[RESOLVED] Form select values as WHERE (column) in query

[InfoBot]

I need to pass values from a form that includes this drop-down selection:

<select name="[COLOR=Sienna]$searchby[/COLOR]">
<option value="null"> Select one
<option value="[COLOR=DarkRed]a.name[/COLOR]"> Name
<option value="[COLOR=DarkRed]a.addr[/COLOR]" > Address
<option value="[COLOR=DarkRed]a.zip[/COLOR]" > Zip code
</select>

to a php/mysql querry where it will serve as the column identifier in a WHERE statement:

SELECT a.name, a.addr, a.zip from table
WHERE [COLOR=Sienna]$searchby[/COLOR] like '$search%'

I have the "$search%" variable passing fine, thanks to folks on this board. When I try to make the column variable, though, I'm getting "Check for... right syntax to use near 'like '%' "

I've read manuals, searched Web and tried several other approaches to get the column variable passed but to no avail. It seems I need to set up an array, but I get an sql error that tells me something like there is "no column named array".

I'm sure this is elementary. Any ideas?

[crazychris]

Drop the $ off the select name

<select name="searchby">
<option value="null">Select one</option>
<option value="a.name">Name</option>
<option value="a.addr"> Address</option>
<option value="a.zip">Zip code</option>
</select>

when the script recieves the form, the searchby field becomes $searchby (actually, better to use $_POST['searchby'])

[InfoBot]

Thanks. I had figured this one out by the time I checked back here, but your reply reinforces what I'm learning.

I snuck in a couple of other errors during the process, so I iterated back from my last stable version and, with the name properly declared, it now works.

Gracias

[crazychris]

hehe, no worries.

you can flag this thread as resolved then?