PHP - Dynamic Stylesheet - Is There A Way To Do This?

...in a different folder?

For instance, I have a stylesheet in my  C:\xampp\htdocs\rcm folder, and I want to link it to files in the C:\xampp\htdocs\ folder. Is there a way of doing this?

Thanks in advance

Hi all

I need to combine these two scripts:

Firstly, the following decides which out of the following list is selected based on its value in the mySQL table:

<select name="pack_choice">
<option value="Meters / Pack"<?php echo (($result['pack_choice']=="Meters / Pack") ? ' selected="selected"':'') ?>>Meters / Pack (m2)</option>
<option value="m3"<?php echo (($result['pack_choice']=="m3") ? ' selected="selected"':'') ?>>Meters / Pack (m3)</option>
<option value="Quantity"<?php echo (($result['pack_choice']=="Quantity") ? ' selected="selected"':'') ?>>Quantity</option>
</select>

Although this works OK, I need it also to show dynamic values like this:

select name="category">
<?php
$listCategories=mysql_query("SELECT * FROM `product_categories` ORDER BY id ASC");
while($categoryReturned=mysql_fetch_array($listCategories)) {
echo "<option value=\"".$categoryReturned['name']."\">".$categoryReturned['name']."</option>";
}
?>
</select>

I'm not sure if this is possible?

Many thanks for your help.

Pete

Hello!

If you search Google, you'll notice the URL: http://www.google.com/webhp?hl=en#hl=en&source=hp&q=php+freaks&aq=f&aqi=g10&aql=&oq=&gs_rfai=CtdaTdtdRTNnPD5HuzASZtMWiCgAAAKoEBU_Q0ZT5&pbx=1&fp=19d754eee0b4f223

You can copy that URL anywhere you like... and the user will still see the same results.

So basically the URL dynamically does an action, and accesses the database.

How on do I make a URL like this? Meaning when you change the URL parameter values, it request a slightly different database query?

Does that make sense? Similarly with NexTag.com: http://www.nextag.com/serv/main/buyer/ProductCompare.jsp?search=camera&page=0&node=500001&psort=%2FDigital-Cameras--zzcameraz500001zB6z5---html&zipcode=&cptitle=657166355&cptitle=656751324&cptitle=620051906&cptitle=705150048

That was a Dynamically made URL (I selected from check boxes which products to compare) and it makes that URL so anyone can see those products.

Any ideas on how to do this?


More examples on the URL: http://www.cars.com/go/compare/modelCompare.jsp?myids=9721,11439  (i select the cars, it generates that URL. Notice the IDs 9721,11439)
 

Okay, so I have a page with a heap of fields and buttons on it (and want to keep it like that really) inside one form.

Layout
Box 1 Button 1
Box 2 Button 2
Box 3 Button 3
Box 4 Button 4
etc...

Box 1 is called 'amount1$id'
Box 2 is called 'amount.$id' where $id is the $id of the data being displayed etc

So a sample form may look like
Box: "amount22" Button: "Pay22"
Box: "amount25" Button: "Pay25"
Box: "amount32" Button: "Pay32"
Box: "amount39" Button: "Pay39"
Box: "amount420" Button: "Pay420"

Its working fine up to that point.

My problem now is that I can't work out how to get that id number from the form once submitted. If I could get the button name somehow I could do it, although I don't know the button or box names when writing the processing script. I could use a loop and check if button1 to buttonX was posted etc, although if there is a large number of entries in the database, then that starts to become a problem. Thoughts?

Is there a way to create dynamic content without the use of a Content Management System?  I simply want to click on a link in my navigation menu and have that display the content in a div section and keep the content in a separate folder on the server.  I didnt want to have to use a CMS to accomplish that because every other feature of it will be unused as I just simply want to display the content in a div when I click on the link.  I use a simple test server on my Home PC that I use to format the articles in html from documents I have wrote in LibreOffice, so I dont really need all the features and sophistication of full blown CMS.  But I also dont want a bunch of different files in my root folder if I can avoid it.  I would rather keep the content in a separate folder and just pull it from there and display it on the main page when the link is clicked.

Anyway, I dont know if its even possible or if its the best way, so I am open to suggestions as well.  Some of the documents are quite long (like several chapters) and some are much shorter (just a few pages).  Any help and ideas are much appreciated.

So we just started php after learning a month of HTML, and we need to create a website with a page that allows the admin to change the colors and font size, header, etc on the webpage itself via forms.

E.g. I'm on the site and I can use a drop down menu to change the background to "blue" for example. No need to go edit the CSS file or anything.

How can be this done?

My second and final question is how would I write code that allows my website to have users "log in" to their account so they can edit their page?

Thanks a lot, I'm not good at PHP yet I've just started so please keep that in mind:)

- Kranti

Hey All,

Im having trouble tring to figure out how to save dynamic radio buttons.

I have a project where i have to use radiobuttons instead of checkboxes.

Basically, there is a database table with "roles" in it ... i.e administrator, user, etc etc etc and these are updateable but the user.

On a specific page, these get spit out with a radio button next to them, like so:

<input name="wfa" type="radio" id="wfa" class="radio" value="1" /><label for="wfa">administrator</label> 

These go on down the page, for as many entries there are in the database.

What i need to do is save the checked ones to the database, but being dynamic i cant save each value to a individual row ..... any help?

Someone mentioned serialize, i had a look at the php documentation but it didnt make much sense to me.

Cheers.

trying  to be able to store client information in mysql db for reference later in an admin area.  The code I have so far allows me to list the client names, but once I "select" the name I want it to show the rest of the database information stored for that user (email, height, weight, phone number, etc...) which it currently is not doing.  Any and all help will be greatly appreciated, tired of struggling through this blindly.

Code: [Select]
<?
// Connect database
mysql_connect("localhost","","");
mysql_select_db("mydb");

if(isset($select)&&$select!=""){
$select=$_GET['select'];
}
?>
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
</head>
<body>
<form id="form1" name="form1" method="get" action="<? echo $PHP_SELF; ?> ">
Client Name :
<select name="select">
<option value="">--- Select ---</option>
<?
// Get records from database (table "users").
$list=mysql_query("select * from users order by id asc");


while($row_list=mysql_fetch_assoc($list)){
?>
<option value="<? echo $row_list['id']; ?>" <? if($row_list['id']==$select){ echo "selected"; } ?>><? echo $row_list['name']; ?></option>
<?

}
?>
</select>
<input type="submit" name="Submit" value="Select" />
</form>
<hr>
<p>
<?

if(isset($select)&&$select!=""){

// Get records from database (table "users").
$result=mysql_query("select * from users where id='$select'");
$row=mysql_fetch_assoc($result);
?>
Information about <strong><? echo $row['name']; ?></strong> client...</p>
<p><? echo $row['email']; ?><br>
........................................<br>
........................................
<?
// End if statement.
}

// Close database connection.
mysql_close();
?>
</p>
</body>
</html>

Hi,
I am using php dynamic listing from database. it is working but I wanna integrate to a horizontal css menu how can i integrate my codes to any css horizantal dropdown menu.

thans.

here my dynamic list.

<?php

function sinirsiz_kategori($parent) {

 
    $sql = mysql_query("SELECT * FROM kategoriler ORDER BY id DESC");

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

         $diziler[$row['id']] = array('baslik' => $row['baslik'],'parent' => $row['parent']);

    }

 
    $has_childs = false;

 
    foreach($diziler as $key => $value)  {

            if ($value['parent'] == $parent) {       

 
                    if ($has_childs === false){

                            $has_childs = true;

                            echo "\t<ul>";

                    }

 
                    echo "<li><a href=\"".$value['baslik'].".php\">".$value['baslik']."</a>";

                    sinirsiz_kategori($key);

                    echo "</li>\n";

            }

    }

 
    if ($has_childs === true) echo "</ul>";

 
}

 
 
?>
<?=sinirsiz_kategori(0)?>

How can I go about making dynamic pages?

eg. user types
"phpfreaks.com/index.php?page=name"

I really can't get my head around this dynamic drop box. What I want is for a drop box to be populated with values in my database.

I have this that connects to my SQL and picks out the table required:

function displayUsers(){
   global $database;
   $q = "SELECT username,"
       ."FROM ".TBL_USERS." ORDER BY userlevel DESC,username";
   $result = $database->query($q);    ...

I then have this to pick out any errors, and also using the num_rows to get the number of rows (values) there a

$num_rows = mysql_numrows($result);
   if(!$result || ($num_rows < 0)){
      echo "Error displaying info";
      return;
   }
   if($num_rows == 0){
      echo "Database table empty";
      return;
   }

From here, I guess I want the num_rows to keep 'adding on' the number of <option value=""> in my selection box according the number of values I have in my database.

At this point, I can pull out the values into a dynamic table ... but I want it into a drop box -- but I'll but up the code for the dynamic table so you can get an idea:


/* Display table contents */
   echo "<table align=\"left\" border=\"1\" cellspacing=\"0\" cellpadding=\"3\">\n";
   echo "<tr><td><b>Username</b></td></tr>\n";
   for($i=0; $i<$num_rows; $i++){
      $uname  = mysql_result($result,$i,"username");
      echo "<tr><td>$uname</td></tr>\n";
   }
   echo "</table><br>\n";
}

I hope you can use the code above to help me develop this darn drop box!

Thanks, Ollie!

Holy smoke, it has been a looooooong time since I have been here. Love the new look of the site btw. Well I am working on a new project for my church. The idea is an email to SMS program. So one use of this would be that if there is a youth activity I could send a message to the entire youth group with one message. I am trying to figure out something specific with it. I need to create a form that grabs info from the database. So more specifically what I need is one drop down that will have "Youth Group", "Parents", "Deacons", etc. now based on what I select I need the second drop down to populate with the corresponding people in that group. Now I want an "All" to be in the drop down to which would let me send the message I type to ALL people in the desired group but to also have the individual names come up just in case I need to do a specific person.

How would I go about doing something like this? I have done much research and cant quite find what I would need to do. Any help would be GREATLY appreciated.

Im working on a site and particularly the photo gallery section. I want to display 9 images on 3 rows (3 per row).

And im stuck. Te only way i can get the images to show up not one on top of the other is by using a table. but how can i tell the table to only put 3 <td>'s then to <br>

Heres my code so far...
<?php
$sql_images = mysql_query("SELECT * FROM photos ORDER BY upload_date DESC LIMIT 9");

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

$photo_id = $row["id"];
$location = $row["location"];

$display_pics .= '
<td width="125px">
<a href="image.php?id='. $album_id .'" class="black"><img src="images/uploads/thumbs/' . $location . '" /></a> 
</td>';
}
?>

Then to display it ...
Code: [Select]
    <table width="top" cellpadding="10" cellspacing="20">
        <tr>
            <?php echo $display_pics; ?>
            </tr>
        </table>

My goal is to do it as simple as possible!

Can anyone help?

If you don't think tables are the way to go im open to any suggestions.

I've thought about <divs> but then the images show up one on top of the other and not beside each other.

By the way all the images are a max of 100px heigh and 100px wide

Thanks for any help or suggestions!

Hi all, completely new at this and have got myself stuck!

I am trying to display a thumbnail image by connecting to my database and retrieving the URL that points to the image.

I know that the connection is working but I can't seem to get it to point to the image. Probably something really simple. Here's my code:

Code: [Select]
<!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=iso-8859-1" />
<title>Image Swap Using CSS</title>
<style type="text/css">

body {
margin: 0;
padding: 0;
background-color:#000000;
}

img {
margin: 0;
padding: 0;
border: none;
}

.test {
margin: 0;
padding: 0;
width: 99px;
height: 130px;
}

.test a:hover img {
visibility:hidden;
}

</style>
</head>

<body>
<?php
// connect to the database
mysql_connect("********", "****") or die(mysql_error()); 
mysql_select_db("eliteescorting") or die(mysql_error());
$start=0;

// expand on the searches
$data=mysql_query("SELECT * FROM escorts WHERE base = 'manchester' ORDER BY RAND()");
$num_results=@mysql_num_rows($data);

for($ii = $start;$ii < $num_results; $ii++){
?>
<div class="test"><a href="#"><img src="<?php echo $data[$ii]['thumb'];?>" /></a></div>
<?php } ?>
</body>

</html>

In my mySQL field 'thumb' contains the full URL of the image ie http://www............

This is just for testing, I'll get round to sorting out security issues later.

Your help will be greatly appreciated!

Regards,
Nortski.

I'm having trouble getting the dynamic data from my <select> menus to write into my MYSQL database.  Can anyone see what I'm doing wrong here?  First post btw

The output looks like this, which is obviously wrong: 

Code: [Select]
<html>
<head> </head>
<link rel="stylesheet" type="text/css" href="./css/newuser.css" />
<body>
<?php session_start();
require 'default.inc.php'; ?>

<?php if (isset($_POST['amount'])): 
$host = 'localhost';
                $user = 'user';
                $pass = 'password';
                $conn = mysql_connect($host, $user, $pass);
                if (!$conn) {
                        exit('<p>Unable to connect to the database 
server</p>');
                }
                if (!@mysql_select_db('spikesusers')) {
                        exit('<p>Unable to locate the database</p>');
                }
        $locationname = $_POST['donor'];
        $donorid = mysql_query("SELECT id FROM donors WHERE locationname='$locationname'");

$amount = $_POST['amount'];
        $year = $_POST['year'];
        
$type = $_POST['type'];
        $typeid = mysql_query("SELECT id FROM donationtype WHERE type='$type'");

$player = $_POST['player'];
//$playerid = mysql_query("SELECT id FROM players WHERE 

        $sql = "INSERT INTO donations SET donorid='$donorid', amount='$amount', yearofdonation='$year', 
typeid='$typeid'";
mysql_query($sql);
?>
<div class='standard'>
<h1>Donation Management</h1>

<?php
if ($sql) {
                echo "New donation added ";
                echo "<p></p>";
echo "<a href=managedonations.php>Back to donation management</a>";
                exit();
        } else {
        echo "Error adding new donation";
        echo "<a href=adddonation.php>Try again</a>";
        exit();
        }
        
?></div>
<?php else:
$host = 'localhost';
                $user = 'user';
                $pass = 'pass';
                $conn = mysql_connect($host, $user, $pass);
                if (!$conn) {
                        exit('<p>Unable to connect to the database
server</p>');
                }
                if (!@mysql_select_db('spikesusers')) {
                        exit('<p>Unable to locate the database</p>');
                }
$donor=@mysql_query('SELECT id, locationname FROM donors');
?>
<div class='standard'>
<h1>Donation Management</h1>
<form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="post">
    <label>Donor:
    <select class="text" name="donor">
<option value=new>Add a new donor...</option>
<?php
while ($donors=mysql_fetch_array($donor)) {
$donorname=$donors['locationname'];
echo "<option value='<?php echo $donorname;?>'>$donorname</option>";
//echo "<option value='hi'>hi</option>";
}
?>
</option>
</select>
</label><br />
<label>Amount:
      <input class="text" type="text" name="amount" class="text" /></label><br />
<label>Year of donation:
      <select class="text" name="year">
<option value='2011'>2011</option>
<option value='2010'>2010</option>
<option value='2009'>2009</option>
      </select>
</label><br />
<?php $player=@mysql_query('SELECT id, firstname, lastname FROM players'); ?>
<label>Player:
<select class="text" name="player">
<option value="player" selected="selected"></option>
<?php
while ($players=mysql_fetch_array($player)) {
$playerfirstname=$players['firstname'];
$playerlastname=$players['lastname'];
echo "<option value=player>$playerfirstname $playerlastname</option>";
}
?>
</select>
</label><br />

<?php $type=@mysql_query('SELECT id, type FROM donationtype'); ?>
<label>Donation type:
<select class="text" name="type">
<?php
while ($types=mysql_fetch_array($type)) {
$donationtype=$types['type'];
echo "<option value=type>$donationtype</option>";
}
?>
</select>
</label><br />

<input type="submit" value="SUBMIT" class="buttons"/>
    <input type="button" name="Cancel" value="CANCEL"
onclick="window.location = 'managedonations.php'" class="buttons"/>
</form>
</div> <?php endif; ?> </body>
</html>