show me the way please!

jwlnewsome

can some one show me the best way to show images from a database in a table

ive got a database called myFiles and a table below

create table myBlobs
(
blobId int auto_increment not null,
blobTitle varchar(50),
blobData longblob,
blobType varchar(50),
primary key(blobId),
unique id(blobId)
);

showfiles.php below

<?php 
$sConn = mysqli_connect("localhost","root", "admin","myFiles")or die("Couldn't connect to database server");
$dbQuery = "SELECT blobId, blobTitle, blobType,blobData ";
$dbQuery .= "FROM myBlobs ";
$dbQuery .= "ORDER BY blobTitle ASC";
$result = $sConn->query($dbQuery) or die("Couldn't get file list");
//$fileType = $row["blobType"]; 
//header("Content-type: $row['blobType']");
?>
<table border="1" cellpadding="0" cellspacing="0" bordercolor="#111111" width="100%">
<tr>
<td width="25%" bgcolor="#FF9900" height="21">
<p style="margin-left: 10"><b><font size="2" face="Verdana" color="#FFFFFF">Description</font></b></td>
<td width="25%" bgcolor="#FF9900" height="21">
<p style="margin-left: 10"><b><font face="Verdana" size="2" color="#FFFFFF">Type</font></b></td>
<td width="25%" bgcolor="#FF9900" height="21">
<p style="margin-left: 10"><b><font face="Verdana" size="2" color="#FFFFFF">File</font></b></td>
<td width="25%" bgcolor="#FF9900" height="21">
<p style="margin-left: 10"><b><font face="Verdana" size="2" color="#FFFFFF">photo</font></b></td>
</tr> 
<tr>
<?php while($row = mysqli_fetch_array($result)){?>
<td width="25%" bgcolor="#FFDCA8" height="21">
<p style="margin-left: 10; margin-right: 10">
<font face="Verdana" size="1">
<?php echo $row["blobTitle"]; ?>
</font>
</td>
<td width="25%" bgcolor="#FFDCA8" height="21">
<p style="margin-left: 10">
<font face="Verdana" size="1">
<?php echo $row["blobTitle"]; ?>
</font>
</td>
<td width="25%" bgcolor="#FFDCA8" height="21">
<p style="margin-left: 10"><font face="Verdana" size="1">
<a href="downloadfile.php?blobId=<?php echo $row["blobId"]; ?>">
Download now
</a></font>
</td>
<td width="25%" bgcolor="#FFDCA8" height="21">
<p style="margin-left: 10">
<font face="Verdana" size="1">
<?php
header("Content-type: $row[blobType]");
echo $row["blobData"]; ?>
</font>
</td>
</tr>
<?php
}
echo "</table>";
?>

this outputs the raw data for the image file in collum 4, but when i click the link in collum 3 the image gets shown in another page downloadfile.php below what i an tring to do is show the image in collum 4 not the raw data and get rid of the download link altogether.

<?php
global $blobId;
if(!is_numeric($blobId))die("Invalid blobId specified");

// Database connection variables

$sConn = mysqli_connect("localhost","root", "admin","myFiles")or die("Couldn't connect to database server");

$dbQuery = "SELECT blobType, blobData ";

$dbQuery .= "FROM myBlobs ";

$dbQuery .= "WHERE blobId = $blobId";

$result =$sConn->query($dbQuery) or die("Couldn't get file list");

if(mysqli_num_rows($result) == 1)
{
$row =$result->fetch_row();
//$fileType = $row[0];
//echo $fileType;
//$fileContent = $row[1];
header("Content-type: $row[0]");
echo $row[1];
}else{echo "Record doesn't exist.";}
?>

ive been on this for days but and going nowhere all the tutorials on the web just seem to show the link method can any body help a newbee (stressed)

thanks in advance
john

ps tutorial link

mrbill501

I dont know how much help this will be seeing as I'm trying to do the same thing and aren't really getting anywhere fast either.
My code for showing Blob images is:
<?php
set_magic_quotes_runtime(0);
ini_set("include_path","/home/fhlinux192/s/studentart.streamlinenettrial.co.uk/user/htdocs/includes/");
include ('pics.inc');
if (isset($GET['$image_id'])) {
$id = $GET['$image_id'];
$connection = mysql_connect($host,$user,$password)
$db = mysql_select_db($database, $connection)
$query = "SELECT image, image_type FROM tbl_images WHERE image_id = $id";

$result = mysql_query($query);

$data = mysql_result($result,"image");
$type = mysql_result($result,"image_type");

Header( "Content-type: $type");
$data = stripslashes($data);
echo $data;

};
?>[/COLOR]

To call an image I use the syntax : <img src="getdata.php3?image_id=334">
Of course this doesnt work but if anything here helps let me know what you figure out. Two heads are better than one and all that.
cheers

mrbill501

Ok nailed it. This code works now:
This is a file called getdata.php3
<?php
ini_set("include_path","/home/fhlinux192/s/studentart.streamlinenettrial.co.uk/user/htdocs/includes/");
include ('pics.inc');
if (isset($GET['image_id'])) {
$id = $GET['image_id'];
$connection = mysql_connect($host,$user,$password)
or die(mysql_error());
$db = mysql_select_db($database, $connection)
or die(mysql_error());
$query = "SELECT image, image_type FROM tbl_images WHERE image_id = '$id'";
$result = mysql_query($query);
$data = mysql_result($result,0,"image");
$type = mysql_result($result,0,"image_type");
header("Content-type: $type");
//print(mysql_result($result, 0, "image"));
echo $data;
};
?>

Using the image code: <img src="getdata.php3?image_id=336"> I can get images to show up in html pages. Hope this helps you out, I know how annoying it can be trying to do this.

jwlnewsome

thankyou,
by reading your code and a little reverse enginnering i think i know where im going wrong

time to play (again)#

thanks again
john

mrbill501

No worries, it's taken me days to get this far so I know how annoying it can be. The only problem I still have is with the echo '<img src="getthumb.php3?thumb_id=334">'; code.
Do you know how to get that code to display the last image uploaded to a database? Ive tried a couple of variations on it with ?image_id=image_id or ?image_id=$image_id and I'm not getting anywhere with either.
Any ideas?