Morgan,
You're the second one I've seen using "do". What "do"? Is it in the php
manual and I missed it?
I changed your code slightly (to major if it's your baby). It might puke if
the $row is empty for the While loop. If it does, try an @ sign before the
while. I didn't test it but it looks like should go. Or, generate some
interesting error messages.
Hope this helps,
Hugh
$db = mysql_connect("localhost", "user", "pass");
mysql_select_db("photo",$db);
$tablei = 0; //this is the little counter....
$result = mysql_query("select photos.filetype,
photos.bin_data,
photos.id_files,
information.id,
information.desc,
photos.fulldesc
from photos,
information
order by information.id DESC",$db);
if (!$result)
{
print "So sorry, no results.";
}
else
{
print ("<table width=100%><tr>"); //added <tr>
while ($row = mysql_fetch_array($result))
{
// Produce the column for each record....
print "<td width=25% align=center valign=top>
<table width=100% border=0 cellspacing=2 cellpadding=2>
<tr>
<td>
image will go here
</td>
</tr>
<tr>
<td >
(this is id number):".$row[id]."
</td>
</tr>
<tr>
<td>
(this is description): ".$row[fulldesc]."
</td>
</tr>
</table>
</td>";
$tablei = $tablei + 1;
if ($tablei == "4")
{
print ("</tr> <tr>");
$tablei = '0';
}
}
print "</tr></table>":
}
----- Original Message -----
From: "Morgan" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>
Sent: Thursday, March 14, 2002 7:49 PM
Subject: [PHP] PHP Is Inserting (the same) Database Record Multiple Times In
My HTML Output
> (sorry if this shows up twice!)
>
> hi, i'm new to php and am having a problem with a script i'm working on.
> i posted this to alt.php a few days ago and got part of my problem
> fixed, but i still have one problem which i can't figure out and haven't
> gotten an answer to there.
>
> here is the code:
>
> <?PHP
> $db = mysql_connect("localhost", "user", "pass");
> mysql_select_db("photo",$db);
> print ("<table width=100%>");
> //this is the little counter....
> $tablei = 0;
> //CONNECT TO DATABASE
> $result = mysql_query("select photos.filetype, photos.bin_data,
> photos.id_files, information.id, information.desc, photos.fulldesc from
> photos, information order by information.id DESC",$db);
> if ($row = mysql_fetch_array($result)) {
> do {
>
> // Produce the column for each record....
> print ("<td width=25% align=center valign=top><table width=100%
> border=0 cellspacing=2 cellpadding=2><tr><td>image will go
> here</td></tr><tr><td >(this is id number):
> ".$row[id]."</td></tr><tr><td>(this is description):
> ".$row[fulldesc]."</td></tr></table></td>");
>
> $tablei = $tablei + 1;
>
> if ($tablei == "4") {
> print ("</tr> <tr>");
> $tablei = '0';
> }
> } while($row = mysql_fetch_array($result));
> } else {print ("No Records");}
>
> ?>
>
> an example of the output can be found here:
> http://www.thosenetworkguys.com/test.php
>
> the problem i am having is that the php code is printing each record in
> the database multiple times. the number of times the record is printed
> is equal to the number of records i have in the database. if you look
> at the url above, you'll see what i'm getting at. for instance, i now
> have 5 records in the database and for some reason each individual
> record is being inserted into the page 5 times now.
>
> the code above was suggested as i was trying to create a page that would
> print an html table and start a new row at four columns. i assume the
> problem i am having now has something to do with the math going on in
> the script to format the html table, but i have tried numerous changes
> to the value and keep getting the same thing.
>
> could someone be kind enough to help out a php newbie and help me figure
> out how to fix this? i just can't figure it out.
>
> thanks.
>
>
>
>
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