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MySQL SELECT QUESTION & THUMB DISPLAY

Posted: Wed Nov 13, 2002 9:37 am
by cooler75
hello,
I have been working on an upload script, and now it's time to display both orginal and thumbnail on the broswer.

This is my SELECT queries:

Code: Select all

$query = "SELECT * FROM tbl_Files WHERE prop_num = $propnum";
$result = mysql_query ($query) or die(mysql_erro().'<p>'.$query.'</p>');
$query2 = "SELECT * FROM tbl_thumb WHERE prop_num = $propnum";
$result2 = mysql_query ($query2) or die(mysql_error().'<p>'.$query2.'</p>');
$num_images = 0;

while ($image_row =mysql_fetch_array ($result) )
  {

<center>
    <a href='image.php?Id=$image_rowїid_files]' target="_new">
    <img src='$path_to_thumb/$thumb_file_name' height="$displayheight" width="$displaywidth" border="1">
    </a>
</center>

  }
I have no problem getting the original image displayed on the broswer, however, i am not able to pull the thumbnail image from the directory to show as a link to the original image, it only show a broken link

please advise
and thank you

Posted: Wed Nov 13, 2002 9:46 am
by volka
uh? question is: are in or out a <?php ?> block?
try

Code: Select all

&lt;?php
$query = "SELECT * FROM tbl_Files WHERE prop_num = $propnum";
$result = mysql_query ($query) or die(mysql_erro().'&lt;p&gt;'.$query.'&lt;/p&gt;');
$query2 = "SELECT * FROM tbl_thumb WHERE prop_num = $propnum";
$result2 = mysql_query ($query2) or die(mysql_error().'&lt;p&gt;'.$query2.'&lt;/p&gt;');
$num_images = 0;

while ($image_row =mysql_fetch_array ($result) )
{ ?&gt;

&lt;center&gt;
    &lt;a href="image.php?Id=&lt;?php echo $image_row&#1111;id_files]; ?&gt;" target="_new"&gt;
    &lt;img src="&lt;?php echo "$path_to_thumb/$thumb_file_name" ?&gt;" height="&lt;?php echo $displayheight; ?&gt;" width="&lt;?php echo $displaywidth; ?&gt;" border="1"&gt;
    &lt;/a&gt;
&lt;/center&gt;
&lt;?php
}
?&gt;

Posted: Wed Nov 13, 2002 10:37 am
by cooler75
hi,
that wasnt the problem, i just forgot to put echo in my last message.

I was wondering if there is anything to do with this line here:

Code: Select all

while ($image_row =mysql_fetch_array ($result) )


Can you tell what the problem is? Is there anything else i forgot to do in order to show the thumbnail from the directory?

thanks again

Posted: Wed Nov 13, 2002 10:42 am
by volka
open "view source" in your browser and take a look what really was sent to the client. Does the src-property really contain what you expected?

Posted: Wed Nov 13, 2002 10:59 am
by cooler75
hi,
that's a good way to check :lol:

what i see from 'view source' is:

<img src='http://www.example.com/images/listing_images/' height="108" width="98" border="1">

which is wrong obviously.

what happen to

Code: Select all

$thumb_file_name
?

could you please lead me to the right direction?
thank you again

Posted: Wed Nov 13, 2002 1:10 pm
by volka
where does $thumb_file_name come from? It's in the script only once - when it shall be printed.
Is it a db-field or do you want to create it from a db-field's value (e.g. original image's name is pic1.gif so the thumb must be called pic1_thumb.gif) ?
However there's something missing ;)

Posted: Wed Nov 13, 2002 1:32 pm
by cooler75
hello guru,
definitely something is missing.

$thumb_file_name is a db field with value, your assumption is totally right, pic.jpg is the original image, pic_thumb.jpg is the thumbnail image.

by looking at the database, i see something like that:

Code: Select all

ID  owner  filename  thumb_file_name  description  prop_num  
  1    2      bg.jpg    thumb_bg.jpg                    1
So I know the image was inserted properly, and the thumb has been created successfuly because when i browse my folder, i see the thumbnail there (thumb_bg.jpg).

$thumb_file_name is only appeared once because i have already select it from the database.

my knowledge to this is limited, would probably need some more help.
i will try to think carefully at the same time :oops:

thank you

Posted: Wed Nov 13, 2002 1:58 pm
by volka
take a look at the line

Code: Select all

&lt;a href="image.php?Id=&lt;?php echo $image_row&#1111;id_files]; ?&gt;" target="_new"&gt;
and how your script access' the field in there. Do the same with the thumbnail field
btw. it should be $image_row['id_files'] ;)