PHP - Assigning Weapons To Soldiers

I just have a quick question, what happens if I don't assign a visibility element to methods?

In other words, if I just have:

[php]
class Article_model
{
   public $data;

   function index()
   {
      return $data;
   }
[php]
...instead of declaring public, private, static, and so on before the function. Will php just consider it public?

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,
i am updating records from database using ajax and javascript on php page.
The result is displayed inside a div (<div id="show"></div>).
Now i want to assign the content of the above div(say y) to php variable for further calculations.

How can i assign the value displayed in div tag to a php variable?

Thanks.

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 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, 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".

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

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 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'>";}?>

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

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.

Hello Everyone,

I am new to forum and could use some help with some php code that isn't working.  I am very new to php/html/javascript and all of what I have learned, I learned from forums like this one so first....thank you!

I am trying to assign a value from a php variable to the value of my form element.  I'm sure there must be a way to do this but I can't seem to get the syntax right.  here is my code...

first I set the value of $loginname elsewhere in the script like so...

<?php
$loginname =strtolower(htmlspecialchars(strip_tags($_GET["loginname"])));
?>
This part works fine..

Then I try to set the value of my hidden text field inside the form to the value of $loginname to be passed to a javascript program.  Everything works except that the value passed ends up being <?echo and not the expected user name inside of $loginname.

<?php

echo '<form name ="currentactivity" Id="currentactivity" action="<?php'.htmlspecialchars($_SERVER['PHP_SELF']).'?>"      method="post">';

echo '<fieldset><legend><b>Your Current Activity Information</b></legend>';
echo '<input type="text" name="loginnm"  style="visibility: hidden" value="<?php echo $loginname;?>">';

echo "<label for='myactivities'>Activity Name:</label>";

 
echo "<select name='myactivities' Id='myactivities' onchange=\"showdetails(this.form)\" value=''>";
echo "<option value = 'Select an activity'>Select an activity</option>";
for ($i = 1; $i <=$rowcount; $i++) {
   echo"<option value=$row[activity_name]>$row[activity_name] </option>";
   
   $row = mysql_fetch_array($result);
}
       
echo "</select>";
 
echo '</fieldset>';

echo '</form>';
?>

Please note..the rest of the code is working perfectly, it is just this one value I can't seem to get.
Any help you can give will be greatly appreciated. 

Hello,

I am creating a class which contains one private function that deals with connecting to a SQL Server and executing the sql that is passed in to that function.
I have defined a class variable which is assigned the value of sqlsrv_query like so.

$this->QueryResult = sqlsrv_query($conn,$sql);
I have placed private $QueryResult at the top of my class.

The other public methods call this private function to assign the result set to the class variable, the private method returns and then the public methods loop through the results array like so.

while($row = sqlsrv_fetch_array($this->QueryResult))

.. but the while loop never gets entered.

If I declare everything in the same method then it works, however there are going to be several public methods that will use this private method and I don't want to duplicate all the database coding.

Hope someone can help