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    Junior Member
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    I've got a form with several select Menus which contain options that I retrieve from a sybase database. I thougt everytime I need to print the menu I wrap the option tag in php and println it out. But I recognized that the second php-Block is not executed when I tested an echo in that block. Thank you very much for the input.
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    Gödelian monster
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    You can have as many separate blocks of PHP code in a page as you want as long as you enclose each instance in php script tags. You can even drop in and out of PHP parsing inside a function or a loop:

    <BLOCKQUOTE><font size="1" face="Verdana,Arial,Helvetica">code:</font><HR><pre>

    <SELECT name="whatever">


    while($r = mysql_fetch_array($result)){

    ?><option value="<? echo $r["column1"] ?>"><? echo $r["column2"] ?></option>




    (Should work the same for Sybase)

    However, AFAIK, PHP doesn't have a println function, just echo, print, and printf.

    [This message has been edited by rycamor (edited October 13, 2000).]

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