Hi,
I have created a block and pasted the following code with a change in your code "removing ." WHERE segment='yes'"" but it is not showing me any result. I also set content type to "PHP Script" but no gain.
I used the following table,

create table segments_table(segment1 varchar(2), segment2 varchar(2),segment3 varchar(2));

inserted values,

insert into segments_table values ('a','b','c');

It is not even showing the block?

Will really appreciate all of you in resolving my issues.

Thanks


global $xoopsDB, $xoopsUser;
echo $xoopsDB;

if ($xoopsUser)
{
$lognom = $xoopsUser->uname();
}

$result = mysql_query("SELECT * FROM ".$xoopsDB->prefix("segments_table") );
while($row = mysql_fetch_array($result))
{

$segment1 = mysql_result($result, "t_row1");
$segment2 = mysql_result($result, "t_row2");
$segment3 = mysql_result($result, "t_row3");

echo "

$segment1 = mysql_result($result, "segment1");
$segment2 = mysql_result($result, "segment2");
$segment3 = mysql_result($result, "segment3");

?

WHERE segment='yes'... this is your_row=something_you_want_to_select

Hi,
I have inserted only 3 records and I want to display all 3 so no need to specify where at the moment. Yes you are right later I will use it but right now I just want an output.

I am also getting error,

Parse error: syntax error, unexpected '<' in c:\apache\Apache\htdocs\html\class\xoopsblock.php(146) : eval()'d code on line 1


Thanks

Hi All,
I am still unable to resolve my issue. Well let me write my requirement again in a more simple way,

Say I have a table "mytable" having one field "myfield" and the table contains only one record. I want to show this one record in the XOOPS block on the main page. Please guide me how to do this?

Thanks

Pl let me know if I have to post/move this requirement in any other forum?