it's the mysql_result()... try this...
echo "<img src='http://www.website.com/index_files/main_files/$planid/"; .
$result["FLRPLAN_2"] . "'>'";
mysql_result($result, 0,'FLRPLAN_2')
^----------- you're always looking at the first row
rf: http://www.php.net/manual/en/function.mysql-result.php
-----Original Message-----
From: markbm [mailto:[EMAIL PROTECTED]]
Sent: Tuesday, July 23, 2002 2:29 PM
To: [EMAIL PROTECTED]
Subject: Re: [PHP] Problem with Not displaying HTML if data not found
When I changed this line, it still gives me the same problem. My db field
is a varchar(125)...could that have anything to do with it? Basically, the
first time, without data in the field...it works fine and the <TR> is not
displayed, but when I add data back into the field, nothing changes...its
like it does not recognize that something is back in the field. I did a
commit on the db each time, and cleared the browser cache, etc.
Mark
"Martin Towell" <[EMAIL PROTECTED]> wrote in message
6416776FCC55D511BC4E0090274EFEF5034CFB9D@EXCHANGE">news:6416776FCC55D511BC4E0090274EFEF5034CFB9D@EXCHANGE...
> what about
> if (strlen($result["FLRPLAN_3"]) > 0)
>
>
>
> -----Original Message-----
> From: markbm [mailto:[EMAIL PROTECTED]]
> Sent: Tuesday, July 23, 2002 2:02 PM
> To: [EMAIL PROTECTED]
> Subject: [PHP] Problem with Not displaying HTML if data not found
>
>
> I am trying to develop a page that, in certain places, I only want it to
> display the <TR> and <TD>s if data is found for a given field. For
example,
> if I only have an image for the first floorplan of a house, I do NOT have
> an image for the second floorplan of a house, then show the first
floorplan
> image only.....skip over that next row because the field in the db is
empty.
>
> The problem that I'm having is that the syntax that I've been using to
> define "empty" in the db field does not always work. The code reads:
>
> if ($result[FLRPLAN_2]<>" ") {
> printf ("<tr>");
> printf ("<td colspan=3 bgcolor=#F0F8FF>");
> printf ("<font size=5> <b>Second Floor Plan:<b>
> </td></font>") ;
>
> printf ("</tr>") ;
> printf ("<tr>");
> printf ("<td colspan=3 >") ;
> echo "<img
> src='http://www.website.com/index_files/main_files/$planid/"; .
> mysql_result($result, 0,'FLRPLAN_2')."'>'";
> printf (" </td>") ;
> printf ("</tr> ");
> };
>
>
>
> if ($result[FLRPLAN_3]<>" ") {;
> printf ("<tr>");
> printf ("<td colspan=3 bgcolor=#F0F8FF><font size=5>
> <b>Third Floor Plan:<b> ");
> printf ("</td></font> ");
> printf ("</tr> ");
> printf ("<tr> ");
> printf ("<td colspan=3 >");
> echo "<img
> src='http://www.website.com/index_files/main_files/$planid/"; .
> mysql_result($result, 0,'FLRPLAN_3')."'>'";
> printf ("</td> ");
> printf ("</tr> ");
> };
>
>
> I have tried:
>
> 1. if ($result[FLRPLAN_3]<>" ") {;
> 2. if ($result[FLRPLAN_3]<>"") {;
> 3. if ($result[FLRPLAN_3]<>'') {;
> 4. if ($result[FLRPLAN_3]<>' ') {;
> 5. if ($result[FLRPLAN_3]<>NULL) {;
>
> and none seem to work correctly (i.e. blank field in db, but HTML row
shows
> up with missing image......or image is in db, but no HTML row shows up.
If
> it helps, each of the "image" fields are varchar(125)....basically I just
> have the name of the image file, and then a pointer to the web server
> directory structure....
>
> Any ideas would be greatly appreciated. Thanks.
>
> Mark
>
>
>
> --
> PHP General Mailing List (http://www.php.net/)
> To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
