PHP - Assign Value To Cookie

How do I assign the entire $_POST array to a varible?

This code isn't working...

<?php
echo 'FormHandler';

class FormHandler{
private $formValues = array();

$this->formValues = $_POST;
}
?>



TomTees

hey everyone

will you please assist.

I'm getting data from a mysql database.

And I want to assign the data to a single value, if possible

For example, in my database I have 2 columns, name and group

Lets say John, Peter and Cathy belongs to group 555, where Sally and Marcus belongs to group 777.

So my sql query outputs all users in group 555

Is there a way to assign the search result to a vlue, for example $result = ???

and If I said echo $result, it should show the users who belongs to group 555 as an example.

I tried to use row[Name][1]; but it doesnt work.

The thing is, I can display the data the way I am familiar with, but I'm sending an automated email to myself, containing the users. So if I would used a for each statement, the mail will trigger and send a copy for every count, in this case, 3 times. And thats my huge problem.

The mail must trigger only once, containing the users in my search query, in this case, group 555.

Any help will be appreciated

Thank you

Hey guys, bascially I'm developing a simple news script, and part of it has a "quick news" section, which will display the latest 3 (or any number I want) news items, each seperated by a line (<hr> in the example below).

What I want to acheive is the top 2 news items have a set LI class assigned to them, and then the 3rd LI in the list to have a different style attached to it (one without a bottom border basically), so it'll bascially repeate out the normal <li> until it hits the last item to be output, at which point it uses a LI with a class attached to it.

What I don't know how to do, is tell the PHP that once it's out putted 2 LI's, then the 3rd LI is to have a different class on it.

eg

Code: [Select]
<li>News Item 1</li>
<li>News Item 2</li>
<li class="last">News Item 3</li>
I'm pretty new to PHP so not sure how to achieve this and would appreicate any help you can offer.

Here's my code so far...

Code: [Select]
<?php
$latestnews_sql = "SELECT * FROM news ORDER BY date DESC LIMIT 3";
$latestnews_query = mysql_query($latestnews_sql) or die (mysql_error());
$list_latestnews = mysql_fetch_assoc ($latestnews_query);
?>
<ul>
<?php do { ?>
<li><p><strong><?php echo $list_latestnews['title']; ?></strong></p>
<?php echo substr($list_latestnews['body'],0,80)."...";?> <a href="news-detail.php?ID=<?php echo $list_latestnews['ID']; ?>">view more</a></p>
<p><span style="font-size:13px; font-style:italic;"><?php echo $list_latestnews['author']; ?> - <?php echo date("l, jS F Y", strtotime($list_latestnews['date'])); ?></span></p>
<hr />
</li>
<?php } while ($list_latestnews = mysql_fetch_assoc ($latestnews_query)) ?>
</ul>

Running a simple query:

SELECT team, rank from table ORDER BY rank ASC

returns 8 records in order from 1 to 8

I then want to a assign a new variable to each that I can use for later queries ie:

team with rank 1 = $team1
team with rank 2 = $team2
team with rank 3 = $team3

etc....


Thanks in advance.

Hey guys, im trying to make a dropdown box get a selected value based on the value from multiple rows.


Here is what im trying to do:

<?php
// select the value from the database
// $value = ...

?>
<form action="...">
<option value="1"<?php if ($value == 1) { echo ' selected="selected"'; }  ?>>First</option>
<option value="2"<?php if ($value == 2) { echo ' selected="selected"'; }  ?>>Seccond</option>
<option value="3"<?php if ($value == 3) { echo ' selected="selected"'; }  ?>>Third</option>
<input type="submit">
</form>
The thing is I have no idea what to put at the top and how to assign the $value part.


Here is my current code.
<?php

mysql_connect("localhost", "", "")or die("cannot connect");
mysql_select_db("test")or die("cannot select DB");
$tbl_name="test_mysql";
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
$count=mysql_num_rows($result);

// MOVED this code to the top before you output anything

if (isset($_POST['Submit'])) {

    for($i=0;$i<$count;$i++){
        $month   = $_POST['month'];
        $date    = $_POST['date'];
        $message = $_POST['message'];
        $title   = $_POST['title'];
        $id      = $_POST['id'];
        $icon      = $_POST['icon'];         // ILYA
        $monthday= $month[$i]."<br>".$date[$i];
        $sql1="UPDATE $tbl_name 
                    SET monthday='$monthday', 
                        month='$month[$i]', 
                        date='$date[$i]', 
                        message='" . mysql_real_escape_string($message[$i]) . "', 
                        title='" . mysql_real_escape_string($title[$i]) . "',
                        icon='$icon[$i]'                        
                    WHERE id=".$id[$i];                        // ILYA

    if(!($result1 = mysql_query($sql1))){                   
        "<BR>Error UPDATING $tbl_name ";                              
        exit();                                             
        }                                                   
    }
}


$result=mysql_query($sql);
$count=mysql_num_rows($result);

?>



Here is the selection dropdown box im trying to make work:

<select name="icon[]" style="display:blolck; width:200px" class="mydds">
<option value="1" title="icon/icon_phone.gif">Phone</option>
<option value="2" title="icon/icon_sales.gif">Graph</option>
<option value="3" title="icon/icon_faq.gif">Faq</option>
</select>


The thing is, it gets multiple rows at once and I want to make it select each one differently based on the rows value.

I dont want it to like interfere.


Do you kinda know what I mean?

Also, heres hows the database looks like.



Picture of webpage.





Here is the full code if anyone would like to see:

<?php

mysql_connect("localhost", "", "")or die("cannot connect");
mysql_select_db("test")or die("cannot select DB");
$tbl_name="test_mysql";
$sql="SELECT * FROM $tbl_name";
$result=mysql_query($sql);
$count=mysql_num_rows($result);

// MOVED this code to the top before you output anything

if (isset($_POST['Submit'])) {

    for($i=0;$i<$count;$i++){
        $month   = $_POST['month'];
        $date    = $_POST['date'];
        $message = $_POST['message'];
        $title   = $_POST['title'];
        $id      = $_POST['id'];
        $icon      = $_POST['icon'];         // ILYA
        $monthday= $month[$i]."<br>".$date[$i];
        $sql1="UPDATE $tbl_name 
                    SET monthday='$monthday', 
                        month='$month[$i]', 
                        date='$date[$i]', 
                        message='" . mysql_real_escape_string($message[$i]) . "', 
                        title='" . mysql_real_escape_string($title[$i]) . "',
                        icon='$icon[$i]'                        
                    WHERE id=".$id[$i];                        // ILYA

    if(!($result1 = mysql_query($sql1))){                   
        "<BR>Error UPDATING $tbl_name ";                              
        exit();                                             
        }                                                   
    }
}


$result=mysql_query($sql);
$count=mysql_num_rows($result);

?>



<!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=utf-8" />
<title>Image Dropdown</title>
<link rel="stylesheet" type="text/css" href="dd.css" />
<script type="text/javascript" src="jquery-1.3.2.min.js"></script>
<script type="text/javascript" src="jquery.dd.js"></script>

</head>
<body>



<form name="form1" method="post" action="updat5.php"><!--ILYA-->
<tr>
<td>
<table width="100%" border="0" cellspacing="1" cellpadding="0">

<tr>
<td align="center"><strong>ID</strong></td>
<td align="center"><strong>Month - Date</strong></td>
<td align="center"><strong>Message</strong></td>
<td align="center"><strong>Title</strong></td>
<td align="center"><strong>Icon</strong></td>

</tr>
<?php
while($rows=mysql_fetch_array($result)){
?>
<tr>
<td align="center"><input style="border: 1px solid #C3C3C3;height: 20px; width: 20px;" id="id" name="id[]" value="<?php echo $rows['id']; ?>"></td>
<td align="center"><input style="border: 1px solid #C3C3C3;height: 20px;" name="month[]" MAXLENGTH="3" size="3" type="text" id="month" value="<?php echo $rows['month']; ?>">
<input style="border: 1px solid #C3C3C3;height: 20px;" name="date[]" MAXLENGTH="2" size="2" type="text" id="date" value="<?php echo $rows['date']; ?>">
</td>
<td align="center"><input style="border: 1px solid #C3C3C3;height: 20px;" name="message[]" size="70" type="text" id="message" value="<?php echo $rows['message']; ?>"></td>
<td align="center"><input style="border: 1px solid #C3C3C3;height: 20px;" name="title[]" size="70" type="text" id="title" value="<?php echo $rows['title']; ?>"></td>
<td align="center">
<select name="icon[]" style="display:blolck; width:200px" class="mydds">
<option value="1" title="icon/icon_phone.gif">Phone</option>
<option value="2" title="icon/icon_sales.gif">Graph</option>
<option value="3" title="icon/icon_faq.gif">Faq</option>
</select>
</td>
</tr>

<?php
}
?>
<tr>
<td colspan="4" align="center"><br><input type="submit" name="Submit" value="Submit"></td>
</tr>
</table>
</td>
</tr>
</form>

<script language="javascript" type="text/javascript">
function showvalue(arg) {
    alert(arg);
    //arg.visible(false);
}
$(document).ready(function() {
try {
        oHandler = $(".mydds").msDropDown().data("dd");
        oHandler.visible(true);
        //alert($.msDropDown.version);
        //$.msDropDown.create("body select");
        $("#ver").html($.msDropDown.version);
        } catch(e) {
            alert("Error: "+e.message);
        }
})    
</script>
</body>
</html>

Why doesn't this code work...

Code: [Select]
// Initialize variables.
$form_value = '';
$form_value = $_POST['form_value'];


I get this error...

Quote

Notice: Undefined index: form_value


Thanks,


Debbie

I am trying to convert a php4 app to php5 and am getting this error (Fatal error: Cannot re-assign $this in ) when I call this class.

$this = $this->createUnique($new); is the line that generates the error.

I try to rename $this to $_this but it does not help. I think I lose the reference to the object later on in the code.

Then I will get an error like (Fatal error: Call to a member function createUnique() on a non-object in)

Is there another way to reference the current instance of an object in php5?

Any help appreciated...



Code Example:
Code: [Select]
class imageobject{

var $handle;
var $height=0;
var $width=0;
var $directory;
var $filename;


//constructor
function imageobject($directory,$filename,$width=0,$height=0,$color="FFFFFF")
{
$this->directory = $directory;
$this->filename = $filename;

if ($filename=="" && $width>0 && $height>0){

$new = $this->createImage($width,$height);
$this = $this->createUnique($new);

}elseif (file_exists($directory.$filename)){
$size = GetImageSize($directory.$filename);
if ($size) $this->handle = $this->getHandle($directory.$filename,$size[2]);
$this->width = $size[0];
$this->height = $size[1];

}


function createUnique($imgnew)
{
$this->type = substr($this->output,0,3);

$unique_str = $this->uniqueName();
switch ($this->type){
case "png":
imagepng($imgnew,RES_DIR.$unique_str);
break;
default:
imagejpeg($imgnew,RES_DIR.$unique_str,$this->quality);
break;
}

$this->handle && imagedestroy($this->handle);
$newobject = new imageobject(RES_DIR,$unique_str,$this->type);
return $newobject;
}

)//end class

Hello All,

I have two php files original.php and return.php
retrun.php file returns a variable after I pass $channel_name to it
Code: [Select]
return $return_variable;
I want to assign this return variable to a variable in original.php
Something like this
Code: [Select]
$results =  include ('return.php?channel='.$channel_name.'');I am not sure how to do the above code. When I call the return.php from a  browser it works properly.

Thanks for any help

basically, I have one calculation that relies on the value returned by an if statement, but i can;t for the life of me figure it out correctly... Here is what i am trying to accomplish:

Here is the code for the IF i need to get the value of

if ($loadedMiles < 500) { 
(500 - $loadedMiles) * .25 * $mainRate;
}


And here is the reason why i need it:

if ($loadSize == "Full") {
$loadedmiles + thevalueoftheif;
}


basically i will need to have a way to get the values of the functions in the IF statements if that makes any sense...

thanks in advance!

Completely new to PHP here,

so on my website I'd like a specific image to appear according to whether or not the user in my DB has a row called 'premium', with a value of 1. I've tried figuring it out on my own but I'm stuck. I watched a video of someone doing something similar so this is what I have currently, but I'm aware it's totally incorrect.

//init.php file

<?php
if (session_status() == PHP_SESSION_NONE) {//php >5.4
  session_start();
}
$_SESSION['user_id'] = 1;   
$db = new PDO('mysql:host=127.0.0.1;dbname=loginsystem', 'root', '');
$userQuery = $db->prepare("
	SELECT idUsers, uidUsers, emailUsers, pwdUsers, premium
	FROM users 
	WHERE idUsers =:user_id
	");
	$userQuery->execute(['user_id' => $_SESSION['user_id']]);
	
	$user = $userQuery->fetchObject();

//webpage in which the image will appear
 

<?php
require_once '../includes/init.php';
?>
 <?php if($user-> premium): ?>
      <img src="../../img/premium.png" alt="Swiftify+"></a>
<?php else: ?>
      <img src="../../img/swiftify.png" alt="Swiftify+"></a>
<?php endif; ?>

 

I have a list of pictures and I want to control the order.


Here is the picture format:
$pic = http://www.picture.biz/incoming/w_2GTEK13T961240561_1.jpg
$pic = http://www.picture.biz/incoming/w_2GTEK13T961240561_2.jpg
$pic = http://www.picture.biz/incoming/w_2GTEK13T961240561_3.jpg
$pic = http://www.picture.biz/incoming/w_2GTEK13T961240561_4.jpg
$pic = http://www.picture.biz/incoming/w_2GTEK13T961240561_5.jpg

Notice the order 1.jpg, 2.jpg, etc..

Here is where I'm stuck...

// begin stuck :)
$orderPic = explode("_", $pic);

foreach($orderPic as $rank){
     echo "$rank"; // I need $rank to equal 1 then 2 then 3 ...
}


It would be nice actually if I could start the insert with number 2 because number one is always a screwy picture.

I hope this makes sense /  Thank for the help!

Scenario: My page generates several PHP variables. Some of those variables get posted to another page.

Question: If a variable is to be assigned to a Session, is there any reason to assign this value to a regular PHP variable?

Example: When an API generates a "shipping label tracking number, which is better, "A" or "B" or subtle variation "C":

A.) 
$result = json_decode(curl_exec($ch));
$_SESSION["trackingNumber"] = $result->parcelTrackingNumber;

... and then use $_SESSION["trackingNumber"] throughout the rest of my script (like this: echo  "Hello, here's your tracking number: ".$_SESSION["trackingNumber"]).

B.) 
$result = json_decode(curl_exec($ch));
$trackingNumber = $result->parcelTrackingNumber;
$_SESSION["trackingNumber"] = $trackingNumber;

... and then use "$trackingNumber" throughout the rest of my script (like this: echo  "Hello, here's your tracking number: ".$trackingNumber).

C.) 
$result = json_decode(curl_exec($ch));
$_SESSION["trackingNumber"] = $result->parcelTrackingNumber;
$trackingNumber = $_SESSION["trackingNumber"];

... and then use "$trackingNumber" throughout the rest of my script (like this: echo  "Hello, here's your tracking number: ".$trackingNumber).

Thank you! I've always wondered about this. Which is best?

 

 

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=350859.0

i have Created a Session  with this code $_SESSION['USER_NAME'] = trim($_POST['username'])


How can i access the $_SESSION['USER_NAME'] in other Pages ;

and will this code for $var = $_SESSION[USER_NAME] 

can i get the SESSIOn value in " $var"  variable ??

I am trying to figure out the best way to do something if there is a way.

I am pulling values from a MySQL table and using a while loop to display the data, so it looks something like this.


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


  echo "<tr>";

  echo "<td>" . $row['last_name'] . "</td>\n";
  echo "<td>" . ucwords($row['first_name']) . "</td>\n";

  echo "<td>" . $row['Class_Date'] . "</td>\n";
  
  echo "</tr>";




for the class dates, I want the column background to be a different color depending on the date. For instance January would hvae a blue background, February a red one etc.

I was going to try and do that with CSS like

Code: [Select]
echo "<td class="blue">" . $row['Class_Date'] . "</td>\n";

in the CSS...
Code: [Select]
.blue{
background-color:#06c;
}

but not sure the best way to make the changing data values different colors with CSS.

any suggestions?