selecting a row to display in HTML

[csparker1]

Hello,

I want to use a drop down menu that contains the first field data from each row as the menu selections. When you select an option in the menu I want it to bring up a HTML page that displays the data thats in the entire row that was selected.

I am new to PHP and would appreciate any help. Thank you

[matto]

I use a function for generating select menus:

function getsqlselect($selname, $sql, $mark="", $nul=false, $onchange=false)
{
if($onchange)
$odazu = " onChange=\"submit()\"";

    $qsel = db_query($sql);
    $sel = "<select name=$selname size=1$odazu>\n";

    if($nul)
            $sel .= " <option value=0></option>\n";

    for($i=0; $i < count($qsel); $i++)
    {
            $satz = $qsel[$i];
            if($satz[1] == "") $satz[1] = $satz[0];

            if($mark != "" && $satz[0] == $mark)
                    $dazwischen = " selected";
            else
                    $dazwischen = "";

            $sel .= " <option value=\"".$satz[0]."\"".$dazwischen.">".dirt($satz[1])."</option>\n";
    }
    $sel .= "</select>\n";

    return $sel;

}

but let me explain:

• this function uses an db abstraction function named db_query. you give it an sql select statement, it returns a 2-dim array (1 record per element, each record consisting of n fields) you'll have to insert your own db query code here.
• it expects an sql statement like:
  select id,name from persons where ...
  (first field is the option value, second the "nice" name)
• selname = the html name that will be given to the select menu
• mark = the value that will be marked as selected
• nul: if set true, an empty option will be inserted above the others: <option value=0></option>
• onChange: inserts onChange=submit(), if set true

regarding record display:
you put such a select on a html page in a form. when submitted, a second php script generates a html page which displays the record that has eben selected in the select menu.

you've given the select a $selname, say id_person.
in the script that's now being called, build this variable into your select statement, and execute:
$sql = "select * from person where id=$id_person";

now display this data in a table, or whatever, done!

if you're new to php, try to understand the code, experiment a bit ...