PHP - Not Getting <input Type=hidden> Value Thru $_get

This topic has been moved to JavaScript Help.

http://www.phpfreaks.com/forums/index.php?topic=317004.0

This is a cross between HTML, PHP, and MySQL.  I'm not exactly sure which to blame (other than myself of course)

I'm collecting a bid price in a standard form

<input type="number" name="maxbid" value="<?php echo $maxbid ?>"/>

I have recalled the value from the database where I store it and display it in the field if it exists.  If the field has never been entered, the field shows nothing.  It's not required that a bid be entered.  There are other fields of data that may be updated and this field is left blank.

On the page called by the submit button I'm retrieving the fields

$maxbid=$_POST[maxbid];

Then I store the fields in the database either with an INSERT or an UPDATE

if(mysql_num_rows(mysql_query("SELECT id FROM data WHERE id = '$id'")))
{
echo "Updating.....";
mysql_query("UPDATE data SET maxbid='$maxbid', name='$name' where id='$id'");
} else
{
echo "Inserting.....";
mysql_query("INSERT INTO data (id, maxbid,name)
VALUES ('$id','$maxbid',$'name')");
}

I know I'm not handling the Numeric values correctly, and I'm having a hard time finding the correct method for doing so.  Right now, if the name is entered byt the maxbid is left blank, nothing gets inserted or updated.  if I enter both the maxbid and the name it works.  I can't have maxbid show up as a 0, I need it to be blank if nothing is entered.

Also when I check $maxbid, it's not an INTEGER (per INTVAL) when I check it, even though maxbid in the table is an INTEGER.  When I pass it around, it seems to lose it's INTEGER value, but still shows as a number.

I know this sounds confusing, heck I'm confused.  If someone could point me to a good tutorial, I'd figure it all out, but most tutorials don't get into the data type that much, and the combination of the form, passing the variables, reading the values from the database, etc. is what's causing me grief.

Any guidance would be appreciated.

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

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!

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.

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

A few months ago, and a good amount of time before that, I had people telling me to use isset() instead of performing to see if the variable is empty, such as: !$_GET['']

I know the differences in the function and what they do, but when could isset() be used in a situation where it's better/more efficient then: !$_GET['']

I do use isset(), though.

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?