Display image Problem

[Skipjackrick]

Hi again,

 

In my database, (when I upload images through an html form) my URL's to images are stored in the following fashion:

 

/home/ectackle/public_html/images/image-name.jpg

 

 

Okay, now I am trying to create a page that allows the user to view the image they uploaded. 

 

Using the following code, everything works great except the image won't display.

 

When I check the URL to the image it shows this

 

http://www.ectackle.com/home/ectackle/public_html/images/image-name.jpg

 

If I could just get rid of the text in red.  I'd be golden.

 

 

Is there an alternate way to display my image or a way to make the /home/ectackle/public_html/images/image-name.jpg format work to display my image????????

 

<?php
$query_image = "SELECT filename, team_id, MAX(date)
	FROM image 
	GROUP BY team_id
	ORDER BY MAX(date)DESC
	LIMIT 1"; 

$image_result = mysql_query($query_image) or die(mysql_error());


while($row = mysql_fetch_array($image_result))
{
$imagename = $row['filename'];
$imagedate = $row['MAX(date)'];


$display_image .=<<<EOD
<h2><div align="center"><img src="$imagename"></div></h2>
EOD;
}
print $display_image;
?>

[BlueSkyIS]

only store the image name, not the full path. tack the path on as needed.

[Skipjackrick]

only store the image name, not the full path. tack the path on as needed.

 

Sounds like an easy fix.

I grabbed the upload form code from someone else.  I can't find the image name in this code.  Its a bit over my head.

 

 

Can you find the variable that would be the filename only?

 

Would it happen to be this..............      $now.'-'.$_FILES[$fieldname]['name']

 

<?php  

// filename: upload.processor.php

// first let's set some variables

// make a note of the current working directory, relative to root.
$directory_self = str_replace(basename($_SERVER['PHP_SELF']), '', $_SERVER['PHP_SELF']);

// make a note of the directory that will recieve the uploaded files
$uploadsDirectory = $_SERVER['DOCUMENT_ROOT'] . $directory_self . 'uploaded_files/';

// make a note of the location of the upload form in case we need it
$uploadForm = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'upload.form.php';

// make a note of the location of the success page
$uploadSuccess = 'http://' . $_SERVER['HTTP_HOST'] . $directory_self . 'index.php';

// name of the fieldname used for the file in the HTML form
$fieldname = 'file';



// Now let's deal with the upload

// possible PHP upload errors
$errors = array(1 => 'php.ini max file size exceeded', 
                2 => 'html form max file size exceeded', 
                3 => 'file upload was only partial', 
                4 => 'no file was attached');

// check the upload form was actually submitted else print form
isset($_POST['submit'])
or error('the upload form is neaded', $uploadForm);

// check for standard uploading errors
($_FILES[$fieldname]['error'] == 0)
or error($errors[$_FILES[$fieldname]['error']], $uploadForm);

// check that the file we are working on really was an HTTP upload
@is_uploaded_file($_FILES[$fieldname]['tmp_name'])
or error('not an HTTP upload', $uploadForm);

// validation... since this is an image upload script we 
// should run a check to make sure the upload is an image
@getimagesize($_FILES[$fieldname]['tmp_name'])
or error('only image uploads are allowed', $uploadForm);

// make a unique filename for the uploaded file and check it is 
// not taken... if it is keep trying until we find a vacant one
$now = time();
while(file_exists($uploadFilename = $uploadsDirectory.$now.'-'.$_FILES[$fieldname]['name']))
{
$now++;
}

// now let's move the file to its final and allocate it with the new filename
@move_uploaded_file($_FILES[$fieldname]['tmp_name'], $uploadFilename)
or error('receiving directory insuffiecient permission', $uploadForm);

// If you got this far, everything has worked and the file has been successfully saved.
// We are now going to redirect the client to the success page.
header('Location: ' . $uploadSuccess);

[BlueSkyIS]

yes, you've got it.

[Skipjackrick]

yes, you've got it.

 

Hmmm...I got a syntax error from it.  Am I putting the value in correctly?

 

You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'name']','22Feb08','1')' at line 1

 

$q = "INSERT INTO image (image_id,filename,date,team_id) VALUES ('NULL','now.'-'.$_FILES[$fieldname]['name']','".date(dMy)."','1')";
$s = mysql_query($q, $conn);

[BlueSkyIS]

i'd build the filename string outside the sql first:

$filename = $now.'-'.$_FILES[$fieldname]['name'];

$q = "INSERT INTO image (image_id,filename,date,team_id) VALUES ('NULL','$filename','".date(dMy)."','1')";

 

[Skipjackrick]

i'd build the filename string outside the sql first:

$filename = $now.'-'.$_FILES[$fieldname]['name'];

$q = "INSERT INTO image (image_id,filename,date,team_id) VALUES ('NULL','$filename','".date(dMy)."','1')";

 

 

DOH!  You're smart.  Thanks!

[Skipjackrick]

Darn it!!

 

It gave me two entries into the database.

 

The first entry was $now

 

and the second entry was -name.jpg

 

 

$filename = $now.'-'.$_FILES[$fieldname]['name'];

$q = "INSERT INTO image (image_id,filename,date,team_id) VALUES ('NULL','$filename','".date(dMy)."','1')";

[Skipjackrick]

AHAHAHA!!

 

Nevermind.  I found my error.  I need to review my script before posting.  I could have saved precious time!!!!!!!!

 

 

THANKS TO ALL!!!!!!!