On Tuesday 30 October 2001 02:53 pm, you wrote:
> It is starting to make sense.  I see the correlation between the manual
> example and the code you show.  Surprisingly (and equally confusing) the
> book I have also uses a while loop with mysql_fetch_array to display a
> list.

You would use a while loop if/when you have multiple records to retrive.  
>From your original code example, it seemed as though you would only have one 
record.  (though, for safety's sake, you should code a LIMIT into your sql 

> This may be the problem as I now get "supplied argument not valid mysql
> result" on this line.
> $query = "select imglocation,height,width from img where
> imgname='$getimg'";

Two things to try:

1.  Run that exact query on a mysql command line, substituting whatever 
variable name you're trying to pass in the URL.  Often times, I find that I 
tihnk I have a PHP problem when, in fact, I screwed up my query syntax 

2.  Explicitly define $getimg as something you know to be a valid image.  If 
that works, then you know it isn't getting passed from the url properly.  

> It looks like I might be able to use the GetImageSize function to do this
> same thing?  maybe?  All I am doing is getting the location (url) of an
> image and its height and width for an <img src= > statement.....  maybe I
> am making this into a bigger job than necessary??  I think I have been
> working on it too long today and I need a long walk.....

Actually, yes you can.  

$myImgSize = getImageSize('/unix/path/to/my/file.jpg'); should return the 
image sizes in an array.  You should then be able to do something like:

<img src="file.jpg" $myImgSize[3]>

Which should automatically include height/width parameters.  (see this 
function on the php site for more info -- also, that's untested, off-the-cuff 
code I just wrote.  Might not work perfectly. :)


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