Getting there now....... all looks well, very tidy etc.
However my drop down box is empty ?? I know there is data in the table I am running the select on though ??? Any ideas ? Thanks, Matt. -----Original Message----- From: Martin Towell [mailto:[EMAIL PROTECTED]] Sent: 24 February 2002 23:43 To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED] Subject: RE: [PHP] RE: (HTML related) helping to creating dropdown lists from fetched arrays. original line: while ($row = mysql_fetch_array($results); new line: while ($row = mysql_fetch_array($results)) -----Original Message----- From: Matthew Darcy [mailto:[EMAIL PROTECTED]] Sent: Monday, February 25, 2002 10:37 AM To: Martin Towell; [EMAIL PROTECTED] Subject: RE: [PHP] RE: (HTML related) helping to creating dropdown lists from fetched arrays. RE: [PHP] RE: (HTML related) helping to creating dropdown lists from fetched arrays.Thanks Martin. That looks perfect. I have adaprted what you sent to use in my test application and I am getting an error on line 22 which is the $row= line. Take a look and see if you can spot my error as it all looks fine to me (again !!!!) Thanks, Matt. <HTML> <HEAD> <META NAME="GENERATOR" Content="vi"> <TITLE> Search for food</TITLE> </HEAD> <BODY BGCOLOR=#ffffff> <CENTER> <TABLE BGCOLOR=#191970 WIDTH=400 HEIGHT=20> <TR WIDTH=400 HEIGHT=10 BGCOLOR=#ff9900> <TD WIDTH=400> <FACE=arial SIZE=4 COLOR=#191970><CENTER> food Search </CENTER></FONT></TD> </TR> <TR HEIGHT=40 WIDTH=400> <TD><FONT FACE=arial SIZE=3 COLOR=#ff9900> <select name="food_type_drop_down"> <?php $sql_select_food_type="select * from food_type_menu"; $results=mysql_query($sql_select_food_type); while ($row = mysql_fetch_array($results); { echo "<option VALUE=".$row["food_search_field"].">".$row["food_type"]; } ?> </select> </TD> </TABLE> </BODY> </HTML> -----Original Message----- From: Martin Towell [mailto:[EMAIL PROTECTED]] Sent: 24 February 2002 23:13 To: '[EMAIL PROTECTED]'; [EMAIL PROTECTED] Subject: RE: [PHP] RE: (HTML related) helping to creating dropdown lists from fetched arrays. try this: (I don't have experience w/ mysql, but I do with other dbs) <select name="test drop down"> <?php $sql_select = "select * from dropdown_options"; $results = mysql_query($sql_select); while ($row = mysql_fetch_array($results); { echo "<option VALUE=".$row["col1"].">".$row["col2"]; } ?> </select> -----Original Message----- From: Matthew Darcy [mailto:[EMAIL PROTECTED]] Sent: Monday, February 25, 2002 10:06 AM To: [EMAIL PROTECTED] Subject: [PHP] RE: (HTML related) helping to creating dropdown lists from fetched arrays. it does help a little. I know the html is wrong I used this as an example to what I wanted. What I need to know is how to use PHP to generate the options in the list from col1 and show the option 2. The data for these is got from an array. so really what I want to know is how to code in PHP <select name=test drop down> <option value=col1($row1)>col2($row1) <option value=col1($row2)>col2($row2) <option value=col1($row3)>col2($row3) </select> you get the idea ? Thanks, Matt. -----Original Message----- From: Cece [mailto:[EMAIL PROTECTED]]On Behalf Of Heilig (Cece) Szabolcs Sent: 24 February 2002 23:07 To: [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Subject: OT:(HTML related) helping to creating dropdown lists from fetched arrays. > I want to create a dropdown list with options from a table. > <?php > > $sql_select = "select * from dropdown_options"; > $results = mysql_query($sql > _select); > > while ($row = mysql_fetch_array($results); > { > echo "<option VALUE=$row["col1"] NAME=option1>" > } Hi! The problem is simple: wrong HTML You have to make html code similar that: <select name=gender> <option value=F>female <option value=M>male <option value=0>not known :) </select> Hope it helps Cece -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php -- PHP General Mailing List (http://www.php.net/) To unsubscribe, visit: http://www.php.net/unsub.php
