Display image path error

[jpknoob]

Hi, I decided to try and upload images by storing their path in a database, but have run into an error when displaying the images. The form works and the folder is populated whenever I run the script. I have tried a lot of solutions from similar threads, but to no avail. My view.php is;

<?php
include 'connect.php';
$query = "SELECT * FROM upload2 ";
$result = mysql_query($query) or die('Error, query failed');
$num = mysql_fetch_array($result);
$path = $row['path'];

if ($num > 0){
	echo "<p>There is an image in here!;</p>";
						
	while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
		echo "<img src='$path' />";
	}								
}

?>

Can anyone help?

[diafol]

> The form works and the folder is populated whenever I run the script

What form? Which folder?

Look at view source in the browser and get the src data.

Does it refer to the correct folder? Sounds unlikely.

[jpknoob]

Are you saying my view code is right?

When i use my upload from and it echoes "folder updated", the image appears in the 'upload' directory that i set it to.

I will check the path in browser

[jpknoob]

Ok, so I typed the path into the browser and get nothing returned. My upload script is;

<?php
$uploadDir = './upload/';

if(isset($_POST['upload']))
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];

$filePath = $uploadDir . $fileName;

$result = move_uploaded_file($tmpName, $filePath);
if (!$result) {
echo "Error uploading file";
exit;
}

include 'connect.php';

if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
$filePath = addslashes($filePath);
}

$query = "INSERT INTO upload2 (name, size, type, path ) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$filePath')";

mysql_query($query) or die('Error, query failed : ' . mysql_error());

mysql_close($connect);

echo "<br>Files uploaded<br>";
echo "<a href='form.html'>upload another?</a>";

}
?>

If i type in www.mysite.com/upload/image1.jpg i can see the image. Can anyone offer any advice?

[diafol]

Check:

1) DB is updated
2) path for display in src

>If i type in www.mysite.com/upload/image1.jpg i can see the image. Can anyone offer any advice?

Your src should read "/upload/image1.jpg" or "upload/image1.jpg" if you are accessing from public root. If not, your db filepath data may be wrong.

[jpknoob]

Ardav, thanks for taking the time to help me out with this problem.

The form update to the database fine, the path uploaded is "./upload/image1.jpg", its clear that the '.' is causing the issue, however, if I remove that from my upload script, I get an error and nothing is uploaded. This is the first time i have attempted to do this process and am a bit lost

[diafol]

<?php
[B]$uploadDir = '/upload/';[/B]

if(isset($_POST['upload']))
{
$fileName = $_FILES['userfile']['name'];
$tmpName = $_FILES['userfile']['tmp_name'];
$fileSize = $_FILES['userfile']['size'];
$fileType = $_FILES['userfile']['type'];
 
$filePath = $uploadDir . $fileName;
 
[B]$result = move_uploaded_file($tmpName, "." . $filePath);[/B]
if (!$result) {
echo "Error uploading file";
exit;
}
 
include 'connect.php';
 
if(!get_magic_quotes_gpc())
{
$fileName = addslashes($fileName);
$filePath = addslashes($filePath);
}
 
$query = "INSERT INTO upload2 (name, size, type, path ) ".
"VALUES ('$fileName', '$fileSize', '$fileType', '$filePath')";
 
mysql_query($query) or die('Error, query failed : ' . mysql_error());
 
mysql_close($connect);
 
echo "<br>Files uploaded<br>";
echo "<a href='form.html'>upload another?</a>";
 
}
?>

OK I tried to keep your code intact, so bold lines for changes. Hope it works.

[pzuurveen]

In your first script
where is the array

$row['path'];

(line 6)

come from?

[jpknoob]

Ardav, your changes worked and the image uploaded without the '.',the image is going into the right folder and the file path is correct. When trying to view the image, i get a blank screen. I am currently trying to figure out why.

$row['path'];

Is from the table in the database. There are 5 columns; id, name, type, size and path.

[jpknoob]

@pzuurveen i see what you mean, i have edited the code from array to rows;

<?php
include 'connect.php';
$query = "SELECT * FROM upload2 ";
$result = mysql_query($query) or die('Error, query failed');
$num = mysql_num_rows($result);
$path = $row['path'];

if ($num > 0){
echo "<p>There is an image in here!;</p>";
						
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
	echo $row['path']; 
	echo "<img src='$path' />";
								}								
							}

?>

I'm echoed the path to ensure it was correct

[pzuurveen]

at line 6

Row['path']

still doesn't exist
you create it at line 11

#while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){

when php has an error like that u often get a blank screen
to show errors: start your script with

error_reporting(E_ALL);

[jpknoob]

Thank you thank you pzuurveen! That little bit of script was all I needed to debug the issue. I wasn't aware of that before. For those that are interested, my completed script is;

include 'connect.php';
error_reporting(E_ALL);
$query = "SELECT * FROM upload2 ";
$result = mysql_query($query) or die('Error, query failed');
$num = mysql_num_rows($result);

if ($num > 0){
echo "<p>There is an image in here!;</p>";
						
while ($row = mysql_fetch_array($result, MYSQL_ASSOC)){
echo "<img src='$row[path]' />";
}							
}

You will have to include you connection details mind!

Thanks again for the help and advice.