unexpected T_VARIABLE

illusive101 · May 23, 2009

Hey guys, trying to load some images from a database and kind of new to the whole PHP / MYSQL thing.

Here's a code snippet and I keep getting the unexpected T_VARIABLE error on page load.


    <?php
	include 'connect.php'

	   $self = $_SERVER['PHP_SELF'];

			$query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo “<img src=\”viewimage.php?id=$row[id]\” width=\”55\” height=\”55\” /> <br/>”;
                }

                mysql_close($dbconn);
?>

Any help is appreciated, thanks guys.

MadTechie · May 23, 2009

try this

    <?php
      include 'connect.php'
      
         $self = $_SERVER['PHP_SELF'];

            $query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die('Error, query failed');

                while($row = mysql_fetch_array($result)){
                           echo "<img src=\"viewimage.php?id=$row['id']\" width="55" height="55" /> <br/>";
                }

                mysql_close($dbconn);
   ?>

DarkSuperHero · May 23, 2009

include 'connect.php';

	   $self = $_SERVER['PHP_SELF'];

			$query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo “<img src=”viewimage.php?id=$row[id]” width=”55” height=”55” /> <br/>”;
                }

                mysql_close($dbconn);

wasnt he just missing the ; after his include?

illusive101 · May 23, 2009

Update ::

I used the method http://www.techcubetalk.com/2009/01/tutorial-on-how-to-store-images-in-mysql-blob-field/ found there to store my images and show them.

However, I got all the code in place but no images show? The database gets larger as I upload them so I know their uploading correct, however they are not displaying on the page I'd like them to. Anyone know why possibly?

MadTechie · May 23, 2009

view the page "viewimage.php?id=1" in your browser if no image appears double check your code for viewimage.php

illusive101 · May 23, 2009

The requested URL /viewimage.php was not found on this server.

illusive101 · May 23, 2009

<?php
	include 'connect.php';

	   $self = $_SERVER['PHP_SELF'];

			$query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo “<img src=\”viewimage.php?id=$row[id]\” width=\”55\” height=\”55\” /> <br/>”;
                }

                mysql_close($dbconn);
?>

There's the code again, I don't see anything wrong :S The connect.php is just fine too.

illusive101 · May 23, 2009

Okay, I got a little bit further.

The viewimage.php loads, but it's completely white -- there's no error nor is there an image.

...?

illusive101 · May 23, 2009

viewimage.php?id=1 brings up a blank image also, which is weird because it's in the database :S

Ken2k7 · May 23, 2009

Why do the quotes look funny in the echo? If this doesn't work, check your PHP file. Is it sending out the correct data headers?

<?php
      include 'connect.php';
      
         $self = $_SERVER['PHP_SELF'];

            $query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo '<img src="viewimage.php?id='.$row['id'].'" width="55" height="55" /> <br/>';
                }

                mysql_close($dbconn);
?>

illusive101 · May 23, 2009

Viewimages.php

<?
if(isset($_REQUEST['id']))
{
             // get the file with the id from database
			include 'connect.php';

			$id    = $_REQUEST['id'];
			$query = "SELECT 'img_name', 'img_type', 'img_size', 'img_data' FROM img_tbl WHERE id = '$id'";

			$result = mysql_query($query) or die(mysql_error());
			list($name, $type, $size, $content) = mysql_fetch_array($result);

			header("Content-length: $fileSize");
			header("Content-type: $fileType");
			print $content;

}
?>

Showimages.php

<?php
	include 'connect.php';

	   $self = $_SERVER['PHP_SELF'];

			$query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo “<img src=\”viewimage.php?id=$row[id]\” width=\”55\” height=\”55\” /> <br/>”;
                }

                mysql_close($dbconn);
?>

kickstart · May 23, 2009

Hi

Think the variables highlighted should match:-

if(isset($_REQUEST['id']))

{

// get the file with the id from database

include 'connect.php';

$id = $_REQUEST['id'];

$query = "SELECT 'img_name', 'img_type', 'img_size', 'img_data' FROM img_tbl WHERE id = '$id'";

$result = mysql_query($query) or die(mysql_error());

list($name, $type, $size, $content) = mysql_fetch_array($result);

header("Content-length: $fileSize");

header("Content-type: $fileType");

print $content;

}

?>

All the best

Keith

illusive101 · May 23, 2009

After changing them, still nothing has happened.

The table's name is img_tbl and the fields are id , img_name, img_type, img_size, img_data.

Would anyone be able to find me a working script to load images from this table? (Or figure out why my current one doesn't work?)

Thanks!

MadTechie · May 23, 2009

can you post your latest code

Viewimages.php

and

Showimages.php

GingerRobot · May 23, 2009

Why do the quotes look funny in the echo?

They appear to be MS 'smart' quotes. Somewhere along the line, something got typed in Word.

In any case, this should really be over in php help

kickstart · May 23, 2009

Hi

Just noticed that the column names have quotes around them rather than back ticks.

if(isset($_REQUEST['id']))
{
             // get the file with the id from database
            include 'connect.php';

            $id    = $_REQUEST['id'];
            $query = "SELECT `img_name`, `img_type`, `img_size`, `img_data` FROM img_tbl WHERE id = '$id'";
            
            $result = mysql_query($query) or die(mysql_error());
            list($name, $type, $size, $content) = mysql_fetch_array($result);
            
            header("Content-length: $fileSize");
            header("Content-type: $fileType");
            print $content;
            
}
?>

All the best

Keith

illusive101 · May 23, 2009

can you post your latest code

Viewimages.php

and

Showimages.php

showimages.php

<?php
      include 'connect.php';
      
         $self = $_SERVER['PHP_SELF'];

            $query = "SELECT id, img_name, img_type, img_size, img_data FROM img_tbl ORDER BY id";

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo “<img src=”viewimage.php?id=$row[id]” width=”55” height=”55” /> <br/>”;
                }

                mysql_close($dbconn);
?>

viewimages.php

<?
if(isset($_REQUEST['id']))
{
             // get the file with the id from database
            include 'connect.php';

            $id    = $_REQUEST['id'];
            $query = "SELECT 'img_name', 'img_type', 'img_size', 'img_data' FROM img_tbl WHERE id = '$id'";
            
            $result = mysql_query($query) or die(mysql_error());
            list($name, $type, $size, $content) = mysql_fetch_array($result);
            
            header("Content-length: $size");
            header("Content-type: $type");
            print $content;
            
}
?>

Still can't figure this out, and sorry for posting in wrong section :-\

chaking · May 23, 2009

the viewimages.php file is using single quotes around the columns instead of back ticks

singlequote: ' ... go around the strings... '$id' ... not needed for integers

backtick: ` ... go around columns

illusive101 · May 23, 2009

Thanks guys for the heads up with the single quotes and back ticks, fixed the problem!

Now how would I go about loading these images onto a separate page automatically when I upload them? So when I upload it, it automatically appears on that page.

Thanks

kickstart · May 23, 2009

Hi

Not sure what you mean.

You already appear to have a script to generate a page to display the images.

All the best

Keith

illusive101 · May 23, 2009

Parse error: syntax error, unexpected '?'

I get that error for line 8, however here is line 8... I don't see a ?


    <?php
	include 'connect.php'


			$query = ("SELECT `id`, `img_name`, `img_type`, `img_size`, `img_data` FROM img_tbl ORDER BY id");

                $result = mysql_query($query) or die(’Error, query failed’);

                while($row = mysql_fetch_array($result)){
                           echo “<img src=\”viewimage.php?id=$row[id]\” width=\”55\” height=\”55\” /> <br/>”;
                }

                mysql_close($dbconn);
?>

Nevermind, now I'm getting a T_Variable error again on the $query line.

kickstart · May 23, 2009

Hi

You are missing the ; at the end of the include line above it.

Not sure about the ? issue. However as mentioned by Ken2k7 the quotes in a few places look rather strange, both around the error message ’Error, query failed’ and also in the echo statement. Especially on the echo statement where you have strange double quotes to delimit the string, but also contained within the string.A guess but not sure if that may be confusing php.

Try this:-

<?php
include 'connect.php'
$query = ("SELECT `id`, `img_name`, `img_type`, `img_size`, `img_data` FROM img_tbl ORDER BY id");
$result = mysql_query($query) or die('Error, query failed');
while($row = mysql_fetch_array($result)){
echo "<img src='viewimage.php?id=$row[id]' width='55' height='55' /> <br/>";
}
mysql_close($dbconn);
?>

All the best

Keith

thebadbad · May 23, 2009

@Keith - still missing a semicolon after the include.

This should work (*fingers crossed*):

<?php
include 'connect.php';
$query = "SELECT `id`, `img_name`, `img_type`, `img_size`, `img_data` FROM `img_tbl` ORDER BY `id`";
$result = mysql_query($query) or die('Error, query failed');
while ($row = mysql_fetch_array($result)) {
echo '<img src="viewimage.php?id=' . $row['id'] . '" width="55" height="55" /> <br />';
}
mysql_close($dbconn);
?>

illusive101 · May 23, 2009

Anyone know why my upload only accepts .gif's?

<label for="">Image:</label> <input type="file" name="imgfile"><br>

// Loading images into Database

                if(isset($_REQUEST[submit]) && $_FILES[imgfile][size] > 0)
                {
                                $fileName       = $_FILES[imgfile][name]; // image file name
                                $tmpName     = $_FILES[imgfile][tmp_name]; // name of the temporary stored file name
                                $fileSize           = $_FILES[imgfile][size]; // size of the uploaded file
                                $fileType         = $_FILES[imgfile][type]; // file type

                                $fp                    = fopen($tmpName, 'r'); // open a file handle of the temporary file
                                $imgContent  = fread($fp, filesize($tmpName)); // read the temp file
                                fclose($fp); // close the file handle

                                $query = "INSERT INTO img_tbl (`img_name`, `img_type`, `img_size`, `img_data` )
                                                                 VALUES ('$fileName', '$fileType', '$fileSize', '$imgContent')";

                                mysql_query($query) or die('Error, query failed');
                                $imgid = mysql_insert_id(); // autoincrement id of the uploaded entry

                    echo "<br>Image successfully uploaded to database<br>";

                }else die('You have not selected any image');

// End loading image

fantomel · May 23, 2009

i've read the whole thread.. i'm gonna try to fix your code or look into it but since you are using mysql to store images my first question if would work with images i would ask somewhere or here on phpfreaks what would be the best solution for storing images for a gallery or something similar. and everyone would answer me the same thing a folder on hdd no database

in the end my advice is use folders to store images and mysql to store their for outputting them on your site.

P.S i think everyone will agreed with me

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