PHP - Remove Recently Type From Input

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!

hidden input field type doesnt display the full value i get from table, say if it is a first name and last name i get only the first name. here is my input field
echo "<input type=hidden name=proid value=$result_row[project_id]></include>";
echo "<input type=text value=$result_row[project_name] readonly></include>";

my sql query returns the full name but when i input into the text field it returns an incomplete project name like say if it is fans of soccer, i just get the value "fans" in my display.....

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 Guys,

I have a script that shows the user of my shop what they have looked at. I was wondering if someone can tell me how I can limit this so it only lists 10 and no more.

Here is the code:

Code: [Select]
if(RECENTVIEWED ==1)
{
if(isset($_COOKIE['jimbeam'])){
$productUrl = SITE_URL."ecom/index.php?action=ecom.pdetails&mode=";
$productUrl = $this->libFunc->m_safeUrl($productUrl);

foreach ($_COOKIE['jimbeam'] as $name => $value)
    {
    $this->obDb->query="SELECT vTitle FROM ".PRODUCTS." WHERE vSeoTitle = '".$value."'";
$rsProd = $this->obDb->fetchQuery();
    $this->obTpl->set_var("TPL_VAR_PRODUCTTITLE",stripslashes($rsProd[0]->vTitle));
$this->obTpl->set_var("TPL_VAR_PRODUCTURL",$productUrl.$value);
$this->obTpl->parse("recent_blk","TPL_RECENT_BLK",true);
    }
$this->obTpl->parse("mainrecent_blk","TPL_MAINRECENT_BLK");
}

}
Thanks,
Jordan

I want to be able to grab any new images uploaded and display them right away on the front page, at the moment I can grab the id of each but that won't update the gallery. Once 4 images are uploaded, the next one to be uploaded pushes the last out of the gallery.   At the moment, I only have 2 images on the db, but I want it ready to be used by a community

<form action="<? echo $_SERVER['PHP_SELF']; ?>" method="post"  
	enctype="multipart/form-data"> 
	Upload:<br><br> 
	<input type="file" name="image"><br><br> 
	<input type="submit" name="submit" value="Upload"> 
	</form>
	
	<?php
	
	if(isset($_POST['submit']))
	{
		mysql_connect("localhost","____","____");
		mysql_select_db("moduni_images");
		
		$imageName = mysql_real_escape_string($_FILES["image"]["name"]);
		$imageData = mysql_real_escape_string(file_get_contents($_FILES["image"]["tmp_name"]));
		$imageType = mysql_real_escape_string($_FILES["image"]["type"]);
		
		if(substr($imageType,0,5) == "image")
		{
			mysql_query("INSERT INTO `images` VALUES('','$imageName','$imageData')");
			echo "Image uploaded!";
		}
		else
		{
			echo '<br>O<font color="#8B0000">nly images are allowed!</font>';
		}
		
	}
	
	?>

<?php 

	mysql_connect("localhost","____","____");
	mysql_select_db("moduni_images");
	
	if(isset($_GET['id']))
	{
		$id = mysql_real_escape_string($_GET['id']);
		$query = mysql_query("SELECT * FROM `images` WHERE `id`='$id'");
		while($row = mysql_fetch_assoc($query))
		{
			$imageData = $row["image"];
		}
		header("content-type: image/jpeg");
		echo $imageData;
	}
	
	else
	
	{
		echo "Error!";
	}
?> 

<div id="user-gallery">
			<h2>Gallery</h2>
			
						
			<div class="img">
			  <a href="scripts/show_image.php?id=2" data-lightbox="image-1">
				<img src="scripts/show_image.php?id=2" width="125px" height="71px">
				</a>
			</div>
			<div class="img">
			  <a href="scripts/show_image.php?id=2" data-lightbox="image-1">
				<img src="scripts/show_image.php?id=2" width="125px" height="71px">
				</a>
			</div>
			<div class="img">
			 <a href="scripts/show_image.php?id=2" data-lightbox="image-1">
				<img src="scripts/show_image.php?id=2" width="125px" height="71px">
				</a>
			</div>
			<div class="img">
			  <a href="scripts/show_image.php?id=2" data-lightbox="image-1">
				<img src="scripts/show_image.php?id=2" width="125px" height="71px">
				</a>
			</div>
		</div>

This topic has been moved to Other Libraries and Frameworks.

http://www.phpfreaks.com/forums/index.php?topic=307986.0

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!

First problem fixed.
My second problem is if the result is 0X.XX,0 or 0X.XX,1 I would like to remove the first 0
The X.XX are numbers, but the 0 is not always there.

Can anyone help please
Thanks

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