PHP - Image Path

hi i want to store url to images in database for logged in users (where id = $id)

and recall the image hopefully using
---------------------
<img src="<?php echo row['link']; ?>" /> or similar and need help with the sql update string any ideas please help i been stuck with this for some time and now decided to ask around in this forum for help, please help if you can.

So, I am your typical newbie to php. I am usually doing more design stuff, but now I am diving into PHP. The script in question is made for a user to upload an image to a respective directory. Also the path, category, and 2 tags representing the image will be uploaded to a mySQL database.

My problem is that I have pieced together some freeware scripts that I have found in order to accomplish my ultimate goal. I need the form to be processed on the same page, because I don't want to redirect users. I also need to upload the respective information to the database. Right now, when you upload an image it only seems to be processing a part of the form, and does not upload any of the information at all. Can someone spot the many things that I am doing wrong and provide me with some direction?

I have two files in question

image_upload.php (Sorry about all of the code)

<?php
require_once 'upload_config.php';

function VerifyForm(&$values, &$errors)
{

// Do all necessary form verification
    if($clear_folder_before_upload){
        $mydirectory = myUploadDir();
        EmptyDir($mydirectory);
    }

    $uploaded_file_counter=0;
    $UploadLimit = $_POST['counter'];
    
    for($i=0;$i<=$UploadLimit;$i++){
        $file_tag='filename'.$i;
        $filename=$_FILES[$file_tag]['name'];

            if($filename!=null) {
            $rand=time();
            $str="$rand$filename";

            // set folder name in here.
            $filedir= myUploadDir();

            //change the string format.
            $string= $filedir.$str;
            $patterns[0] = "/ /";
            $patterns[1] = "/ /";
            $patterns[1] = "/ /";
            $replacements[1] = "_";
            $dirname=strtolower(preg_replace($patterns, $replacements, $string));
            //end of changing string format

            //checking the permitted file types
                            if($check_file_extentions)    {
                            $allowedExtensions = allowedfiles();
                                foreach ($_FILES as $file) {
                                                    if ($file['tmp_name'] > '') {
                                                      if (!in_array(end(explode(".",
                                                            strtolower($file['name']))),
                                                            $allowedExtensions)) {
                                                            $fileUploadPermission=0;
                                                      }
                                                      else
                                                      {
                                                          $fileUploadPermission=1;
                                                      }
                                                    }
                                                  }
                            }
                            else{
                                $fileUploadPermission=1;
                            }
            //end of checking the permitted file types

                if($fileUploadPermission){
                    if(move_uploaded_file($_FILES[$file_tag]['tmp_name'],$dirname)) {
                            echo "<img src='$dirname'>";
                            $uploaded_file_counter+=1;
                        }
                }
                
            }

    }
 
if($uploaded_file_counter==0){
     echo "<br /> <b style='font-weight:bold;color:red'>Oops! Please select an image file</b>";
}else{
     echo "<br /> <b>You requested ".$i." image files to upload and ".$uploaded_file_counter." files uploaded sucessfully</b>";
echo $filename0;
$contentdiv = 'Upload succesful! Upload Again';
}


}

function DisplayForm($values, $errors)
{
?>

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.1//EN" "http://www.w3.org/TR/xhtml11/DTD/xhtml11.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>JayGilford.com :: Name form example</title>
    <script type="text/javascript" src="js/jquery-1.4.2.js"></script>
    <script type="text/javascript" src="js/jquery-ui-1.8.custom.min.js"></script>

    <style type="text/css">
        div#upload_box_wrapper{width:350px;text-align:center; float: left;font-family: verdana; font-size: 12px;padding: 5px 5px 5px 5px; color: black;}
        div#upload_box_wrapper #upload_title_text{text-align:left;width: 100%;display: block;border: 0px solid green; margin-bottom: 5px;}
        div#upload_box_wrapper #FileUploadDiv{display: block;width: 98%;border: 0px solid green; text-align: left;padding: 0px 0px 0px 0px;}
        div#upload_box_wrapper p.add-new-upload-box{float: left;width: 100%;text-align: left;}
        div#upload_box_wrapper a.add-new-upload-link{font-weight: bold;font-size: 13px;padding-right: 3px;text-decoration: none; color: black;}
        div#upload_box_wrapper .upload-button{border: 0px solid lightgreen;height:30px; width:120px;cursor: pointer;background:url(./images/upload_button.png) no-repeat center left; color: white;} 
    </style>
</head>
<body>
<form method='post' action='' enctype='multipart/form-data'>
        <div id="upload_box_wrapper">

                <div id="upload_title_text">Image Upload:</div>
                    <div id="FileUploadDiv" DivValue="0">
                        <input name='filename0' type='file' id='filename'>
                        <select name="img_category">
<option value="Vacation">Vacation</option>
<option value="Holiday">Holiday</option>
</select>
                        <input name="tag1" type="text" id="tag1">
                        <input name="tag2" type="text" id="tag2">
                    </div>
                    <input type="hidden" value="0" id="counter" name="counter">
                    <div align='left'>
                        <input type='submit' name='submit' value='Upload Now!' class="upload-button">
                    </div>
</div>
</form>

</body>
</html>


<?php
}

function ProcessForm($values)
{

$filename0 = $_POST['filename'];
$filetype = $_POST[''];
$img_category = $_POST['img_category'];
$tag1 = $_POST['tag1'];
$tag2 = $_POST['tag2'];

////////
//figure out the image??
/////

session_start();
$img = $_POST[ 'imglink' ];
$filetypes = array( 'gif','jpg','jpeg','png' );
++$filetypes;
$filetype = strstr( $img, '.' );
if( $filetype != $filetypes )
{ echo 'not supported'; }
session_register( 'Image' );


$Img = $_SESSION[ 'Image' ];
mysql_query( "INSERT into `images` ( `image_id`, `filename`, `filetype`, `url`, `img_category`, `tag1`, `tag2` ) 
 VALUES ( '', '$filename0', '$filetype', '$Img', '$img_category', '$tag1', '$tag2' )" );
session_destroy();
mysql_close( $DB );
 
// Replace with actual page or redirect :P

}
 
if ($_SERVER['REQUEST_METHOD'] == 'POST')
{
$formValues = $_POST;
$formErrors = array();
 
if (!VerifyForm($formValues, $formErrors))
DisplayForm($formValues, $formErrors);
else
ProcessForm($formValues);
}
else
DisplayForm(null, null);
?>


and the upload_config.php file


<?php
///////////////////////////////////////////////////////////////////////////
//establish mysql connection (I have this set correctly on the original file)
///////////////////////////////////////////////////////////////////////////
$db = mysql_connect('RESPECTIVE', 'MYSQL', IDANDPW');
 
if (!$db) {
    echo "Unable to establish connection to database server";
    exit;
}
 
if (!mysql_select_db('chad_images', $db)) {
    echo "Unable to connect to database";
    exit;
}

///////////////////////////////////////////////////////////////////////////
// set your upload directory name in here.
///////////////////////////////////////////////////////////////////////////
function myUploadDir(){
    $myUploadDir = "./upload/";
    return $myUploadDir;
}

///////////////////////////////////////////////////////////////////////////
// if you want to imposs file type rescriction then
// $check_file_extentions=1  or if you want to off this option then
// $check_file_extentions=0
///////////////////////////////////////////////////////////////////////////

$check_file_extentions = 1;

////////////////////////////////////////////////////////////////////////////
// set your allowed type in the array.
// examples: zip, pdf, psd and many more.
//
////////////////////////////////////////////////////////////////////////////
function allowedfiles()
                {
                        $allowed_file_extensions=array("png","jpg","gif","bmp");
                        return $allowed_file_extensions;
                }

///////////////////////////////////////////////////////////////////////////
// if you want to delete all the files in uploaded directory
// $clear_folder_before_upload = 1  or if you want to off this option then
// $clear_folder_before_upload = 0
///////////////////////////////////////////////////////////////////////////

$clear_folder_before_upload = 1;

function EmptyDir($dir) {

    $handle=opendir($dir);
    while (($file = readdir($handle))!==false) {
    @unlink($dir.'/'.$file);
    }
    closedir($handle);
}

?>


Any help would be greatly appreciated!
 

I have a PHP script that uploads images to a folder on my server (attachments folder). Currently the folder sits within my webroot and is publicly accessible (I have to use chmod 777 due to permissions issue). So, I created the "attachments" folder outside of my webroot (so that it is not publicly accessible), but I do not know how to set the path in the PHP code to upload it to that "attachments" folder outside of the webroot. As you see in the snippet of PHP code below, the code currently uploads the the "attachments" folder within the www (webroot) directory. How do I make it upload to the "attachments" folder OUTSIDE of the www (webroot) directory?

foreach($files[$form] as $file){
         $str = $file[1];
         if (eval("if($str){return true;}")) {
            $_values[$file[0]] = $_FILES[$file[0]]["name"];
            $dirs = explode("/","attachments//");
            $cur_dir =".";
            foreach($dirs as $dir){
            $cur_dir = $cur_dir."/".$dir;
            if (!@opendir($cur_dir)) { mkdir($cur_dir, 0777);}}
            $_values[$file[0]."_real-name"] = "attachments/".date("YmdHis")."_".$_FILES[$file[0]]["name"]."_secure";
            copy($_FILES[$file[0]]["tmp_name"],$_values[$file[0]."_real-name"]);
            @unlink($_FILES[$file[0]]["tmp_name"]);
         }else{
          $flag=true;
          if ($_isdisplay) {
              //$ExtFltr = $file[2];
              //$FileSize = $file[4];
            if (!eval("if($file[2]){return true;}")){echo $file[3];}
            if (!eval("if($file[4]){return true;}")){echo $file[5];}
              $_ErrorList[] = $file[0];
          }
         }
      }

Hello freaks!

Im new to this forum, but im not all that new to PHP and MySQL.
Although there's been some years since the last time I used it, so don't go all freaky on me if I dont do this right

Let's go on-topic:
Im in progress of making an internal web-page for me and my colleagues to make things a bit easier for us.
I am making an database of our different projects, and I need some help with the input form - as I need to upload an image to the server, and store the path in the MySQL database.

In my input form, I need to store information from text fields, and I need to upload an image to the server and store the path in the database.
Before I can even start to code this (although I have coded the input forum without the upload), I need to know what would be the best way to do this.
I guess there are several ways.. What would the expert do (That's you right?)?

Should I have the information input, and image upload in the same form, or should I make a second form (maybe on a different page) for the upload?
Is it necessary with two tables, one for the info and one for the image path, and then tie them together with the imageID, or is it fine to use just one table?

Any thoughts would be appreciated!

<!-- TechThat -->

I need help making the uploaded image file name, that's chosen to be uploaded, be displayed on the html page with the path /upload/ added to the beginning of the displayed file name like so:
../upload/test.png
Any help/improvements will be appreciated.

<html>
 <head>
  <title>PHP Test</title>
 </head>
  <body>
 <?php
 if ($form_submitted == 'yes') {
 $allowedExts = array("gif", "jpeg", "jpg", "png");
 $temp = explode(".", $_FILES["file"]["name"]);
 $extension = strtolower( end($temp) );
 if (  $_FILES["file"]["size"] < 200000
       && in_array($extension, $allowedExts) )
 {
     if ($_FILES["file"]["error"]!= 0)
     {
         echo "Return Code: " . $_FILES["file"]["error"] . "<br>";
     } else {
         $length = 20;
         move_uploaded_file($_FILES["file"]["tmp_name"], "upload/" . $newfilename );
         $file_location = '<a href="http://../upload/' . $newfilename . '">' . $newfilename . '</a>';
     }
 } else {
   echo "Invalid upload file";
 }
?>
<label for="file">Filename:</label>
<input type="file" name="file" id="file">
</body>
</html>

Hi guys,

Happy New Year

http://www.phpfreaks.com/forums/php-coding-help/?action=post

How can i get the path

http://www.phpfreaks.com/forums


thanks,

Directory structu

/home/~test/public_html/soap/
    .../xml/
    .../bin/
    .../products/browser/

In file /home/~test/public_html/soap/bin/products/browser/test.php, I include another file and call it's function:
Code: [Select]
require("../../xml/processXML.php");

$xmlData = generateXML(....);

In the /home/~test/public_html/soap/xml/processXML.php, I have the following code:
Code: [Select]
function generateXML()
{
     exec("../bin/code.cur");
}

The problem is the exec("../bin/code.cur") fails of the relative path issue.  I did a getcwd() before the exec() and it returned /home/~test/public_html/soap/bin/products/browser.  I have tried using dirname() and no help.  I hope this is clear enought to understand where I am coming from.

Thank you.
AM

hey guys im still having a issue with using the root path when requiring external files. So i can use one path and never have to worry about this issue.

require("/functions/function_battle.php");
the file is located here
Code: [Select]
C:\Software\XAMPP\xampp\htdocs\System_Lords\functions\function_battle.php
i dont understand (include_path='.;C:\Software\XAMPP\xampp\php\PEAR')
im suppose to be including the php folder or something?


Code: [Select]
Fatal error: require() [function.require]: Failed opening required '/functions/function_battle.php'
 (include_path='.;C:\Software\XAMPP\xampp\php\PEAR') in
 C:\Software\XAMPP\xampp\htdocs\System_Lords\include\battle.php on line 2

Hi, I am trying to setup cakephp cake console in win7 and run cake bake as in the link below.I set up the env variable in win7 as below with a php and cakeapp entry in this Path.I have easyPHP 12.1 and php works fine. I get an error with php in failing to load. dynamic library but the php dir is in the path below so I have no idea . I cant set my php env var is my problem . The php.ini is in this dir specified below.   http://www.nurelm.co...ols-in-windows/   E:\Program Files\EasyPHP-12.1\php\php546x121019214357\;E:\AA-website design\acl\app\Console\;    

Hi,
I want to get root path and use on links even if i'm in sub to sub folder etc...
suppose my site name is fitness.com and having two subfolder. so url will become

http://www.fitness.com/dir/dir_sub

here i want to put a link to go to root directory file. there are different ways to do that. e.g:

Code: [Select]
<a href='../../filename.php'>go to that page</a>

but i want to use some constant that always shows to root directory. as i defined here

Code: [Select]
define("SERVER_NAME" , $_SERVER['HTTP_HOST']);
//and used it like this
<a href='<?php echo SERVER_NAME;  ?>/filename.php'>go to that page</a>

it works when we're on root directory. but when we go to sub directory it added sub directory name with it. which i don't want.

is there any way?

Thanks

Hi,

I have a dynamic variable like this:

/def/g/qaz/pol/cxz/cba/abc

I only wish to keep this:

/def/g/qaz/pol/cxz

How would I do this?

Thanks,

- mme

Hello,
I would like to know how to include a root file in a 4 deep folder using dots.
Is it good : include "../../../config.php" or i can simple include "./config.php" ?
How does this works? Is it better to write all the path?
Thank you

Hey guys! I just joined this forum today, and I'm wondering about something...

I see a lot of sites, and in the code of the site, style sheets and javascript includes are references in a manner similar to this...

"href='http://www.somesite.com/css/css.php/someDirectory/style.css'"

How is this done? The path following the "css.php" is what confuses me.  Thanks for any information you have!

In NetBeans - in what I think is the Web Root - I have...

config.inc.php
secure/checkout.php

In "checkout.php", when I use this code it works...

Code: [Select]
<?php require_once "../config.inc.php"; ?>
but when I try to use an Absolute Path, it errors out...

Code: [Select]
<?php require_once "/config.inc.php"; ?>

I thought "/" moved you up to the Web Root and then you drill down from there?

So the second example should also work in my mind?!

What is wrong?



Debbie

I am trying to find the URL to a directory TWO levels above the script level (if you know what I mean).

For example.. the script sits at
Code: [Select]
http://www.domain.com/TestArea2/members/index.php
BUT I need to find the path to 
Code: [Select]
http://www.domain.com/TestArea2/

the nearest I get is...


$domain = $_SERVER['HTTP_HOST'];
$domain .=  $_SERVER['REQUEST_URI'];
echo $domain; // outputs www.domain.com/TestArea2/members/index.php 



Is there a simple method to do this or have I got to split the string and work backwards?

Many thanks