]]> PHP tanha http://www.daniweb.com/forums/thread124409.html http://www.daniweb.com/forums/thread124360.html Thu, 15 May 2008 13:06:19 GMT

After logging in, main.php is loaded and instead of displaying the correct included page, it displays Main Page.

So what is wrong w/ my code?

Thanks in advance for any and all help...

this is placed inside the "main.php"

<?php
    require('core/db.php');
    require('core/check.php');
?>
 
This is placed inside the check.php

<?php
    if (level == "1")  { include('page_1.php'); }
    else
    if (level == "2") { include('page_2.php'); }
    else
    if (level == "3") { include('page_3.php'); }
?> 
Inside the pages I have
<?php
  echo "Page_Name";
?>



]]> PHP Vai http://www.daniweb.com/forums/thread124360.html http://www.daniweb.com/forums/thread124355.html Thu, 15 May 2008 12:47:13 GMT

i have created this piece of code to retrieve the path to a file from a mysql db, the file exists and can be displayed directly from the sql query. the code looks like this

$query = "SELECT broad1 FROM images_broad";
$result=mysql_query($query);

while($row=mysql_fetch_array($result)){
$image_path="{$row['broad1']}";
}

i now want to display this image inside an email. so the next part of the code use a variable to hold the first part of the path as only the relative path is stored in the db. they are the appended together to give the absolute path of the image to be displayed. the code looks like this

$path= 'http://www.acmeart.co.uk/mercury';
$broad_img1='<img src="' . $path."/".$image_path .'" />';

when the source code of the email is viewed on the $path is showing in the source, would the $image_path still hold a value inside.

]]> PHP kevin wood http://www.daniweb.com/forums/thread124355.html http://www.daniweb.com/forums/thread124347.html Thu, 15 May 2008 11:25:20 GMT ok lets make this a little simpler than my last topic.

I found this site which everyone who follows it seems to be getting results from, though it says "You are expected to be knowledgeable in the UNIX shell." which i have got no idea what that even is.

I tried to follow the instuctions as best as I could with no luck so could someone please explain that to me in simpler/noob terms.

Also i can not use .htaccess files so is there anyway i can do this without them?

Thanks,
Scottmandoo

btw im asking here because it also states "Support for these instructions is not available from DreamHost tech support."

]]> PHP Scottmandoo http://www.daniweb.com/forums/thread124347.html http://www.daniweb.com/forums/thread124344.html Thu, 15 May 2008 11:13:12 GMT Hi I want to split a variable to check what type of user they are. If they are user type 1 their username will be in the form "LLL" where L is a letter. If they are user type 2 their username will be in the form "LLNNNNN" where L is a letter and N is a number. I thought I could use PHP split and then is_numeric function to check the username. Would this be the best way to do it? If so what is the code I need to use to split the variable?

Many thanks for your help
NH

]]> PHP twelvetwelve http://www.daniweb.com/forums/thread124344.html http://www.daniweb.com/forums/thread124329.html Thu, 15 May 2008 09:33:40 GMT i have store an image on the server from a file upload with a random number attached to it. i have then stored the random number in a mysql db for use on other pages to display the images. the problem i am having is that the image will not display on other pages when i try to recall it from the... i have store an image on the server from a file upload with a random number attached to it. i have then stored the random number in a mysql db for use on other pages to display the images.

the problem i am having is that the image will not display on other pages when i try to recall it from the server.

i have stored the image in a folder on the server using this code

"image/thumbs/thumb_".$image_name . $rand;

$image_name being the name of the file to be stored and $rand being the random number. i have then stored the random number in the db using

$rand= rand(0, 100);
$sql="INSERT INTO random_number (random1) VALUES ('$rand')";
$query = mysql_query($sql);

to get the file from the server to be displayed the code i have used is this

$image = "";
$broad_img1= "";
$path= 'http://www.acmeart.co.uk/mercury/';
$web_image_folder = 'image/thumbs';
$exts = array('jpg', 'png', 'gif', 'jpeg');
$image_name = 'thumb_image1';


// check for each extension
foreach($exts as $ext) {
  if (file_exists($web_image_folder.'/'.$image_name.'.'.$ext . $rand)) {
    $image = $image_name.'.'.$ext . $rand;
  }
 
}


// check if we have an image
if ($image != "") {
    // ok we have the image
    $broad_img1='<img src="' . $path."/".$web_image_folder."/".$image . '" />';
} else {
    // error, we can't find the image.
}

this code worked fine before i was adding the random number. it retrieved the image and displayed the cached version (when cached cleared displayed new upload image).

i have got the random number out of the db using this code

$query = "SELECT random1 FROM random_number";
$result=mysql_query($query);

while($row=mysql_fetch_array($result)){
$rand="{$row['random1']}";

the number is then placed into the variable and then tried to be used to display the image but the image will not display. it may be the way i have written the code to retrieve the image but my eyes are tired of looking so i dont think i would see a mistake if it slapped me.

i will post the second half of the code in full below.

]]> PHP csshacker http://www.daniweb.com/forums/thread124327.html http://www.daniweb.com/forums/thread124326.html Thu, 15 May 2008 09:14:58 GMT Hi guys,

I did a quick tutorial on the internet on using php and mysql. It is my first ever time using it so forgive me if im being a bit of a "NOOB".

anyways ill post the script below and can you give me a few pointers about where im going wrong?

Bare in mind its my first time using these scripts and phpmyadmin.
--------------------------------------------------------------------------------------------------------------

<?php
$db=mysql_connect("grafax.co.uk", "dbOne", "*password*") or die("Could not connect to localhost."); mysql_select_db("people", $db) or die("Could not find database.");

// The above lines establishes a connection with the database. Keep localhost as is unless something different is mentioned by your sql host. usrnm is user name and pswd is password. What I want to say is, copy these lines as they are and just replace the required fields and it should connect.

$querySQL = "insert into people (d_Name, d_Birthday,
d_Birthday) values ($Name, $Birthday, $Telephone)";
if(!$querySQL) error_message(sql_error());

// The above statement generates an error if you have setup the table in such a way that there should not be a duplicate entry.
?>

<form name="toSave" method="post" action="save_it.php">
  <p>name
    <input type="text" name="Name" size="100" />
    <br>
birthday
<input type="text" name="Birthday" size="50" />
<br>
telephone
<input type="text" name="Telephone" size="50" />
</p>
  <p>
    <input type="submit" name="Submit" value="Submit" />
  </p>
</form>

--------------------------------------------------------------------------------------------------------------

Go on this link to see a screen shot of my database's settings etc.
http://www.grafax.co.uk/db.bmp

--------------------------------------------------------------------------------------------------------------
As far as i know, my database is called either web163-dbone (or as I named it dbone), the table is people, the website is http://www.grafax.co.uk.

If you need any more information just post it in this thread.

Thanks again guys :D
Max

]]> PHP crissrusso http://www.daniweb.com/forums/thread124272.html http://www.daniweb.com/forums/thread124269.html Wed, 14 May 2008 23:37:08 GMT

Here is what I am trying to accomplish: When a user logs in, their username will be displayed... Trouble is I am unsure how to co-relate the username w/ the id... thanks for any and all help....

<?php
  require_once('core/db.php');
  require_once('core/secure.php');

  $query = mysql_query("SELECT username FROM admin WHERE id = 1");
  $username = mysql_fetch_array($query);
 ?>
<?php echo $username[username]; ?>

]]> PHP Vai http://www.daniweb.com/forums/thread124269.html http://www.daniweb.com/forums/thread124262.html Wed, 14 May 2008 21:52:21 GMT Hello I have been working on this script for a little while. All is working good(get all information into the xml). But I am needing to have the php code add an image to the xml for download. If anybody as any ideas i would greatly appreciate it. I have this pulling testing info from the MYSQL... Hello
I have been working on this script for a little while. All is working good(get all information into the xml). But I am needing to have the php code add an image to the xml for download. If anybody as any ideas i would greatly appreciate it.

I have this pulling testing info from the MYSQL db. There will be an image that will need to be added to this document.

if(isset($_POST['file_name'])) {
$xml_dec = "<?xml version=\"1.0\" encoding=\"UTF-8\" ?>";
$rootElementStart = "<user>";
$rootElementEnd = "</user>";
$xml_doc = $xml_dec;
$xmlfilename = $_POST['file_name'];
$userinfo = mysql_query("select userid, firstname, lastname, email from users") or die(mysql_error());
while($row = mysql_fetch_array($userinfo))
{
$xml_doc .= $rootElementStart;
$xml_doc .= "<id>";
$xml_doc .= $row['userid'];
$xml_doc .= "</id>";
$xml_doc .= "<name>";
$xml_doc .= $row['firstname']." ".$row['lastname'];
$xml_doc .= "</name>";
$xml_doc .= "<email>";
$xml_doc .= $row['email'];
$xml_doc .= "</email>";
$xml_doc .=$rootElementEnd;
}
$default_dir = "testing/";
$default_dir .= $xmlfilename.".xml";
$fp = fopen($default_dir,'w');
$write = fwrite($fp,$xml_doc);
//header($default_dir);
echo $default_dir;
echo "Document has been written successfully!";
}
?>

]]> PHP ProfessorPC http://www.daniweb.com/forums/thread124262.html