PHP - Assigning Array Key Name By Variable

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. 

Okay, really newbie question, but for this code...
Code: [Select]
<!-- Gender -->
<label for="gender">Gender:</label>
<select id="gender" name="gender">
<option value="">--</option>
<option value="F">Female</option>
<option value="M">Male</option>
</select>

1.) How do I assign a variable to this?

2.) How do I make this "sticky"?


Here is how I have usually done other form types...
Code: [Select]
<!-- First Name -->
<label for="firstName">First Name:</label>
<input id="firstName" name="firstName" type="text" maxlength="30"
value="<?php if(isset($firstName)){echo htmlentities($firstName, ENT_QUOTES);} ?>" /><!-- Sticky Field -->
<?php
if (!empty($errors['firstName'])){
echo '<span class="error">' . $errors['firstName'] . '</span>';
}
?>


Oh, by the way, at the top of my PHP file I have this code...
Code: [Select]
if ($_SERVER['REQUEST_METHOD']=='POST'){
// Form was Submitted (Post).

// Initialize Errors Array.
$errors = array();

// Trim all form data.
$trimmed = array_map('trim', $_POST);

Thanks,


Debbie

hey guys,
I'm quite new to PHP and i was wondering if someone would be able to help me out with this.
The code i have checks the database to see if a user has provided a Vimeo ID. If they haven't, $video_check will equal a line of html that says the user hasn't added any videos to their portfolio. If the a vimeo ID is present, i want $video_check to equal a bunch of html and css with a foreach function inside that is used to display the users videos from vimeo.
The foreach function works fine when its not assigned to the $video_check variable. How do i structure it so that it displays correctly?

If you click on videos on this page you might get a better idea of what i'm talking about.
http://www.myfilmportfolio.ie/profile.php?id=33

and here is the code i'm having the problem with:

///////  check if user has provided vimeo id  //////////////////////////
$vimeoID = $row["vimeoID"];   
$video_check='';
if (empty($vimeoID)){
     $video_check = '<h3>'.$firstname .' has not added any videos to their portfolio</h3>';
     }
     else{
     $video_check = '<div id="stats">
                    <div style="clear: both;"></div>
               </div>
               <div id="wrapper">
               <div id="embed"></div>
                         <div id="thumbs">
               <ul>
               <?php foreach ($videos->video as $video): ?>
                    <li>
                                        <h4><?=$video->title?></h4>
                    <a href="<?php echo $video->url ?>">
                    <img src="<?php echo $video->thumbnail_medium ?>" class="thumb" /></a>
                    <p><?=$video->description?></p>
                                             <br />
                    </li>
                    <?php endforeach ?>
               </ul>
               </div>
            </div>';
   }


if anyone could help me out id really appreciate it.
Cheers,
G

I am facing problem to execute query by assigning NULL value to a variable and then executing query.In MySQL DB four fields Mobile,landline, pincode,dob are set as integer and date(for dob) respectively.The default is set as  NULL and NULL option is selected as yes.All these fields are not mandatory.The problem is that when I edit the form my keeping the value as empty in DB these are saved as 0, 0 , 0 & 0000-00-00 inspite of Null. I have tried everything but still the defect persist. Please help me to come out of the problem 

The code, I have used:

<?php
//require_once 'includes/config.php';

      $dbusername = $_POST['email'];
        $dbfirstname = $_POST['first_name'];
        $dblastname = $_POST['last_name'];
        //$dbmobile_number = $_POST['mobile'];
      if (isset($_POST['mobile']))
      
      {
         $dbmobile_number = $_POST['mobile'];
         
      }
       
      else
      
      {
         $dbmobile_number = "NULL";
         
      }
            
      $dblandline_number = $_POST['landline'];
        $dbdob = $_POST['dob'];
      
      if(isset($_POST['is_email']))
      {
         $dbSubscribe_Email_Alert = '1';
      }
      else
      {
         $dbSubscribe_Email_Alert = '0';
      }
      
      if(isset($_POST['is_sms']))
      {
         $dbSubscribe_SMS = 0;
      }
      else
      {
         $dbSubscribe_SMS = 0;
      }
       
      
        $dbAddress_firstname = $_POST['shipping_first_name'];
        $dbAddress_lastname = $_POST['shipping_last_name'];
        $dbAddress = $_POST['shipping_address'];
        $dbcity = $_POST['shipping_city'];
        $dbpincode = $_POST['shipping_pincode'];
        $dbstate = $_POST['shipping_state'];
        $dbcountry = $_POST['shipping_country'];
      
      echo "Welcome".$dbusername;

//if($_POST['btnSave'])

      //if ($_POST['btnSave'])

//{
   
   //echo "Inside query loop";
$connect = mysql_connect("localhost","root","") or die("Couldn't connect!");
mysql_select_db("salebees") or die ("Couldn't find DB");

//$query = mysql_query("SELECT * FROM users WHERE username='$username'");

$query = mysql_query("update users set firstname = '$dbfirstname', lastname = '$dblastname', mobile_number = '$dbmobile_number', landline_number = '$dblandline_number', dob = '$dbdob', Subscribe_Email_Alert = '$dbSubscribe_Email_Alert', Subscribe_SMS = '$dbSubscribe_SMS',  Address_firstname = '$dbAddress_firstname', Address_lastname = '$dbAddress_lastname', Address = '$dbAddress', city = '$dbcity', pincode = '$dbpincode', state = '$dbstate', country = '$dbcountry'      where username = '$dbusername' ");

header("location:my_account.php");


//}

//else

//{
//die();

//}


?>

I need to assign unknown array keys and their values to strings.

The code I am using is:
Code: [Select]
$cart = $_SESSION['cart'];
$items = explode(',',$cart);
$qty = array();


foreach ($items as $item) {
$qty[$item] = $qty[$item] + 1;
}
print_r($qty);
The keys change and are not ordered. I do not know what the key or it's value may be.  So as an example, I might have:
Code: [Select]
Array ( [2] => 5 [4] => 8 ) or in a different scenario, I might have:
Code: [Select]
Array ( [1] => 6 )
What those numbers represent is a product id (the key), and the quantity (the value of the key).

Can someone help me to assign these to a string?  I have been reading all morning, and not making any progress. 

Thank you!

OK,

When the user fills the info in the form out it goes in the DB fine.  I can then array them on the "showroom page" fine.

When they upload a picture it goes into the /images/ folder fine.

Problem is... On each array on the showroom page I need the image they uploaded to be displayed.

Cant work it out.

Help would be GREAT!!!!!

I get the following error when I try to pass a value to a methiod in a loop:

Warning: Attempt to assign property of non-object in /Users/staceyschaller/Sites/dev_zone/ckwv2/classes/class.php on line 670

This one has me very baffled. It will work the first time, and seems to work every other time, so I have no clue what is wrong. Here is the code:

This code is part of my "display" class:

function display_partner ($type,$loc,$rand=0,$narrow=0) {
$this->partners = new partner($this->cxn);
$display = '
<div id="cont_info" class="partner-list">
<div>
<h3 class="settings">'.ucfirst($loc).' '.ucfirst($type).last_letter($type).'s</h3>
</div>
<div class="settings-value" style="height:12px;padding:0;margin:0;text-align:right;padding-right:10px;">
<a href="" class="trunc">Add your organization to this list</a></p>
</div>
<div style="height:2px;padding:0;margin:0;">
<hr class="account" />
</div>
';
$ids = $this->partners->get_partners_list($type,$loc,$rand);
for ($b=0;$b<sizeof($ids->id);$b++) {
$this->partnerID = $ids->id[$b];
$display .= ($narrow)? $this->card_partner_narr():$this->card_partner();
if ($b!=(sizeof($ids->id)-1)) { $display .= '<hr class="account" />'; }
}

if (sizeof($ids->id)==0) {
$display .= '<div style="color:#999999;display:line;text-align:center;height:20px;">No Partners found for '.ucfirst($loc).' '.ucfirst($type).'</div>';
}

$display .= '
</div>';

return $display;

}

function card_partner () {
$this->partners->set_partner_id($this->partnerID);
$part_info = $this->partners->get_partner_info();
if ($part_info) {
$display .= '
<table class="settings">
<tr>
'.$this->show_if($part_info['partLogo']['val'],'<td class="settings-value" rowspan="2"><img src="'.LOGO_FOLDER.$part_info['partLogo']['val'].'" '.resize_img(LOGO_FOLDER.$part_info['partLogo']['val'],175).'alt="'.$part_info['partName']['val'].'" /></td>').'
<td class="settings-value" colspan="2"><h5>'.$part_info['partName']['val'].'</h5></td>
</tr>
<tr>
<td class="settings-value">
<span style="color:999999;">'.$part_info['partAddress']['val'].'<br />
'.$part_info['partCity']['val'].', '.$part_info['partST']['val'].' '.$part_info['partZIP']['val'].'<br />
'.$part_info['partPhone']['val'].'</span><br />
<a href="'.$this->form->show_href($part_info['partWeb']['val']).'" target="_blank">'.$part_info['partWeb']['val'].'</a>
</td>
<td class="settings-value">'.$part_info['partInfo']['val'].'</td>
</tr>
</table>
';
}

return $display;
}



This code is part of my "partners" class:

function set_partner_id($partID) {
echo '<p>partID: '.$partID.' '.gettype($partID).'<br>
$this->partner->id: '.$this->partner->id.'</p>';
$this->partner->id = $partID; ///*** ERROR HAPPENS HERE ***/
echo '<p>id set: '.$this->partner->id.'<br>
$this->partner->id: '.$this->partner->id.'</p><hr>';
}

function get_partner_id() {
return $this->partner->id;
}

// gets user info at login
function get_partner_info() {
$this->partner = $this->cxn->proc_info('partner','partID',$this->partner->id);//$this->partner->id
return $this->partner;
}


The following is the output generated:

partID: 24 string
$this->partner->id:
id set: 24
$this->partner->id: 24

partID: 26 string
$this->partner->id:
Warning: Attempt to assign property of non-object in /Users/staceyschaller/Sites/dev_zone/ckwv2/classes/class.php on line 670
id set:
$this->partner->id:

partID: 17 string
$this->partner->id:
id set: 17
$this->partner->id: 17


partID: 25 string
$this->partner->id:
Warning: Attempt to assign property of non-object in /Users/staceyschaller/Sites/dev_zone/ckwv2/classes/class.php on line 670
id set:
$this->partner->id:

As you can see, the value passes to $this->set_partner_id($partID) each time. It is formatted as a string. When it assigns the value to $this->partner->id, however, sometimes it works, and sometimes it doesn't. It's probably something obvious, but I've racked my brain to see what it is. Any ideas?

Hi guys im in the middle of optimizing code..

Code: [Select]
$sql = "SELECT * FROM sa_enemystats WHERE username='$username'";
$res = mysql_query($sql);
while ($row = mysql_fetch_assoc($res)) {

$enemy['current_health'] = $row['current_health'];
$enemy['current_skill'] = $row['current_skill'];
$enemy['level'] = $row['level'];
$enemy['damage'] = $row['damage'];
$enemy['evade'] = $row['evade'];
$enemy['accuracy'] = $row['accuracy'];
$enemy['speed'] = $row['speed'];
$enemy['luck'] = $row['luck'];

echo 'debug: variables synced with db table';
}I know that the setting of these variables are messy  and can be done in a better way... but how? I have tried foreach and copying other's code for an array copy but it lead no where. Then again my syntax could be wrong... I used something like
Code: [Select]
foreach ( $enemy[$value] as $row => $value) {
$enemy[$value] = $row[$value];
}

I need a help in the following :
I have an admin page from where the admin attaches an gif image.This attached image should be shown to the user as a scrolling image.
The code for the scrolling image is done through javascript.
So my actual problem is getting the name of the attached gif image to the javascript array which is used for scrolling horizontally.

hi all,

I have an language pack for example:
languages/en.php:
Code: [Select]
$en['mail']['letter closing'] = "regards,\n your friend!";
and in my config:
Code: [Select]
$language = "en";
$include_language = @include("languages/".$language.".php");
if(!($include_language))
{
    $try_default_language = @include("languages/nl.php");
    if(!($try_default_language))
    {
        echo "kan de taalpakket niet vinden<br>";
        echo "Could not find the language pack.<br>";
        echo "example on error: ".$test." shows nothing";
        exit;
    }
}
In my function I want to include the language pack for example i have $language = 'en' so I want to include $en['general']['letter closing']

I will do this:
Code: [Select]
global $language,${$language}['general'];
But that gives an error unexpected '[' blah blah.

How can i call the variable variable array in the valid php way?

Code: [Select]
$word = 'numbers';
$numbers= array('1', '2', '3', '4');

echo $$word[0];
I expected the output to be '1'. It ended up being nothing... 

Why does this not work? Is it not possible to have a variable variable array? And if not, is there a workaround?

Cheers,
Joe

Hi,

I need help with this, I am quite new to arrays. This one works fine:

echo array_sum(array(34,19,22,17,7,22,8,23,17,31,21,32,7,29,13,7))/count(array(34,19,22,17,7,22,8,23,17,31,21,32,7,29,13,7));
which gives "19..."

it gets average number. What I want to put the values in the array in a variable so I do this:

$t = "34,19,22,17,7,22,8,23,17,31,21,32,7,29,13,7";

so why can't I use array like this:

echo array_sum(array($t))/count(array($t));

I need to use variable in array, please someone help.

Hi,

I am basically trying to apply a different amount to $defineMe for certain numbers of an array .. in this case it's array numbers 11 - 16

Code: [Select]
<?php
 
if ($defineNumber > 11 and $defineNumber < 16) {
$defineMe = 35.00;
} else {
$defineMe = 50.00;
}

$defineNumber[01] = array('name'=>'Product 1');
$defineNumber[02] = array('name'=>'Product 2');
$defineNumber[03] = array('name'=>'Product 3');
$defineNumber[04] = array('name'=>'Product 4');
$defineNumber[05] = array('name'=>'Product 5');
$defineNumber[06] = array('name'=>'Product 6');

$defineNumber[11] = array('name'=>'Product 11');
$defineNumber[12] = array('name'=>'Product 12');
$defineNumber[13] = array('name'=>'Product 13');
$defineNumber[14] = array('name'=>'Product 14');
$defineNumber[15] = array('name'=>'Product 15');
$defineNumber[16] = array('name'=>'Product 16');
  
?>

Hey, im struggeling with a small piece of code....

Thats my code:

Global $user_ID;
$user_ID = get_current_user_id();

   echo"User number $user_ID is loggedin";

Echo gives me this:
"User number 2 is loggedin"

 

Now i want to add it to my array:
 

$atts['href'] .=  $user_ID . 'abc';

'href' is set to "https://test.com/"

What im getting now ist

https://test.com/abc

 

Can someone tell my why the '2' is missing??

 

 

hello all,

I have an array value in my post array ($_POST['check0'] ) that I am trying to break up into php variables. For some reason if I do this:
$var1 = $_POST['check0'][0];
all I get is the first letter of the string from the first value of the array?