[SOLVED] Pull an image out of a database

[mikemessiah]

Hey there all. Excuse me as I am new to PHP and exceptionally dof.

 

I am trying to pull an image out of an mssql database and display it. The problem is it keeps coming out as a bunch of garbage. What I gather is that i need to have a separate page that checks if its an image and then prints it or something to the main page ? Anyone have an example of this.

 

Thanks

[mikemessiah]

OK, no one answered so I will try elaborate some more. Basically I need to pull out related images and display them in my main page. The problem is that I either get no image coming back ( just the image placeholder ), or I can get a bunch of hex garbage to come back. Here is the mini.php page i use :

 

<?php #mini.php

// Used to locate record based on ID number

// Should pass image back and display

require_once( "./common_db.php" );

 

if ( ! $link_id = db_connect($default_dbname) ) {

die( "connection to db failed" );

}

 

$id = $_GET['id'];

$q = "select Photo1 from $user_tablename where ID = $id";

 

if ( ! $image_result = mssql_query($q,$link_id) ) {

die( "error with query: $q" );

}

 

if ( mssql_num_rows( $image_result ) < 1 ) {

die( "no results found" );

}

 

$image = mssql_fetch_assoc( $image_result );

 

//echo $image;

 

header("content-type image/jpeg");

echo $image['Photo1'];

 

//Print $image;

//echo($image);

 

?>

 

Here is the part of code that calls mini.php

 

 

while($query_data = mssql_fetch_array($result))

{

$ID = $query_data["ID"];

$StreetAddress = $query_data["Street_Address"];

$Price = $query_data["Price"];

$Zone1 = $query_data["Zone1"];

//$Photo1 = $query_data["Photo1"];

//$Photo2 = $query_data["Photo2"];

//$Photo3 = $query_data["Photo3"];

//$Photo4 = $query_data["Photo4"];

 

echo "<TR>\n";

echo "<TD WIDTH=\"111\" ALIGN=\"LEFT\"><font size=\"2\">$ID</font></TD>\n";

echo "<TD WIDTH=\"128\" ALIGN=\"LEFT\"><font size=\"2\">$StreetAddress</font></TD>\n";

echo "<TD WIDTH=\"121\" ALIGN=\"LEFT\"><font size=\"2\">$Price</font></TD>\n";

echo "<TD WIDTH=\"67\" ALIGN=\"LEFT\"><font size=\"2\">$Zone1</font></TD>\n";

echo "<TD WIDTH=\"46\" ALIGN=\"LEFT\"><img src='mini.php?id=$ID'></TD>\n"; // Calls mini.php which should locate the image by ID number and pass it back with the correc header    // echo "<TD WIDTH=\"46\" ALIGN=\"LEFT\"><img src='D:\Mike Misc\Graphics\Wallpapers\1.jpg'></TD>\n";

// echo "<TD WIDTH=\"46\" ALIGN=\"LEFT\"><img src=\"$Photo1\"/></TD>\n";

// echo "<TD WIDTH=\"48\" ALIGN=\"LEFT\"><font size=\"2\">$Photo2</font></TD>\n";

// echo "<TD WIDTH=\"93\" ALIGN=\"LEFT\"><font size=\"2\">$Photo3</font></TD>\n";

// echo "<TD WIDTH=\"243\" ALIGN=\"LEFT\"><font size=\"2\">$Photo4</font></TD>\n";

 

echo "</TR>\n";

}

 

 

Any  ideas why it doesnt work. I am lost.

[Shazbot!]

Hi,

I had the same problem you are having of uploading an image to a mssql database and then downloading it for viewing.

The issue was resolved and you can find my solution here.

http://www.phpfreaks.com/forums/index.php/topic,146750.0.html

 

The problem is the way you upload the data, there is an article in the thread that explains this.

Hope this helps!