Load Photo from DB

northview

I am trying to load a photo that was uploaded to to the DB. I can see the photo in the database...but it just will not display. Here is my code:

<?php
$result=mysql_query("Select value from site_setup where name='PHOTO_RIGHT'");
$row=mysql_fetch_array($result);
?>
<img src ="<?=$PHOTO_RIGHT?>" width="200" height="180" alt="<?=$CHURCH_NAME?><br><?=$MAILING_ADDRESS?>">

Also, once I upload the photo, the photo will not display on the sample page:

list($photo) = mysql_fetch_row(mysql_query(" select value from site_setup where name = 'PHOTO_RIGHT' "));

$title = "Photo Right Side";	
$heading = "Photo Right";
include("header.php");

<? if(_GET("er", false) == "done") error("Successfully Done"); ?>
<? if(_GET("er", false) == "not") error("Not Done"); ?>

<? if($photo) { ?>
<IMG SRC="..photo/<?=$PHOTO_RIGHT?>" width="200" height="180""><BR>
<A HREF="photo-right.php?mode=del">remove</A>
<P> </P>
<? } ?>

Thank you for your help.

Krik

There is a lot of problems with your code but I will try to focus on just getting the photos name out of the database.

First I think your MySQL query is part to blame.

mysql_query("Select value from site_setup where name='PHOTO_RIGHT'")

Let put that query into plain english.

SELECT the value of the "value" column FROM the "site_setup" table WHERE the "name" column equals "PHOTO_RIGHT:"

Note that it is not retrieving the value of the "name" column. So I hope the name of the photo is in the value column. If so hopefully only one row in your table has "PHOTO_RIGHT" as the value of the "name" column, otherwise you may not be able to get the photo you want consistently, without adding a bit more code.

Next the value of any column does not become a php variable. PHOTO_RIGHT can not become $PHOTO_RIGHT without you directly setting the php variable $PHOTO_RIGHT to PHOTO_RIGHT. But as noted above you didn't retrieve the value of the name column so with your current code you can't even set $PHOTO_RIGHT to PHOTO_RIGHT. However you can set $PHOTO_RIGHT to equal the value of the "value" column.

$PHOTO_RIGHT = $row['value'];

Note that all values you retrieved in your query are stored in the $row array.

Basically all the above was written to point out that you do not have the $PHOTO_RIGHT variable defined anywhere. Hopefully everything above will help you figure out how to get it defined so you can see your image.

I would ask if you have your error reporting set to report all php errors. If not turn it on, at least while you are in the development process, it will help you find some of these errors.

northview

Krik:

I tried what you said and KABOOM...the whole page blew up. I know it had something to do with my syntax...but I am not sure what. I am such a newbie...that I am not even sure how to plug it all in.

I will try again. Thank you.

Krik

If your getting errors feel free to post them here with the relevant lines in the error and we can try to help. PHP and MySQL errors can sometimes be more informative than an explanation of the problem.

If you tried the

$PHOTO_RIGHT = $row['value'];

I suspect it went crazy because the name of the file is not in the "value" column but the "name" column (that's a guess). To get the name just change the "value" in your query to a "".

mysql_query("SELECT * FROM `site_setup` WHERE `name` = 'PHOTO_RIGHT'")

and then do

$PHOTO_RIGHT = $row['name'];

If that fails add this line after the $row variable line

print_r($row);

This will output a list of the items in the $row array look for the item that has the files name and whatever is in the square brackets ([]) in front of it put in the place of the "name" in the above example.

I hope that is bit more helpful as my previous post was more to point out the major missteps in your code and not so much provide the answer.

northview

Thanks for you help...I will give those suggestions a try and get back to you.