Re: Need help to create HTML table with 2 columns.

second column should print next record that was returned from the query. Thanks. John

From: "Jeff Lewis" <jeff@hyrum.net> Reply-To: "Jeff Lewis" <jeff@hyrum.net> To: "John Bass" <php_mysql21@hotmail.com>, <php-general@lists.php.net> 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" <php_mysql21@hotmail.com> To: <php-general@lists.php.net> 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" <jeff@hyrum.net> Reply-To: "Jeff Lewis" <jeff@hyrum.net> To: "John Bass" <php_mysql21@hotmail.com>, <php-general@lists.php.net> Subject: Re: [PHP] Need help to create HTML table with 2 columns. Date: Thu, 23 Aug 2001 10:41:08 -0400 John, Try this: <TABLE> <?php $query = "SELECT image_link, web_url FROM testdata"; $result = mysql_query ($query) or die ("Query Failed"); while ($row = mysql_fetch_object ($result)) { printf ("<tr><TD>".$row->image_links."<br>".$row->web_url."</TD> <TD>".$row->image_link."<br>".$row->web_url."</tr>\n"); } ?> </TABLE> ----- Original Message ----- From: "John Bass" <php_mysql21@hotmail.com> To: <php-general@lists.php.net> 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. <TABLE> <?php $query = "SELECT image_link, web_url FROM testdata"; $result = mysql_query ($query) or die ("Query Failed"); while ($row = mysql_fetch_object ($result)) { printf ("<tr><TD>%s<br>%s</TD> <TD>%s<br>%s</tr>\n",

$row->image_link,

$row->web_url,$row->image_link, $row->web_url); } ?> </TABLE> Regards, John Bass _________________________________________________________________ Get your FREE download of MSN Explorer at

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