Miss-Ruth

WOW!!! that was pretty quick! Thanks! I'll give it a try.

Server side - PHP receives a path to an image, how how do I send the image through instead of the path? Hope that makes scence. I've got a folder on the server with several images. I need to send the selected one (upon what the viewer selects) from PHP. How to code?

Ggg!!! Thanks DW!!!!

how to seperate the display results? I hope it makes sence. $query = "SELECT * FROM dyomails WHERE identity = '$SomeVar'"; $results = mysql_query($query); if($line = mysql_fetch_assoc($results)) { echo $line["email"]; echo $line["logo"]; }

Thanks Bricktop. One more question. is ["image"] my db field name? is "Some Image" the file name of the image? { echo $line["email"]; echo '<img src="'.$line["image"].'" alt="Some Image" />'; }

Hello! you mean to have 2 echos? { echo $line["email"]; echo '<img src="'.$line["image"].'" alt="Some Image" />'; }

My Apologies ILMV! I was tired in trying to figure this out since I'm NOT a php girl. On the other hand Bricktop, THANKS!!! that's the sort of answer I was waiting for. I'll have a go and come back to mark this as ANSWERD. Thanks Guys!! You are always helpful.

ILMV ---- DID YOU READ MY REQUEST? If you want me to say in other words "how to call an image from db? This is my script. I need to add my requirement in to it"

how to bring in the relevant image with it (from db)? <?php $link = mysql_connect("localhost","aabbcc","aabbcc1"); mysql_select_db("abcd"); $query = 'SELECT * FROM aabb'; $results = mysql_query($query); $query = "SELECT * FROM aabb WHERE identity = '$SomeVar'"; $results = mysql_query($query); if($line = mysql_fetch_assoc($results)) { echo $line["email"]; } mysql_close($link); ?>

in my code above, can I alter the script like this? is it correct? { echo $line["email","logo"]; } or like this { echo $line["email"],["logo"]; }

sorry I forgot to mention, that's variable passed which is sort of a personal ID. This script work well. But How to add the image thing so it match the $fin (code -ID)to the image

How to pull an image by the ID from the mysql db? (image field name: logo) My current code: <?php $link = mysql_connect("localhost","abc","abc1"); mysql_select_db("abcdefg"); $query = 'SELECT * FROM abcmails'; $results = mysql_query($query); $SomeVar = $fin; $query = "SELECT * FROM dyomails WHERE identity = '$SomeVar'"; $results = mysql_query($query); if($line = mysql_fetch_assoc($results)) { echo $line["email"]; } mysql_close($link); ?>

Thanks Matt. I'll give it a try right away and get back to you.

Thanks ILMV. This is the script I've done so far. when a image is been sent from flash ($output), I need it to display the image + the other data as a HTML email using PHP. I dunno whether it's too much But please get started and then there will be a limit. <?PHP $to = "abca@mail.com"; $subject = "the subject"; $headers = "From:" .$email."\r\n"; //Capture Post data $data = explode(",", $_POST['output']); $width = $_POST['width']; $height = $_POST['height']; $message = "Name: " . $thename; $message .= "\nCompany or Club Name: " . $thecompany; $message .= "\nAddress: " . $theaddress; //Output image and clean header( "Content-type: image/jpeg" ); ImageJPEG( $output ); imagedestroy( $output ); ?>

:'( Atleast could someone give a try. $output is the file which is been send by flash. PHP $to = "abc@cba.com"; $subject = "subjesct"; $headers = "From:" .$email."\r\n"; $message = "Name: " . $name; $w = 'width'; $h = 'height'; $img = output($w, $h); $message. = "<img src=\"output\" >";

Hello!! Any PHP guy with a bit of a knowledge in Flash ? When I send a JPEG from flash how to get it to email in PHP.