SOLVED: Display image stored in MySQL

bluefireco · January 26 '10, 03:51 AM (#1)

Hi everyone,

I'm having a slight problem with displaying an image stored in my mysql database. The data is being stored ok, I tested this by outputting the image data to a file. When I went to open the file, there was my image. However, when I go to display the image in the browser I get problems. The following is the code in my image loader file...

PHP Code:


			
<?php
//mysql connect & retrieve stuff would go here

header("Content-Type: image/jpg"); // The image I am loading is in fact a .jpg

//$fp=fopen("test_image.jpg","w+");
//fwrite($fp,$image_data); // this resulting image appears fine
//fclose($fp);

echo $image_data; // this one does not
?>

Any ideas what I am doing wrong? Any help is appreciated.

**mlseim** · January 26 '10, 10:00 AM (#2)

If you're storing your images in MySQL,
what is the fwrite portion for? I'm confused.

bluefireco · January 26 '10, 01:03 PM (#3)

It was the test to make sure the data was being stored properly, which it is. Which is also why it is commented out. I guess I shouldn't have included it.

**mlseim** · January 26 '10, 01:09 PM (#4)

Show us the script then, where you're displaying the JPEG from your MySQL database.

EDIT:
And I assume your image variable is a BLOB?

.

bluefireco · January 26 '10, 01:27 PM (#5)

The code I posted above is most of it, but here's the entire thing. Yes, it is stored as blob, again, the data outputs to a file perfectly. It has problems being output in the browser.

PHP Code:


			

<?php

$mysql_con = mysql_connect("127.0.0.1","root","***")
mysql_select_db("pic_db");

$result = mysql_query("SELECT * FROM pics where id = ".$_GET['id']." LIMIT 1");
$data = mysql_fetch_row($result);


// This tells the browser that the content should be a jpg image...
header("Content-Type: image/jpg");

// This is where the data is being given to the browser
echo $data[2];

?>

**mlseim** · January 26 '10, 01:39 PM (#6)

Let's say that script you show is called "image.php" ...

On another page, you do this:

<html>
<body>
<img src="image.php?id=235" />
</body>
</html>

You're saying that doesn't work?

smoseley · January 26 '10, 01:52 PM (#7)

It should work anyway.

Make sure you have no white space (spaces or return carriages) outside your <? PHP ?> block. That could corrupt the output.

Also, isn't the correct mime type JPEG (with an "E")? so it shoudl be header("Content-Type: image/jpeg");

smoseley · January 26 '10, 01:53 PM (#8)

If the above doesn't work, try adding a Content-Length header (just do a strlen() on the image data to get that value).

**mlseim** · January 26 '10, 02:44 PM (#9)

Steven ...

I missed this one: image/jpeg (that might be it).

bluefireco · January 26 '10, 07:43 PM (#10)

Tried those with no luck. I think it may have to do with the way the text is formatted when it shows up on the page. When I just display the raw data, all the new lines disappear. Is there any way to keep that from happening?

On a side note, the reason I had "content-type: image/jpg" is because when I checked the header information for other images (on my own site) it was image/[file ext] depending on the file type.

**mlseim** · January 26 '10, 08:35 PM (#11)

Show us the actual page ... where the image is supposed to appear.
And also the PHP scripting that creates that page?

bluefireco · January 26 '10, 11:25 PM (#12)

Here's a link to the page,

http://clan.bluefirecode.com/system/..._pic.php?id=28

The script that creates the page is almost exactly what's in my third post. All the page is doing is loading an image to be called later. No html or anything, it's simply an image.

bluefireco · January 27 '10, 12:20 AM (#13)

Ok, I've given up on this storage strategy. Instead I'm going to store references to the image files (as means of security). I've sniffed around the internet for hours now and apparently I'm not the only one with this problem (and still no solution).

Thanks for your help anyway.

Dorky · January 27 '10, 04:12 AM (#14)

ok dude. your not actually echoing the image like this

Quote:

$image_data

your just trying to echo the data. this can be done but not with echo. instead use this
http://www.php.net/manual/en/function.image2wbmp.php
if you want to use echo then you will need to echo the image in an image tag and src= to the image file you wrote to.

bluefireco · January 27 '10, 08:45 AM (#15)

Yea, I suppose GD would work, however that's not installed on my server (right now anyway). The problem is that the way I was doing it should work, you don't need the img tag or anything (and believe me I tried that at one point). The idea is that you tell the browser that it's about to display an image, using the headers, and then just output the image data and it should display the image (the equivalent of right-clicking on an image and hitting 'View Image').

smoseley · January 27 '10, 09:42 AM (#16)

Echo works just fine... you're storing the image data as a BLOB in the DB, right? You should be able to just echo it to the page.

GD is used to actually CREATE an image, or to alter an existing image.

smoseley · January 27 '10, 09:48 AM (#17)

There's some funky stuff going on in that image. Can you attach the original so I can compare?

smoseley · January 27 '10, 09:52 AM (#18)

Can you do me a favor and post the script you're using to insert the image in the DB? I think that may be where the problem is.

**mlseim** · January 27 '10, 10:22 AM (#19)

And also give us the listing for this script: get_pic.php

scratch-out any MySQL login stuff.

Dorky · January 27 '10, 11:06 AM (#20)

or imagejpg(); i have had the same issue, like you said, as have others. i just think there are too many other ways to do it then butting you head on a buggy approach.

Dorky · January 27 '10, 11:36 AM (#21)

are the other images on the page or is this being tested as standalone?

bluefireco · January 27 '10, 06:35 PM (#22)

Link to image we're trying to display:
http://clan.bluefirecode.com/system/...aded_image.jpg
Link to the image loader page (get_pic.php?id=28):
http://clan.bluefirecode.com/system/..._pic.php?id=28
...and it's contents:

PHP Code:


			
<?php

$mysql_con = mysql_connect("127.0.0.1","root","***")
mysql_select_db("pic_db");

$result = mysql_query("SELECT * FROM mypic_pics WHERE id = ".$_GET['id']." LIMIT 1");
$data = mysql_fetch_row($result);

$image = $data[2]; // data[2] is the image contents

header("Content-Length: ".strlen($image));
header("Content-Type: image/".$data[3]); //data[3] is the file type (jpg/jpeg/png/gif)

echo $image;
?>

here is the upload script...

PHP Code:


			
// these two lines are specifically for finding the file extension if the name has more than one '.' in it.
$name_array = explode(".",$_FILES['picture_upload']['name']);
$last_index = sizeof($name_array) - 1; 

$image_data = file_get_contents($_FILES['picture_upload']['tmp_name']);    
$image_data = mysql_real_escape_string($image_data);

$query = "INSERT INTO mypic_pics (user,data,ext,caption,password,album,public) VALUES ('admin','";
$query .= $image_data."','".$name_array[$last_index]."','A test caption.',' ','0','0')";

$result = mysql_query($query);

My image data is being stored as a long blog, which I can't imagine is the problem.

Again, I don't believe that the storage is the problem because I can pull the data out of mysql and write it to a normal jpg file just fine; the image appears without error. The error occurs when I try to give the data directly to the browser.

If you want to try your own upload, I'm having the script tell you your id (it wil appear right above 'home'... here is the link to the page:

http://clan.bluefirecode.com/page/?page=home

**mlseim** · January 27 '10, 07:48 PM (#23)

How do you know that $data[2] is the image contents?
$image = $data[2]; // data[2] is the image contents

I would expect it be like this ....

Say for instance, that the name of your MySQL image field is 'data' ...

$image=$data['data'];
$extension=$data['ext'];

You didn't give us the names of your columns, but I think you should reference the name of the column.

EDIT ... yes, you showed us the names .. use 'data' instead of 2.
also the other ones.

bluefireco · January 27 '10, 10:34 PM (#24)

Quote:

Originally Posted by mlseim

How do you know that $data[2] is the image contents?
$image = $data[2]; // data[2] is the image contents

I would expect it be like this ....

Say for instance, that the name of your MySQL image field is 'data' ...

$image=$data['data'];
$extension=$data['ext'];

Yea, sorry, id is set to auto-increment and was not included in sql script. Here are all the columns...

id,user,data,ext,caption,password,album,public

Rest assured that $data[2] is in fact the data (I've checked it's output to a file). But thanks for the advice for naming, probably much easier than using the numbers.

**mlseim** · January 27 '10, 11:05 PM (#25)

What happens if you do this (manually force a jpeg content-type) ... since I have no way to test it myself ...

save your original in a safe place and try this in it's place ...

PHP Code:


			
<?php



$mysql_con = mysql_connect("127.0.0.1","root","***")

mysql_select_db("pic_db");



$result = mysql_query("SELECT * FROM mypic_pics WHERE id = ".$_GET['id']." LIMIT 1");

$data = mysql_fetch_row($result);



$image = $data['data'];



header("Content-Type: image/jpeg");



echo $image;

?>