Displaying Binary BLOB - OLE Object

sgtpepper · May 14, 2003

I am trying to display a binary blob from a mysql table. The blob was saved as a OLE Object. My database was originally a mssql7 database that I imported into a mysql database.

It is stored in a field called 'medium_photo'

How can I display the image? You will notice I attempted to display it in my code below...to no avail..

Thank you for any help you can give....

My current code is as follows....

mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

// Below will display all listings///
//$query="SELECT * FROM listingsDB WHERE Title like '%'";

// Below is an example on how to setup multiple searches for different fields//

$query="SELECT * FROM residential WHERE town like '%$town_listing%'";

//$query="SELECT * FROM listingsDB WHERE Title like '%$title%' AND user_ID = '$user'";

//$query="SELECT * FROM listingsDB WHERE Title like '%$search%'";

$result=mysql_query($query) or die (mysql_error());
$num=mysql_numrows($result);
?>

<?
if(mysql_numrows($result) == '0') {
echo "Sorry, no results were found...";
} else {
// display results here
}
?>

<?
while ($row = mysql_fetch_array($result)) // this will grab the results from query
{
?>
<table border="1" width="100%">
<tr>
<td width="25%"><b><? echo $row['town']; ?></b></td>
<td width="25%">Listing created on:<? echo $row['user_ID']; ?></td>
<td width="25%"><? echo $row['year_built']; ?></td>
</tr>
<tr>
<td width="25%"><? echo $row['medium_photo']; ?></td>
<td width="25%"><? echo $row['lister_office_name']; ?></td>
<td width="25%"> </td>
</tr>
<tr>
<td width="25%"> </td>
<td width="25%"> </td>
<td width="25%"> </td>
</tr>
</table>

<?
}
?>
<?

goa103 · May 14, 2003

Hi,

OLE object... I don't know how these objects are stored but you may try this:

$row['medium_photo'] -> <img src="medium_photo.php?photo_id=<?= $row['id'] ?>">

$row['id']: primary key of your town table

medium_photo.php:
Get $row['medium_photo'] from $_POST ['photo_id']
header ("Content-type: image/jpeg");
echo ($row['medium_photo']);

It will work if your OLE object is stored as a JPEG... What the OLE object look like ?

JM

sgtpepper · May 14, 2003

Where should

$row['medium_photo'] -> <img src="medium_photo.php?photo_id=<?= $row['id'] ?>">

$row['id']: primary key of your town table

Be placed?

No matter where I place it I get the following error

Parse error: parse error, expecting T_STRING' orT_VARIABLE' or '{'' or'$'' in /home/httpd/virtual/century21highview.com/htdocs/century21/test.php on line 43

sgtpepper · May 14, 2003

The files are JPG's 320x208 stored as BLOB's

sgtpepper · May 14, 2003

I am able to get the first part of the code to display broken images.

medium_photo.php?photo_id=430729

I do not think I have medium_photo.php set correctly. Can you give me a little more detail on how to setup this file?

Thank you for the help.

sgtpepper · May 14, 2003

Hmmmm....I am still having issues with this. I am pulling out my hair...what little I have left! :p

So someone look at this code and offer me a suggestion on how to get my binary blob images to show?

I would appreciate it!

test.php (Main Script)

<?
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

// Below will display all listings///
//$query="SELECT * FROM listingsDB WHERE Title like '%'";

// Below is an example on how to setup multiple searches for different fields//

$query="SELECT * FROM residential WHERE town like '%$town_listing%'";

//$query="SELECT * FROM listingsDB WHERE Title like '%$title%' AND user_ID = '$user'";

//$query="SELECT * FROM listingsDB WHERE Title like '%$search%'";

$result=mysql_query($query) or die (mysql_error());
$num=mysql_numrows($result);
?>

<?
if(mysql_numrows($result) == '0') {
echo "Sorry, no results were found...";
} else {
// display results here
}
?>

<?
while ($row = mysql_fetch_array($result)) // this will grab the results from query
{
?>
<table border="1" width="100%">
<tr>
<td width="25%"><b><? echo $row['town']; ?></b></td>
<td width="25%">Listing created on:<? echo $row['user_ID']; ?></td>
<td width="25%"><? echo $row['year_built']; ?></td>
</tr>
<tr>
<td width="25%">

<? echo $row['medium_photo'];?>
</td>
<td width="25%"><? echo $row['lister_agent_name']; ?></td>
<td width="25%"> </td>
</tr>
<tr>
<td width="25%"> </td>
<td width="25%"> </td>
<td width="25%"> </td>
</tr>
</table>
<?
}
?>

medium_photo.php (photo processing script)

<?
mysql_connect(localhost,$username,$password);
@mysql_select_db($database) or die( "Unable to select database");

$query="SELECT * FROM residential WHERE town like '%$town_listing%'";

$result=mysql_query($query) or die (mysql_error());
$num=mysql_numrows($result);
?>

<?
if(mysql_numrows($result) == '0') {
echo "Sorry, no results were found...";
} else {
// display results here
}
?>

<?
while ($row = mysql_fetch_array($result)) // this will grab the results from query
{
?>

<?
Get $row['medium_photo'] from $_POST ['photo_id']
header ("Content-type: image/jpeg");
echo ($row['medium_photo']);
?>
<?
}
?>
<?

BeastRider · May 14, 2003

What is a "broken image".?

medium_photo.php must display absolutely nothing before the content header is displayed. Thus, there can be no spaces, space lines, nada before the first line of php code. Otherwise you get a text content header generated and the image will not work. Ditto for the lines following the php code... you add garbage to the image. In your example, you close the php code "?>" insert a blank line, then start php again "<?" before you issue the content header.

I use output buffering in my image output routine which solved all the problems for me.

getthumb.php

<?php
ob_start(); // turn output buffering on
// open the database and retrieve the record
(!($dbh=mysql_connect ("localhost",
"USERNAME",
"USERPASS")));
(!mysql_select_db("DBNAME",$dbh));
$query = "SELECT thumbnail FROM thumbs
WHERE thumbid = $thumb";
(!($result = @ mysql_query ($query, $dbh)));
$data = @mysql_fetch_array($result);
$image = $data["thumbnail"];
header("Content-Type: image/jpeg");
echo $image;
ob_end_flush();
?>

BeastRider · May 14, 2003

I was looking more at your code, and I don't understand why the medium_photo.php code is doing anything more than retrieving a single row which contains the image data.

I've posted my getthumb.php code which retrieves a single image in the previous message. It gets called by this line

echo "<img src=\"getthumb.php?thumb={$row["thumbid"]}\">";

So, you can see there is a one on one relationship between the thumb requested and the thumb returned. So, I'd guess I'd want to know why you get a whole raft of rows all over again using %town_listings%, when all you want is a single row.

goa103 · May 15, 2003

Hi,

<img src="medium_photo.php?photo_id=<?= $row['Id'] ?>">

<? echo $row['medium_photo'];?>

Remove your echo. Echo is output, so you output your image like you output text... The img tag allows you to insert an image using a PHP script.

I was looking more at your code, and I don't understand why the medium_photo.php code is doing anything more than retrieving a single row which contains the image data.

BeastRider is right. Why don't you just get the image ? That's why I asked you to pass a photo_id to medium_photo. That script is meant to display 1 image using the unique id.

<?php
ob_start(); // turn output buffering on
// open the database and retrieve the record
(!($dbh=mysql_connect ("localhost",
"USERNAME",
"USERPASS")));
(!mysql_select_db("DBNAME",$dbh));
$query = "SELECT thumbnail FROM thumbs
WHERE thumbid = $thumb";
(!($result = @ mysql_query ($query, $dbh)));
$data = @mysql_fetch_array($result);
$image = $data["thumbnail"];
header("Content-Type: image/jpeg");
echo $image;
ob_end_flush();
?>

This is a good example but you don't need the ob_ functions. medium_photo.php is a new PHP script, it's not part of the other script. The idea is to query the DB, create a MIME header (image/jpeg) and output $image using echo. The difference with your echo resides in the header. It indicates that the following data are part of an image.

JM

sgtpepper · May 16, 2003

I want to thank everyone for helping. I still am having an issue and hope maybe you can help me solve it.

Here is what I am doing.....

I have one table called residetial with a field called 'medium_photo'. This field is where I am storing OLE binary blob jpegs. I also have a field called 'Id' which displays an id number for each listing.

I am trying to display the images and having problems. My php code shows a broken images and points to

getthumb.php?thumb=430729 (thumb= changes based on image so this tells me that it is seeing the Id).

Here is the code I am an using

picture.php
///////////////
<?php
include("template/generic/top.html");
?>

<?php
include("test.html");
?>

<?
$username="XXXX";
$password="xxxx";
$database="xxxx";