next link is clicked, i get the error 'you do not have a search parameter' which comes from this code:

Code: [Select]
if (!isset($var))
          {
          echo "<p>We dont seem to have a search parameter!</p>";
          exit;
          }
The next link is defined by the code below:

Code: [Select]
// check to see if last page
          if (!((($s+$limit)/$limit)==$pages) && $pages!=1) {
       
          // not last page so give NEXT link
          $news=$s+$limit;
       
          echo "&nbsp;<a href=\"$PHP_SELF?s=$news&q=$var\">Next 10 &gt;&gt;</a>";
          }
Can anyone see where I have gone wrong with this code without me sending the whole search form? Any help would be very much appreciated. Thanks

I have a simple form which sends results (using php) to my email. It works fine.

I want to make a similar form, and send the results not to email but to another web page, showing a summary of the form results, so user can check if all is entered correctly before continuing.

Can anyone advise if this is possible and how.

I am a real php beginner ... 

EDIT: sorry - new to posting guidelines - using XAMPP with mysql 5.1.44 (I think this is correct!!)

Newbie question he I have successfully created a form that that displays students (based on a selected lesson/group/date) with the ability to enter scores.

I'm now trying to set the form up so that if data already exists then it is displayed in the form.

Eg: if a member of staff choses group 7AS, lesson 1, 24/09/2010 and another member of staff has entered a score already for one student it shows up in the <select> drop down by using
Code: [Select]
<option>" . $scorevalue . "</option>

I had this working at one point but have made a change somewhere and cannot get it working again! can anyone spot an obvious mistake I am making?

The code I am using is as follows:
Code: [Select]
<input type="hidden" name="tutor" value="<?php echo $_GET["tutor"]; ?>" />
<input type="hidden" name="date" value="<?php echo $_GET["date"]; ?>" />
<input type="hidden" name="lesson" value="<?php echo $_GET["lesson"]; ?>" />


<?php

error_reporting(-1);

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }

mysql_select_db("lakeside", $con);

$result = mysql_query("SELECT students.admin, students.fname, students.surname, students.tutor
FROM students
WHERE tutor='$_GET[tutor]' 
ORDER BY students.tutor, students.surname");


echo "<table class='scores'>
<tr>
<th>Admin</th>
<th>Firstname</th>
<th>Surname</th>
<th>Tutor</th>
<th>Score</th>
<th>Code</th>
<th>Comment</th>
</tr>";

while($row = mysql_fetch_array($result))
  {
  
$sqlstatement = "SELECT * FROM scores where admin = '" . $row['admin'] . "' and lesson = '" . $_GET[lesson] . "' and date = '" . $_GET[date] . "' ";
$scorevalue = mysql_query($sqlstatement);  



echo " . $score . ";

while($rowvalue = mysql_fetch_array($scorevalue))
{
$score = $rowvalue['score'];
$comment = $rowvalue['comment'];
$code = $rowvalue['code'];
}
  
  
  echo "<tr>";
  echo "<td>" . $row['admin'] . "</td>";
  echo "<td>" . $row['fname'] . "</td>";
  echo "<td>" . $row['surname'] . "</td>";
  echo "<td>" . $row['tutor'] . "</td>";

  echo "<td><select name='score"  . $row['admin'] . "' value='{$row['score']}' />
   <option>" . $scorevalue . "</option>
   <option>0</option>
   <option>1</option>
   <option>2</option>
   <option>2.5</option>
   <option >3</option>
   <option>3.5</option>
   <option>4</option>
   </select>
   </td>";

Hi Guys.

I have some code which displays the rows of a database table in a html table on my webpage:

Code: [Select]
<?php//to display image from source$dir = "360_covers";echo '<table>';echo '<tr><th>Title</th><th>Cover</th><th>comment</th></tr>';$result = mysql_query("SELECT * FROM xbox_games order by gametitle");while ($row = mysql_fetch_assoc($result)) {    echo "<tr><td>{$row['gametitle']}</td><td><img src=\"$dir/{$row['cover']}\" width='38' height='38'></td><td>{$row['cover']}</td></tr>";}//close out tableecho '</table>';?>

What i want to do now is add a form, possibly at the end of each row of the html table, which will allow a session user to update that record.

Can anybody advise on the best way to do this?

I know the php is basic and not completed yet, but I am working on that. My next step I want the result to show on index.php when a form is submitted to calculator2.php. I just cant seem to get it. Please help!

index.php
Code: [Select]
<div class="post">
        <h2 class="title">Calculator</h2>
<hr />
         <form method="post" action="calculator2.php">
Fireplace Front Width: <input type="text" name="fw">
<br />
Fireplace Back Width: <input type="text" name="bw">
<br />
Fireplace Depth: <input type="text" name="fd">
<br />
<input type="submit" name="Submit" value="Submit">
</form>
            Pounds Of Glass Needed: <?php echo $res1; ?>

      </div>

calculator2.php
Code: [Select]
<?php
$frontwidth = $_POST['fw'];
$backwidth = $_POST['bw'];
$firedepth = $_POST['fd'];
$x = $frontwidth + $backwidth+ $firedepth;
$y = ($x / 3) * .6667; 
$res1 = $y *2;
echo $res1;
?>

Sounds a bit long winded in the name so i'll explain...

I'm coding a site where the user inputs their vehicle reg, and submits the form. The site will then show their car's details. This is being done using an external service that provides an up to date database of vehicle details.

So far, i've got the form and when you submit it, it returns the vehicle data in an XML, but the XML it displays is remote. I need to retrieve the details back to the website to display on the page. Here is what I have already:

Code: [Select]
<html>
<head>
<title>Reg Test Page</title>
</head>
<body>
<form method="post" action="https://www.****.com/UAT/">
<input type="hidden" name="ESERIES_FORM_ID" value="B2INT">
<input type="hidden" name="MXIN_USERNAME" value="****"><br />
<input type="hidden" name="MXIN_PASSWORD" value="****"><br/>
REG<input type="text" name="MXIN_VRM" value="REG NUM"><br />
<input type="hidden" name="MXIN_TRANSACTIONTYPE" value="03"><br />
<input type="hidden" name="MXIN_PAYMENTCOLLECTIONTYPE" value="02"><br />
<input type="hidden" name="MXIN_CAPCODE" value="1"><br />
<input type="submit" />
</form>
</body>
</html>

I've ***'d out some details for obvious reason. The form currently works fine and returns a valid XML response, but it's remote. My original plan was to get the form to load up in another php file that processes the detail and sends a file_get_contents() request, but looking at the results page on the xml, there is no extension to the posted url and if you go directly to it, you get the error "METHOD NOT ALLOWED. ONLY 'POST' IS ALLOWED."

The other issue I have is that i'm not sure how it's handling the request, and the url is not modified, so i'm not sure how i'd set an url with values to retrieve it?

Any help is gratefully received.

Cheers Guys.

Hi,

I'm currently learning PHP through tutorials and I am working on a webpage that filters statistics (crime records) from an SQL table and filters them by an option of years that the user has picked from a dropdown box.

$desiredyear = $_POST['year'] is the year picked from the drop down box on a different page.

I did not encounter any problems simply displaying the entire table until I tried to filter the crimes by year. I keep getting this error message.

Warning: sqlsrv_fetch_array() expects parameter 1 to be resource, boolean given in C:\RDEUsers\NET\400792\1.php on line 40

Here is the error line.

Code: [Select]
while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC))
And here is my code so far

Code: [Select]
<?php 

$server = 'SQL2008';
$connectionInfo = array( "Database"=>"rde_400792");
$conn = sqlsrv_connect($server,$connectionInfo);

$desiredyear = $_POST['year'];


$describeQuery='select year, force, homicide from crime where year = $desiredyear ';
$results = sqlsrv_query($conn, $describeQuery);

echo '<table border="1" BORDERCOLOR=Black>';
echo '<tr><th bgcolor = "LightBlue">Year</th><th bgcolor = "LightBlue" >Force</th><th  bgcolor = "LightBlue">Homicide</th></tr>';


while($row = sqlsrv_fetch_array($results, SQLSRV_FETCH_ASSOC)) 

{           echo '<tr>';
   echo '<td >' .$row['year'].'</td>'; 
   echo '<td>' .$row['force'].'</td>'; 
   echo '<td>' .$row['homicide'].'</td>'; 
   echo '</tr>';
} 

echo '</table>';
sqlsrv_close($conn); 
Please can somebody help me fix this? It's driving me mad and I've exhausted my limited knowledge on the subject. I just don't understand what I'm doing wrong.