>$service_type_insert = $x['${key}service_type']; //returns empty even
>though I know a value is there
change the single quotes to double, thusly:
$service_type_insert = $x["${key}service_type"];
'cause PHP is using the literal "${key}service_type" and not interpreting it
-----Original Message-----
From: Tom Beidler [mailto:[EMAIL PROTECTED]]
Sent: Friday, January 04, 2002 11:00 AM
Cc: php list
Subject: [PHP] Re: how can I add a variable to this sample
So far nothing seems to work which leads me to believe there's something
else wrong.
Here's what I've tried
$service_type_insert = $x['926service_type']; //works fine when I hard code
a value in and returns proper value
$service_type_insert = $x[$$keyservice_type]; //returns empty even though I
know a value is there
$service_type_insert = $x['${key}service_type']; //returns empty even
though I know a value is there
$service_type_insert = eval($x['$keyservice_type']); //returns empty even
though I know a value is there. I added single quotes because without it
returned a parse error.
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