RE: [PHP-DB] Help wanted
| From: | Bart A. Verbeek | Date: | Thu, 12 Oct 2000 13:39:56 +0000 |
| Subject: | RE: [PHP-DB] Help wanted | ||
| References: | 1 | Groups: | php.db |
| Request: | Send a blank email to php-db+get-3552@lists.php.net to get a copy of this message | ||
Thanks for the tips Miles.
After puzzleling with the querie on the MySQL-console I got it to work.
I will try this in the future also before mailing the list..
-----Oorspronkelijk bericht-----
Van: Miles Thompson [mailto:milesthompson@sprint.ca]
Verzonden: donderdag 12 oktober 2000 0:15
Aan: Bart A. Verbeek
CC: PHP-DB mailinglist
Onderwerp: Re: [PHP-DB] Help wanted
Bart,
1. Have you tried these queries interactively at the mysql console and do
they
work?
2. Do you have any duplicate field names that may need qualification?
3. Use mysql_num_rows() afer each query to see how many rows are being
returned.
That should help diagnose the problem.
Miles Thompson
"Bart A. Verbeek" wrote:
> I can't seem to get this code to work.
> I've created three db's (one with mineral_info, one with photo-filenames
and
> one with foto_ID's and min_ID's). I want to display the according photo's
> with the minerals (somethimes there are more photo's) and when there are
no
> corresponding foto_ID or mineral_ID entries in the third db. the last else
> (echo: No picture available yet!) must be executed. But I keep getting the
> following error: "Warning: 0 is not a MySQL result index in
> /home/mineralen/www/php/min_details2.inc on line 14" The strainge part is
> that the echo-comm. for no image is executed, even if I know there is an
> image!! I've tried different queries/code but I can't get it to work.
Maybe
> you can help.
> The script I use is down below. Thanks in advance, Bart
>
> <?php
> $naam_nl = $link;
> $result = mysql_query ("SELECT * FROM min_data WHERE ID LIKE '$naam_nl'");
>
> if ($row = mysql_fetch_array($result)) {
> do {
>
> echo ("Hier moet de naam staan: " . $row["naam_nl"] . " uit eerste
> q");
>
> $result_foto = mysql_query ("SELECT bestand, omschrijving, vindplaats,
> fotograaf
> FROM min_foto, min_data, foto_ID
> WHERE foto_ID.foto_ID='$link'
> AND min_foto.bestand=foto_ID.foto_ID");
>
> if ($row_foto = mysql_fetch_array($result_foto)) { //this line causes
> error??
> do {
>
> echo ("<img border='0' src='images/" .
> $row_foto["bestand"] .
> ".jpg' align='center' valign='middle'>\n
> Hier moet de foto staan images/". $row_foto["bestand"] .
> ".");
>
> } while ($row_foto = mysql_fetch_array($result_foto));
> } else {
> echo ("No picture available yet!");
> }
> // echo ("</td>\n");
>
> } while ($row = mysql_fetch_array($result));
> } else {
> print ("Sorry, de door u opgevraagde gegevens
> zijn niet beschikbaar. Helemaal eind script");
> }
> ?>
>
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