PHP - Help With Input Type Hidden

How can I make it so that the things that the user recently typed in an input is not displayed as a drop down.  I want nothing to be displayed. 

I am trying to set the hidden input in the javascript but it doesn't set it in the php. 

This is the form with the hidden input. It corresponds to a field in the database table with the name of draggables.

<form method="post" accept-charset="utf-8" id="myForm2" action="/mealplans/add">

<input type="hidden" name="draggables" id="drag12"/>

<button id="myId" type="submit">Submit Form</button>

</form>

This is where I change the value in the javascript:

if(data == "draggable3"){
alert("data " + data);
var x = document.getElementById("drag12").value;
x =  "draggable3";
alert(x);
} //end if

                
The alert pops up with draggable3 as it should but the value of null is what gets saved in the database. The javascript doesn't change the PHP value; it only changes the client side value.         

I've also tried getElementsByName but that didn't work either.

var y = document.getElementsByName("draggables").value;    //only changes it on the client

This is actually a cakephp site but this problem is with the PHP so I am publishing it here.

This function is throwing the following error:

invalid input syntax for type timestamp "1301297896"

Code: [Select]
function removeInactiveUsers(){
      if(!TRACK_VISITORS) return;
      $timeout = time()-USER_TIMEOUT*60;
     
$result = pg_query_params($this->connection, "delete from active_users where timestamp < $1", array($timeout));
      $this->calcNumActiveUsers();
   }
I know the problem is with $timeout using a php timestamp vs postgresql timestamp but I'm not sure how to fix that

Hey guys!
I'm making a locally run web based application for entering/storing/editing data.
I am trying to do this thing where I export a 'records.txt' file to a different location which can be selected by the user.
I have a lame <input type='text' value='H:\'> box going on which I was going to use to define the path for the file to be saved, but I want to be able to use a pop up windows (or whatever native OS finder application) just like the one which appears when you use <input type='file'>.

Now that I think about it, I think this is more of a html question, but it's 2.30am and my project is due today...
See! what the hell, I just habitually pressed ctrl-s after working in dreamweaver for hours and the exact function which I am looking for popped up, how do I do that?!

Please thanks guys

I think I understand that there can only be one "submit" button on a page. I want buttons for separate forms to not trigger the empty errors on the other forms...   How do you accomplish specific button actions?   I think I've tried it with javascript but you can do it with PHP right? Thanks for any help.

Hello,

Iv written a small script in JS to add input boxes of the type file to the page.. My problem is when i use multiple of those input boxes and submit the form the $_FILES variable only reads 1 of those boxes.. This is the javascript im using:
Code: [Select]
function addImageBox() {
var imageBoxes = document.getElementById("imageBoxes");
var tr = document.createElement("tr");
var td = document.createElement("td");
var td2 = document.createElement("td");
    var inputFile = document.createElement("input");
   
   
    inputFile.setAttribute("type", "file");
    inputFile.setAttribute("name", "image[]");
    inputFile.setAttribute("style", "width: 450px");

    td.appendChild(document.createTextNode("Image"));
    td2.appendChild(inputFile);
   
    tr.appendChild(td);
    tr.appendChild(td2);
   
    imageBoxes.appendChild(tr);
}

When i add 2 boxes (making it 3 boxes on the page in total) and count the $_FILES var after submiting the form, the count method returns 1, and when i var_dump it only 1 array is in $_FILES. Any idea's? thanks!

Long time lurker, first time poster here.  I am having a problem with a multiple file uploader form I am trying to build.  I am using the CodeIgniter framework, but I believe my problem is with the form itself.

Here is the form:
Code: [Select]
<form action="http://localhost/show/submit" id="sendfiles" method="post" name="sendfiles" enctype="multipart/form-data">   
<input type="hidden" name="doup" value="1" />
    <input type="file" name="userfile1" value="" id="userfile1" size="20"  />    <br />
    <input type="file" name="userfile2" value="" id="userfile2" size="20"  />    <br />
    <input type="file" name="userfile3" value="" id="userfile3" size="20"  />    <br />
    <input type="submit" name="submit_btn" value="Upload"  />    </form>   
   

The form is being passed to this function:


if(empty($_POST['doup'])){
            foreach($_REQUEST as $k => $v)
                print ">>> $k : $v <br />";
                
            $data['form_open'] = form_open_multipart(uri_string(), array(
                'id' => 'sendfiles',
                'method' => 'post',
                'name' => 'sendfiles'    
            ));
            $data['form_flag'] = form_hidden('doup', '1');
            $data['file1'] = form_input(array(
                'id' => 'userfile1',
                'name' => 'userfile1',
                'size' => '20',
                'type' => 'file',                
            ));
            $data['file2'] = form_input(array(
                'id' => 'userfile2',
                'name' => 'userfile2',
                'size' => '20',
                'type' => 'file',                
            ));
            $data['file3'] = form_input(array(
                'id' => 'userfile3',
                'name' => 'userfile3',
                'size' => '20',
                'type' => 'file',                
            ));
            $data['form_submit'] = form_submit('submit_btn', 'Upload');
            $data['form_close'] = form_close();           
            $this->load->view('show/submit', $data);
        } else {
            $config['allowed_types'] = 'mp3';
            $config['max_size']    = '10240';
            $config['upload_path'] = $this->config->item('file_upload_path');
            $uploaded_files = array();
            for($nona=1; $nona<4; $nona++){
                if (strlen($_FILES['userfile'.$nona]['tmp_name']) <=0){
                    echo  "tempname$nona failed";
                    continue;
                }
                
                    
                //print "------ USERFILE $nona ------ <br />";
                //print sizeof($_FILES['userfile'.$nona])."<br />";
                //foreach ($_FILES['userfile'.$nona] as $k => $v){                    
                //    print "$k  :  $v <br />";
                //}
                        
                $this->upload->initialize($config);
                $upload = $this->upload->do_upload('userfile'.$nona);
                if (!$upload){
                    $form_errors = array('form_errors' => $this->upload->display_errors());
                    $this->load->view('show/submit', $form_errors);
                    break;
                } else {
                    $data = array('upload_data' => $this->upload->data());
                    $data = $data['upload_data'];
                    $fp = $data['full_path'];
                    $filename = strtolower(random_string('numeric', 16) . $data['file_ext']);
                    $rel_path = 'show/submissons/uploaded/';
                    $path = $this->config->item('media_root_path') . $rel_path;
                    
                    exec('mkdir -p ' . $path , $a1, $r1);
                    exec('mv ' . $fp . ' ' . $path .  $filename, $a2, $r2);
                    
                    array_push($uploaded_files, $path .  $filename);
                }
                
            }
            $this->session->set_flashdata('img_data',implode("||", $uploaded_files));
           // redirect("show/submit_info");
        }    



The problem is this:

When I choose only one file to upload, it works.  I can choose any of the 3 inputs and it works great.  But as soon as I try to choose 2 or more files, the form fails to post.  When I press submit, the page simply reloads and none of the form variables get posted.  I have checked settings in php.ini and tried building and rebuilding this form.  I'm hoping someone here can provide some insight into what I'm doing wrong?

Hi..

I used this code for displaying formatted number and I notice that instead comma(,) it become point(.):

Code: [Select]
$P28_maxdoz = $row['P28_maxdoz']; 
$P28_maxdoz = number_format($P28_maxdoz, 2, '.', ',');


<table>
<tr>
<td><input type='text' name='P28_maxdoz' value="<?php echo $P28_maxdoz;?>"></td>
</tr>
</table>

the output is : 22.782.20

i need output is : 22,782.20

Thank you

hi there,

i am fairly new to OOPs in php, i get an error when i declare the argument type (as object) in a function and pass the same type (object).



class eBlast {
       public static function getEmail(object $result) {
           return $result->email;
       }
}

$r = mysql_fetch_object($query);
eBlast::getEmail($r);

echo gettype($r); // outputs: object



error is :

Code: [Select]
Catchable fatal error: Argument 1 passed to eBlast::getEmail() must be an instance of object, instance of stdClass given, called in C:\wamp\www\integra\client\pl_eblast\admin\send_emails.php on line 145 and defined in C:\wamp\www\integra\client\pl_eblast\app\app.eBlast.php on line 8
if i remove the type declaration in the function it works, but just would like to know why it shows error when pass the same type, also isnt mysql_fetch_object is the instance of stdclass?

thanks in advance!

Probably a stupid question . . .   The MIME type in my php.ini file is set to text/html:

default_mimetype = "text/html"

But, because I use XHTML 1.0 Strict, the MIME type of my web pages is set to text/xml:

<meta http-equiv="content-type" content="text/xml; charset=utf-8" />

I should therefore change the php.ini MIME type to text/xml, right?  

I hope I can explain what is happening.  I have created two forms in PHP.  The first 'almost' works, i.e. it shows the data.  But I have two problems - 1) the second pulldown menu is always empty and 2) $value from the first pulldown menu ALWAYS equals the last entry thus the last 'if' in the function subdomains ($domains) is always called (but still empty).  The code may explain this better than me:

<!DOCTYPE html>
<html>
<body>
   <!-- processDomains.php is this file - it calls itself (for testing purposes so I can see what is happening) -->
   <form action="processDomains.php" method="post">
   <?php
       // create the domains array (there are actually several entries in the array but I cut it down for testing)
       $domains = array (1 => 'Decommission', 'Migration');
    
       echo "Select Domain:";
       echo "<br>";
    
       // Make the domain pull-down menu - this displays correctly
       echo '<select name="domain">';
       foreach ($domains as $key => $value) {
          echo "<option value=\"$key\">$value</option>\n";
       }
       echo '</select>';

       // input doesn't matter what is 'submitted', always goes to last $value
       echo '<input type="submit" name="submit" value="Submit">';

       // call function subdomains
       subdomains ($value);
    
       function subdomains ($domains) {
           // define values for each array - each array contains available choices for the subdomain pulldown menu
           $migration = array (1 => 'Application Migration', 'Application Patch', 'Application Upgrade');
           $decommission = array (1 => 'Applications', 'Servers', 'Storage');
    
           if ($domains === 'Migration') {
              echo "Select subdomain:";
              echo "<br>";
              // Make the Migration pull-down menu
              echo '<select name="migration">';
              foreach ($migration as $key => $value) {
                  echo "<option value=\"$key\">$value</option>\n";
              }
              echo '</select>';
           } else if ($domains === 'Decommission') {

              /* ===
               * since 'Decommission' is the last entry in the 'Domains' pulldown list, $value ALWAYS equals
               * 'Decommission' and $domains equals $value. So this menu SHOULD work but is always
               * empty.  Thus, two problems - the pulldown menu is always empty and $value isn't based
               * upon user input.
              */
              echo "Select subdomain:";  // this prints so I know I'm in 'Decommission (I eliminated the echo "$domain" to show I'm always coming here)'
              echo "<br>";
              // Make the 'Decommission' pull-down menu
              echo '<select name="decommission">';
              foreach ($decommission as $key => $value) {
                  echo "<option value=\"$key\">$value</option>\n";
              }
              echo '</select>';
              echo '<input type="submit" name="submit" value="Submit">'
           ) // end of 'if-else'
    } // end of function 'subdomain'
     
?>
</form>
</body>
</html>

Let me say thank you in advance and I appreciate the help!  I know I'm doing something (or more than one thing) wrong and I hope someone can tell me what it is.    Best Regards!
Edited by mac_gyver, 19 January 2015 - 09:37 PM.
code tags around posted code please

I am writing a script that will parse my PHP classes and check for things like coupling, visualize my objects and connections, dependencies, check for convention usage, etc.   So, I have a simple file upload.  I'm never saving the files, just get contents and dump the file and work with the string version.   I'm writing it for me, but I figure I might want to open it for others to use in the future, so I may as well write it that way to begin with -- so I need to validate user input.  Problem is, the user input is supposed to be valid PHP code.  I'm thinking that, as long as I'm careful, I shouldn't be executing any code contained in strings, but I'm no security expert and I want a warm fuzzy that my thought on this is correct.  What kinds of things do I need to look out for?  Is it possible to inject when working with strings?   My initial thought is to regex the entire file and replace key portions with known replacements.  So ( and ) would become !* and !^ or $ would become @~ (combinations that -- I think -- don't make sense to php?)  But that may be completely unnecessary processing time if I'm not in any danger, here.  Thanks ahead of time for any help.   PS - as a side question -- what's the best way to verify a file is a php file?  I know of getimagesize for images, but should I just check for <? to verify it's php?  That seems like it would be too easy to fool -- then again, it might not matter much.   -Adam

I have a calendar select date function for my form that returns the date in the calendar format for USA: 02/16/2012.  I need to have this appear as is for the form and in the db for the 'record_date' column, but I need to format this date in mysql DATE format (2012-02-16) and submit it at the same time with another column name 'new_date' in the database in a hidden input field.  Is there a way to do this possibly with a temporary table or something?

Any ideas would be welcome.

Doug