Re: newbie PHP/HTML pull down menu problem

Something like this should work: after mysql_connect() and mysql_select_db() ... $sql = "SELECT color_id,color FROM mytable"; $result = mysql_query($sql); echo '<form blah,blah,blah>'; echo '<select name="color_id">'; while ($row = mysql_fetch_array($result)){ printf("<option value=\"%s\">",$row["color_id"]); printf("%s</option>",$row["color"]); } echo '</select>'; echo '</form>'; HTH, - E PS Check the manual for printf() if you're not sure how it works... http://www.php.net/manual/en/function.printf.php

Hi, I want to pull information out of my mySQL database and use it as a pull down menu ona web page. There are two sets of information that I need, one

is a color, and the other is the color_id, so I want to use it in the way below but without having to physically enter all the data. I'd like to use

color_id as the value and the actual color name as the text in the pull down

button. Can someone please help me on how to pass the info into a useable way to get the info out of PHP and into HTML? I've haad no problems going from HTML to PHP, but can't seem to grasp going the other way, especially with a list of things. I've been banging my head against this for two days,

but something tells me it's extremely easy. Thanks, Brian <form name="tester" action="script.php" method="get"> <select name="color"> <option value="1">red <option value="2">blue <option value="3">black <option value="4">ice <option value="5">purple </select> </form> -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php

_________________________________________________________________ きっと見つかるあなたの新居　不動産情報は MSN 住宅で http://house.msn.co.jp/