return results of a query

[lukeurtnowski]

I'm trying to display the results of a query using the following code.

$query = "SELECT thumb_name, photo_name, Email FROM PHOTO WHERE Email = '$Email'";
$handle = mysql_query($query) or die("Error: " . mysql_error()); //Do the query
echo $handle;
while($row = mysql_fetch_array($handle)){
$thumb_name = $row['thumb_name'];
$photo_name = $row['photo_name'];
echo '<img src="/images/ASP_Images/Thumbnail/"'.$thumb_name.'" />';
} 

But when I run the code, I get
Resource id #5
and no images?
Why does the while loop not run?
Thanks...

[galen]

maybe the query didnt return any results. try the query in phpmyadmin or somethign to make sure it has results

[lukeurtnowski]

If I echo the query, I get
SELECT thumb_name, photo_name, Email FROM PHOTO WHERE Email = 'jon@test.com'
and if I paste it into mysql I get 3 results (see screenshot)

[centered effect]

Remove this line:

echo $handle;

[galen]

did you view source and see if maybe the images are just invalid

[lukeurtnowski]

When I view source, I get
<fieldset style="clear:both">
<legend>Images</legend>
Resource id #5</fieldset>
So its not even seeing the loop
http://www.houstonasp.com/php/show_provider.php?ID=3

[SJH]

What does the following code output?

$query = "SELECT thumb_name, photo_name, Email FROM PHOTO WHERE Email = '$Email'";
$handle = mysql_query($query) or die("Error: " . mysql_error()); //Do the query
$row = mysql_fetch_array($handle);
echo '<pre>';
print_r($row); 

[galen]

next thing i woudl do is check mysql_num_rows() to make sure the query is getting results

[lukeurtnowski]

I'm so confused, I checked mysql_num_rows()

$sql = "SELECT thumb_name, photo_name, Email FROM PHOTO WHERE Email = '$Email'";
$handle = mysql_query($sql) or die("Error: " . mysql_error()); //Do the query
$row = mysql_fetch_array($handle);
echo mysql_num_rows($handle);
echo '<pre>';
print_r($row); 

And it returns 0 (but I ran the query in PHPMyAdmin and it produces 3 results.. see screenshot)

[SJH]

Try adding the pointer to your MySQL connect resource as a second parameter in your mysql_query() function call.

i.e. $handle = mysql_query($sql,$dbconn);

Where $dbconn is your database connection resource, obviously.

[9three]

$query = "SELECT thumb_name, photo_name, Email FROM PHOTO WHERE Email = '$Email'";
$handle = mysql_query($query) or die("Error: " . mysql_error()); //Do the query
echo $handle;
while($row = mysql_fetch_array($handle)){
$thumb_name = $row['thumb_name'];
$photo_name = $row['photo_name'];
echo '<img src="/images/ASP_Images/Thumbnail/".$thumb_name." />';
} 

Assuming you already made a proper connection

Change to:

$query = "SELECT FROM PHOTO WHERE Email = '$Email'";
$handle = mysql_query($query) or die("Error: " . mysql_error()); //Do the query
while($row = mysql_fetch_array($handle)){
$thumb_name = $row['thumb_name'];
$photo_name = $row['photo_name'];
echo '<img src="/images/ASP_Images/Thumbnail/"'.$thumb_name.'" />';
} 

Although you don't need Email = '$Email' because you have it stored already. You could simply just do:

SELECT FROM PHOTO

and for your loop, you could do

$thumb_name = $row['thumb_name'];
$photo_name = $row['photo_name'];
$email = $row['Email'];

Now you can use the variable $email to display the assigned email for each row.

[lukeurtnowski]

k, ill try

[lukeurtnowski]

I get

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near 'FROM PHOTO' at line 1

[wibble wobble]

You gave him a query with no select...

luke: Can you post your full code and I'll help you?

[9three]

$query = "SELECT * FROM PHOTO WHERE Email = '$Email'";

[lukeurtnowski]

Oh, I got it, I accidentally changed the value of $Email.
Thanks for the help.
(I Used)
$query = "SELECT thumb_name, photo_name, Email FROM PHOTO WHERE Email = '$Email'";
$handle = mysql_query($query) or die("Error: " . mysql_error()); //Do the query
while($row = mysql_fetch_array($handle)){
$thumb_name = $row['thumb_name'];
$photo_name = $row['photo_name'];
echo '<img src="/images/ASP_Images/Thumbnail/".$thumb_name." />';
}
but added some code (the lightbox effect)