Re: Problem w/query - again

[Amit Tandon <attand@xxxxxxxxx>]

Dear Ethan

The line you are getting is because the $_POST[fieldname] is blank. So for
the following line
 if ( ! empty( $_POST['field'] ) )
change it to
 if ( ! empty( $_POST["$field"] ) )

Your line : Program is searxchinbg for variable name field
New line : The Program is seacging for varable stored in $field. Rember to
use double quotes

And to verify the value echo $_POST["$field"] before your if line i.e.
 if ( ! empty( $_POST["$field"] ) )

============
regds
amit

"The difference between fiction and reality? Fiction has to make sense."


On Fri, Feb 10, 2012 at 11:04 AM, Ethan Rosenberg <ethros@xxxxxxxxxxxxx>wrote:

> At 12:13 AM 2/10/2012, Amit Tandon wrote:
>
>> Dear Ethan
>>
>> It seems you are trying to build a query.But you are not getting field
>> names. If you required field names then change the following line to
>>
>> foreach ( $allowed_fields AS $field => $_POST['field'])
>> to
>> foreach ( $allowed_fields AS $field)
>>
>> This would convert the variable field to value. In yoyr line the variable
>> field is treated as array index
>> ============
>> regds
>> amit
>>
>> "The difference between fiction and reality? Fiction has to make sense."
>>
>> <snip>
>>
>> >
>> > Advice and help please.
>> >
>> > Thanks.
>> >
>> >
>> > Ethan
>> > PHP Database Mailing List (http://www.php.net/)
>> > To unsubscribe, visit: http://www.php.net/unsub.php
>> >
>> >
>>
>
> Amit -
>
> Thanks.
>
> Tried it.  Still does not work.
>
> This is the query I get:
>
>
> select * from Intake3 where  1
>
> Ethan
>
>