At 12:48 AM 2/10/2012, Amit Tandon wrote:
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"] ) )
  Â
============
regds
amit
"The difference between fiction and reality? Fiction has to make sense."
On Fri, Feb 10, 2012 at 11:04 AM, Ethan
Rosenberg <<mailto:ethros@xxxxxxxxxxxxx>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
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Amit -
Thanks.
Tried your edit. Still does not work.
Ethan
>
>
Amit -
Thanks.
Tried it. Â Still does not work.
This is the query I get:
select * from Intake3 where  1
Ethan
->>>>>> Amit -
SORRY.
Works Perfectly!!!!
I had commented out my output routine!!
Ethan
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