Unexpected return...

[Germaris]

Hi there!

This query :

SELECT img_name FROM $table WHERE email = '$email' AND ok = '1' ORDER BY id DESC LIMIT 0 , 3") or die("&error=Failed !");

is supposed to return a 2 elements array (as I already know the content of $table satisfying to the query statements)

"ok" is a column which content is 1 (valid image) or 0 (invalid image)
there's 3 rows containing the required $email and only two where ok='1'

Statement

if ( mysql_num_rows($sql) > 0) {
    while ($row =mysql_fetch_array ($sql)) {
        $array .= $row['img_name']."|";
    }
    print "&numberOfRows=".mysql_num_rows($sql)."&myArray=".$array;
}
else {
    print "&error=No images found.";
} 

returns a wrong number of rows (3 instead of 2) but a right array !!!???

Why?

Many thanks for your advices, remarks and suggestions!

[bradgrafelman]

First and foremost: Stop Using the MySQL Extension!.

Next, note that your variable named $array is quite a misnomer, since you don't have an array at all. Since arrays are much nicer to work with, however, I suggest you create one. In other words, try changing this:

$array .= $row['img_name']."|"; 

to something more like:

$array[] = $row['img_name'];

Then, after the while() loop, do a [man]print_r/man on $array (since it will actually be an array this time) and show us the output.

[Germaris]

Thank you, bradgrafelman, for your kind response!

Alas, the solution you suggest doesn't work.

Using:

$array[] = $row['img_name']; 

returns to Flash "Array" instead of a string containing the expected image names...

Please, keep in mind that my only problem is:
Why PHP outputs a wrong number of rows (3 instead of 2)?

I know that the output I use in Flash as an "array" is in fact a string with a "|" delimiter

print "&numberOfRows=".mysql_num_rows($sql)."&myArray=".$array;

It gives me the right names of the images and in the correct order as expected and I can easily manipulate it in Flash to obtain the results I want.
So, I do not wish to modify something which is actually working nice...

I see it in Flash in a trace field (it's exactly the same as using print_r() in PHP...).

About, your recommendation for using mysqli, I know that.
But it implies to modify too much scripts and, for the moment, I don't have time to dedicate to this task.
I'll. for sure, do it in the future when my present project will be completed.

Best regards,

Gerard

[bradgrafelman]

Germaris;11022367 wrote:

Alas, the solution you suggest doesn't work.

I haven't suggested any solutions yet - I'm just trying to get you to output more data to provide to us so that we can better understand exactly what's happening.

Germaris;11022367 wrote:

Using:

$array[] = $row['img_name']; 

returns to Flash "Array" instead of a string containing the expected image names...

Well, yes, that would be expected if you tried to directly output the array rather than using [man]print_r/man as I requested above.

Additionally, this is the first time that you've mentioned that the output of this PHP script isn't being viewed directly but is instead being supplied to some Flash object. For debugging purposes, can you try accessing the PHP script directly to view the raw output? If so, try adding the print_r() I suggested above.

Germaris;11022367 wrote:

Please, keep in mind that my only problem is:
Why PHP outputs a wrong number of rows (3 instead of 2)?

I would argue that your only problem is thinking that your only problem is that PHP is outputting wrong information. 😉

There must be a reason that mysql_num_rows() is returning three rather than two. The only reason I can think of is: Your result set contains three rows - not two.

Germaris;11022367 wrote:

I know that the output I use in Flash as an "array" is in fact a string with a "|" delimiter

print "&numberOfRows=".mysql_num_rows($sql)."&myArray=".$array;

Can you view the raw value before you do any manipulation? Are you sure you aren't getting something like "value1|value2||" or "value1||value2" or "|value1|value2", suggesting that there really are three rows being returned by the SQL query?

Germaris;11022367 wrote:

I see it in Flash in a trace field (it's exactly the same as using print_r() in PHP...).

What is "it" ? print_r() on an array will show you a nicely-formatted representation of each element in that array. You haven't shown us anything of that nature (which is why I asked about the "raw value" above).

[Germaris]

After having the proof that my PHP Script was good and returns the expected results, I made a deep investigation of my Flash .fla file.
Debugging shown that Flash had a misinterpretation of the PHP output it received.
The error was in Flash, not in PHP!!!
And the error was due to a human error.
Mea culpa.

Please look at my PHP Test in action and the script at the end of this post.
You can see the results at:
[url]http://www.notre-ann...rict/photos.php[/url]

I want to thank you all very much for your interest and the time you dedicated to my problem
even if I had the sensation some were irritated by my apparent lack of interest for their posts.

Best regards.

PS: English isn't my native language, please excuse my mistakes.

---

My Test script as follows:

<?php
include "key.php";
$link = mysql_connect( $host ,$user ,$pass);
if (!$link) {
        die("&erreur=Connexion impossible avec le serveur : " . mysql_error());
}
$db_selected = mysql_select_db( $db, $link);
if (!$db_selected) {
        die ($db." inutilisable : " . mysql_error());
}
$table = "_files";
$courriel = "[e-mail]"; // here, of course, my real email address
$sql =mysql_query("SELECT * FROM $table WHERE email = '$courriel'AND ok = '1' ORDER BY id DESC LIMIT 0 , 3") or die("&erreur=Échec de la requête !");
if ( mysql_num_rows($sql) > 0) {
        while ($row =mysql_fetch_array ($sql)) {
                $flashstr .= $row['img_name']."|";
        }
        print "&erreur="."&nombre=".mysql_num_rows($sql)."&liste=".$flashstr;
}
else {
        print "&erreur=Pas de photos présentes pour cet utilisateur.";
}
?>