PHP - Php Xml Dom Converts < And > To &lt; And &qt; When Assigning Nodevalue

Hey everyone. I have been trying to assign dates to image so I could sort them, and also display the date they were uploaded. But I need some help. I have a script that uploads an image to a file directory, and also stores the file directory and the upload date in a database. And I also have a script that displays the images in the file directory but I cannot find a way to assign the right date to the right image.
Here is the script that displays the images:

Code: (php) [Select]
if (isset($_POST['submit'])) {
// fill an array with all items from a directory
$dir = "thumbs/";

$handle = opendir($dir);
while ($file = readdir($handle)) {
$files[] = $file;

// This removes the full stops from the array
unset($files[0]);
unset($files[1]);
               
}   
            foreach($files as $fileimg)
{

echo "<img src='". $dir . $fileimg ."'/>\n";
                echo "$date";
}
       
}
and here is the script for calling the date from the database:

Code: [Select]
$query = mysql_query("SELECT imgaddress, uploaddate FROM thumbs WHERE imgaddress='" . $dir . $file . "'");

                if (mysql_num_rows($query) == 1) {
                     $row = mysql_fetch_assoc($query);

                     $date = $row['uploaddate'];
If I use the database statement before the foreach, then it does not get the right file directory, but if I call it inside the foreach, then I cannot use krsort() to sort the array.

Any help would be greatly appreciated, thanks

Good Day Everyone!

I'm having trouble figuring out how to assign weapons from a MySQL table to soldiers.
Say I have the following...
$soldiers = 100;
$damage = 0;
$q = mysql_query("SELECT `item_quantity`, `item_power` FROM `items` WHERE `owner_id`='".$userid."' ORDER BY `item_power` DESC");
while($row = mysql_fetch_assoc($q)) 
{
   // assign weapons to soldiers and add `item_power` * `item_quantity`to $damage
}
I want to assign all the weapons until there's not enough soldiers to assign them to, and I want to assign them in order of `item_power` descending.

I'm sure this is really simple, I just can't for the life of me get my head around it.

Thanks in advance!

Kind Regards,
Ace

Hi all,

I can't seem to designate an array key by using a variable and I was wondering if this is possible. I'm looking to do something like this:

Code: [Select]

<?php
$key = "apple";
$arr = array($key => "fruit");
?>
Any suggestions would be appreciated!

Hi, I cant quite figure out how to  assign an unique id to the picture I upload to my page, all help would be appreciated.

Here's the upload code:
 
 //This is the directory where images will be saved 
 $target = "images/"; 
 $target = $target . basename( $_FILES['photo']['name']); 
 
 //This gets all the other information from the form 
 $name=$_POST['name']; 
 $email=$_POST['email']; 
 $phone=$_POST['phone']; 
 $pic=($_FILES['photo']['name']); 
 
 // Connects to your Database 
 mysql_connect("your.hostaddress.com", "username", "password") or die(mysql_error()) ; 
 mysql_select_db("Database_Name") or die(mysql_error()) ; 
 
 //Writes the information to the database 
 mysql_query("INSERT INTO `employees` VALUES ('$name', '$email', '$phone', '$pic')") ; 
 
 //Writes the photo to the server 
 if(move_uploaded_file($_FILES['photo']['tmp_name'], $target)) 
 { 
 
 //Tells you if its all ok 
 echo "The file ". basename( $_FILES['uploadedfile']['name']). " has been uploaded, and your information has been added to the directory"; 
 } 
 else { 
 
 //Gives and error if its not 
 echo "Sorry, there was a problem uploading your file."; 
 } 
 ?> 

would anyone know how to assign the value (value returned by some function in the class) of variable $temp in the class: source looks like:

class newRecord{
var $temp = listCat('gallery_menu', 'submenu', 'submenu_id'); ////// this line is not working, does not assign?

function listCat($table, $submenu, $submenu_id)
{
$return = '<select name="'.$submenu_id.'" id="'.$submenu_id.'">';
$result = mysql_query("SELECT * FROM $table");
while($row=mysql_fetch_object($result))
{
$return = $return.'<option value="'.$row->id.'">'.$row->$submenu.'</option>';
}
$return = $return.'</select>';
return $return;
}


}

if function is used out of the class and used in index.php directly, returns expected value of it (there is no doubts - works ok)

I have two tables in my MySQL database:

1. user
user_id | firstname | lastname | nickname etc.

and

2. con (for contribution)
con_id | user_id (foreign key) | name | contribution | category etc.


Here's how I wanted to solve the problem of assigning the foreign key to the 2nd table (I'm a beginner so bear with me. )

When logging in, I assigned the user_id to the session variable like this:

// do the while loop

while ($assoc = mysqli_fetch_assoc($rows)) {

// assign database COLUMNS to the variables

$dbuser_id = $assoc['user_id'];

$dbuser_name = $assoc['nickname'];

$dbuser_password = $assoc['password'];

etc.....................


// set a session after login

$_SESSION['user'] = $dbuser_name . $dbuser_id;



This prints out the user_id just like I wanted to:


echo "Your user_id is: " . $_SESSION['user_id'] = $dbuser_id;


Gives:
Code: [Select]
Your user_id is: 35


... and this is how the assignment of the foreign key looks like WHILE posting the contribution:


 $user_id = $_SESSION['user'] = $dbuser_id;

  if (!empty($knuffix_name) && !empty($knuffix_category) && !empty($knuffix_contribution)) {
      // Write the data into the database
      $query = sprintf("INSERT INTO con (con_id, user_id, name, contribution, category, contributed_date) VALUES (' ', '$user_id', '%s', '%s', '%s', now())",

mysqli_real_escape_string($dbc, $knuffix_name),

mysqli_real_escape_string($dbc, $knuffix_contribution),

mysqli_real_escape_string($dbc, $knuffix_category));



When I now do a contribution through the input areas, nothing gets inserted into the database and I automatically get logged out. Obviously it's not working as I thought it would and I'd like to know why is that?
Notice that if I take the user_id part OUT, it's working again, so the rest of the code must be right then.

Another question I have is:
Is this common practice to solve this problem of assigning a foreign key, or is there a better way of doing it?

hi there,
i'm trying to put a message in the footer of a page which welcomes a person who is logged in with his/her name, using sessions of course; when i place this:
Code: [Select]
$username = $_SESSION['valid_user'];
in the footer before the echo:
Code: [Select]
echo "You are logged in as $username";
but the session is also needed before the footer to use the username for other things , such as -checking his credit- so if place in footer the footer shows name in browser but checking credit would not happen as the assignment is at the buttom.

IF i place the assignment above at the top of the file:

everything works for the user and checking credit ..etc...but the footer is not there...

cud not put this any clearer..sorry...hope if someone cud help...thanks

Hi folks,

I was wondering how to do this. I want the if statement to detect if the query string has any of these values. so im trying to assign them all to the same variable. However, this code wont work. Whats the trick here?

<?php
$primary=$_GET['intro'];
$primary=$_GET['port'];
$primary=$_GET['about'];
$primary=$_GET['contact']; 

         if(isset($primary)){
     echo "<img src='graphics/left-a.png'>";}
else {echo "<img src='graphics/leftb.png'>";}?>

So I have a text file "name.txt"

in the text file I have ids and auth keys set up like this

1234564 abcdfhu
3123900 sdkoao

etc

How could I make it to where the ids are $ids and the auth keys are $auth_keys?

I've tried using foreach() but I cant get it

Hi, I have just started creating my first class in php. I'm trying to assign $_SERVER['REMOTE_ADDR']; to a protected variable but I keep getting an error message. I'm still trying to get my head around oop php. 

My current code is "protected $user_ip = $_SERVER['REMOTE_ADDR'];" and the error message is "Parse error: syntax error, unexpected T_VARIABLE".

I have a query that pulls 1 field with 20 rows of data.  I need to assign each row to a different variable so that I can then display them in different locations on a page.
I cannot make each row of data a different field because of other constraints.
My data is very well normalized.
I am using mysqli so something like the old mysql_result would be lovely!
How can this be done without hitting my database 20 times?
Thanks for the help.

Good Day to everyone!

We are developing an e-commerce website as a compliance to our school project. We encountered a problem in manipulating the shopping cart.. My only question is, what are you going to POST from buttons of "Add to cart" to specifically add that particular product to my user cart. We find a hard time dealing with this buttons because it comes up with the same IDs or names. HELP PLEASE T_T

Hi all, creating a MYSQL, PHP & XHTML site designed to support local rugby clubs. Just putting the final touches to the functionality now so thanks for your help so far.

I would like to provide site administrators with the ability to assign photos to a members profile when initially registering their account, I have no experience of dealing with what presumably will be a function that will upload a photo to a location and then linking it some how to data in the database.

Thanks for your help,

Tom

I didn't know how to describe this, so please excuse my title.

As most of you know, this is possible:
foreach($items as $key => $item)
{
${$key} = $item;
}

// I would then be able to access it like so:
$key // or whatever the value of "$key" is.

However, how could I append something extra to the ${$key} variable?

Instance:
foreach($items as $key => $item)
{
$my_{$key} = $item;
        // or 
        ${$key}_arr = $item;
}

// So that the end result of my variable becomes like so:
$my_key 
// or
$key_arr
// Instead of just
$key // or whatever the value of "$key" is.

Is there a known method for this?

Any input is appreciated, thank you.

Hi people.

I am creating a carpet website for a good friend of mine as a favour.  I am storing the carpets info in a MYSQL database and am currently trying to relay the info on a page.  I have the array kind of done but it is not producing the results I want.  Let me explain further:

The columns in my database under the table "carpets" a

id (auto incremented)
colour
type
title
price
description
imageloc

------

under the type I have various types of carpet (6-ish).  They a Twist, Striped etc etc...

Heres the php problem.


I have all the images there and ready on the page and want to link them all with linking commands. i.e:

the main page is at: http://www.ircdirect.co.uk/FTPServers/supremecarpets/index.php

the carpets results page is at: http://www.ircdirect.co.uk/FTPServers/supremecarpets/carpetlist.php

on the index.php page I want to link them so that for example:

If he looks for a striped carpet he clicks "striped" and is shows carpetlist.php but with the striped carpets displayed.

example link: <A HREF="carpetlist.php?type=striped">IMAGE HERE</A>

I have tried various coding on the carpetlist.php and cant seem to get it to work.

here is the snippets:

<?php
// Make a MySQL Connection
$query = "SELECT * FROM carpets GROUP BY type"; 
 
$result = mysql_query($query) or die(mysql_error());


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


echo "<TABLE CELLPADDING=0 CELLSPACING=0 WIDTH=100% />";

echo "<TR />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/topleft.png /></TD />";

echo "<TD BACKGROUND=images/main/search/top.png /> </TD />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/topright.png /></TD />";

echo "</TR />";

echo "<TR />";

echo "<TD BACKGROUND=images/main/search/left.png />";
echo " ";
echo "</TD />";

echo "<TD BACKGROUND=images/main/search/bg.png />";

echo "<FONT FACE=VERDANA SIZE=1 />";
echo $row['title'];

echo "</TD />";

echo "<TD BACKGROUND=images/main/search/right.png />";

echo " ";

echo "</TD />";

echo "</TR />";

echo "<TR />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/bottomleft.png /></TD />";

echo "<TD BACKGROUND=images/main/search/top.png /> </TD />";

echo "<TD WIDTH=23 /><IMG SRC=images/main/search/bottomright.png /></TD />";

echo "</TR />";

echo "</TABLE />";

echo "<BR />";

}
?>

It displays ALL the results which is not what I want.  I want only striped carpets, blue carpets etc....

Help is needed and VERY much appreciated!

Kind Regards,
Ian