second column should print next record that was returned from the query.
Thanks.
John
>From: "Jeff Lewis" <[EMAIL PROTECTED]>
>Reply-To: "Jeff Lewis" <[EMAIL PROTECTED]>
>To: "John Bass" <[EMAIL PROTECTED]>, <[EMAIL PROTECTED]>
>Subject: Re: [PHP] Need help to create HTML table with 2 columns.
>Date: Thu, 23 Aug 2001 11:29:36 -0400
>
>What did you want in the second column? I was just going by what you had
>listed below: $row->image_link,
> > > > $row->web_url,$row->image_link, $row->web_url);
>
>
>Jeff
>- Original Message -
>From: "John Bass" <[EMAIL PROTECTED]>
>To: <[EMAIL PROTECTED]>
>Sent: Thursday, August 23, 2001 11:16 AM
>Subject: Re: [PHP] Need help to create HTML table with 2 columns.
>
>
> > Hi,
> >
> > Thank you for the email. I tried your code, but first column is
>printing
> > image_link & web_url fields correctly and second column is repeating
>values
> > of the first column.
> >
> > Other records or rows are printing same way.
> >
> > Regards,
> >
> > John Bass
> >
> > >From: "Jeff Lewis" <[EMAIL PROTECTED]>
> > >Reply-To: "Jeff Lewis" <[EMAIL PROTECTED]>
> > >To: "John Bass" <[EMAIL PROTECTED]>, <[EMAIL PROTECTED]>
> > >Subject: Re: [PHP] Need help to create HTML table with 2 columns.
> > >Date: Thu, 23 Aug 2001 10:41:08 -0400
> > >
> > >John,
> > >
> > >Try this:
> > >
> > >
> > > > >$query = "SELECT image_link, web_url FROM testdata";
> > >$result = mysql_query ($query) or die ("Query Failed");
> > >while ($row = mysql_fetch_object ($result))
> > > {
> > > printf ("".$row->image_links."".$row->web_url."
> > >".$row->image_link."".$row->web_url."\n");
> > > }
> > > ?>
> > >
> > >
> > >- Original Message -
> > >From: "John Bass" <[EMAIL PROTECTED]>
> > >To: <[EMAIL PROTECTED]>
> > >Sent: Thursday, August 23, 2001 10:30 AM
> > >Subject: [PHP] Need help to create HTML table with 2 columns.
> > >
> > >
> > > > Hi All,
> > > >
> > > > I am trying to generate a table with two columns. Each column
>should
> > >print
> > > > 2 fields named image_link and web_url. Have spent over week looking
> > > > documentations and trying different coding but nothing is working.
>Any
> > >help
> > > > will be greatly appreciated.
> > > >
> > > >
> > > > > > > $query = "SELECT image_link, web_url FROM testdata";
> > > > $result = mysql_query ($query) or die ("Query Failed");
> > > > while ($row = mysql_fetch_object ($result))
> > > > {
> > > > printf ("%s%s %s%s\n",
>$row->image_link,
> > > > $row->web_url,$row->image_link, $row->web_url);
> > > > }
> > > > ?>
> > > >
> > > >
> > > > Regards,
> > > >
> > > > John Bass
> > > >
> > > > _
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