Posts by FlashCreations Page 3

FlashCreations 20 Posting Whiz

15 Years Ago

Alright just a few minor problems:

function ajaxGetInfo(title)
{
	a = ajax();
	if(a!=null)
	{
		a.onreadystatechange = function() { if(a.readyState == 4) document.getElementById('info').value = a.responseText; }
		a.open("GET", "includes/getDescription.php?title="+title, true);
		a.send(null);
	}
	else document.getElementById('info').value = "Error retrieving data!";
}
 
function ajax()
{
	if(window.XMLHttpRequest) return new XMLHttpRequest();
	if(window.ActiveXObject) return new ActiveXObject("Microsoft.XMLHTTP");
	return null;
}

index2.php

<?php
session_start();
// Checks if the user isn't registered in session
if (!isset($_SESSION['user'])) echo header("Location: login.php");
?>
<!doctype html>
<html>
	<head>
<?php
require_once("includes/connection.php");
require_once("includes/functions.php");
include("includes/header.php");
?>
	</head>
    <body>
<?php
include("includes/nav.php"); 
include("includes/upper_content.php"); 
include("includes/getDescription.php"); 
?>
<div id="main_body">
<table width="936" align="center" cellpadding="10">
	<tr>
		<td align="center">
			<p align="center">Enter your addition to one of the books in the drop-down list. There is no length limit.</p>
      		<hr />
      		<table>
            	<tr>
                	<td align="center">
      					<form method="post" action="submit_contribution.php">
        					<fieldset>
          						<legend>Select a Book:</legend><br />
          						<?php $result = mysql_query("SELECT id, book_name FROM iin_books"); ?>
		  						<select name="category" onchange="ajaxGetInfo(this.value)">
		  							<option>Select a book title...</option>
		  							<?php if(mysql_num_rows($result)!=0): ?>
                                    <?php while($nt=mysql_fetch_array($result)): ?>
			  						<option value="<?=$nt['sub_cat_id']?>"><?=$nt['book_name']?></option>";
                                    <?php endwhile;
										  endif; ?>
			  					</select>
          						<p>Contribution:<br /> <textarea name="contribution" id="contribution" cols="50" rows="10"></textarea></p>
          						<p style="color:#F00; font-family:'Lucida Sans Unicode', 'Lucida Grande', sans-serif; font-size:12px;">
            						<input type="checkbox" /> I accept <a href="tou.php" style="color: #09F;">Terms of Use</a> and <a href="privacy.php" style="color:#09F;">Privacy Policy</a>.
								</p>
          						<p><input type="submit" value="   Submit Your Contribution   " /></p>
        					</fieldset>
                        </form>
        			</td>
        			<td valign="top">
      					<p>Book Description</p>
      					<textarea name="info" id="info" cols="20" rows="5" readonly></textarea>
      				</td>
				</tr>
      		</table>
      		<br />
      		<br />
      		<a href="logout.php">Logout</a>
		</td>
	</tr>
</table>
</div>
<?php include("includes/footer.php"); ?>
</body>
</html>

Table styling...*cringe!*

getDescription.php:

<?php 
require_once("includes/connection.php");
require_once("includes/functions.php");

function return_description($title) 
{
	//On the below line, replace * with the name of the row that …

Web Development mysql php

FlashCreations 20 Posting Whiz

15 Years Ago

Re: Dropdown list OnChange PHP/MySQL

Nope you need AJAX (JavaScript to do this). PHP is server side. Javascript is clientside. Serverside code is long completed when you see the final output. There would be no way to perform a query without calling a script on the server with AJAX. Sorry to disappoint. But if you do want to try AJAX:

<html>
 <head>
  <title>Daniweb Ajax Demo</title>
  <script type="type/javascript">
  function ajaxGetInfo(title)
  {
   a = ajax();
   if(a!=null)
   {
    a.onreadystatechange = function { if(a.readyState == 4) document.getElementById('info').value = a.responseText; }
    a.open("GET", "getInfo.php?title="+title, true);
    a.send(null);
   }
   else document.getElementById('info').value = "Error retrieving data!";
  }
  function ajax()
  {
   if(window.XMLHttpRequest) return new XMLHttpRequest();
   if(window.ActiveXObject) return new ActiveXObject("Microsoft.XMLHTTP");
   return null;
  }
  </script>
 </head>
 <body>
  <select name="books" onchange="ajaxGetInfo(this.value)"></select>
  <textarea name="info" id="info" cols="20" rows="5"></textarea>
 </body>
</html>

Then just write a php script called getInfo.php that accepts the title as a GET argument and echoes back the information.

Web Development mysql php

FlashCreations 20 Posting Whiz

15 Years Ago

Re: Pass POST[] Array to Jquery

So you want to reply to send back to AJAX the exact content that AJAX sent through post? That seems kind of pointless to me, unless you want to do something to the data first (like search for it in a DB). In that case do this:

demo_insert.php:

//Use the POST value you sent up with $_POST['content']
//Send data back to JQuery with echo
echo "Hello JQuery!";

Web Development javascript-jquery php

FlashCreations 20 Posting Whiz

15 Years Ago

Re: Get only 4-digits with PHP

Inside a complex syntax brackets ( {$var} ) you can call any function with a return value that can be converted to casted to a string. So using your code that would be:

$message .= "<td>XXXX-XXXX-XXXX-{substr($_GET['cc_number'] , -4, 4)}</td>";

//Or this....
$message .= "<td>XXXX-XXXX-XXXX-".substr($_GET['cc_number'] , -4, 4)."</td>";

//Or even this...
$message .= "<td>XXXX-XXXX-XXXX-";
$message .= substr($_GET['cc_number'] , -4, 4);
$message .= "</td>";

The first or second ones make most sense to me, so I would chose one of them, but it's totally up to you!