The manual says the second parameter needs to be an array. I assume it
is not, but you have not shown us how $type is assigned so we cannot tell.
HTH
Chris
Rw wrote:
>This is a continue from this morning (thanks so much for the responses)..
>yielding a data type mismatch:
>
> $CheckArr = arra
That did it!
Thanks!
- Original Message -
From: "Martin Clifford" <[EMAIL PROTECTED]>
To: <[EMAIL PROTECTED]>; <[EMAIL PROTECTED]>
Sent: Thursday, July 11, 2002 1:32 PM
Subject: Re: [PHP] Newbie continued..wrong datatype
As far as I'm aware, t
As far as I'm aware, the first argument to the in_array() function is the needle (what
you're searching for) and the second is the array to be searched through (the
haystack).
So if $type represents what you're searching for, then it would be written as:
in_array($type, $CheckArr);
If you sup
This is a continue from this morning (thanks so much for the responses)..
yielding a data type mismatch:
$CheckArr = array("Periodic", "Sale", "Return");
IF (SUBSTR($approvalcode,0,1) == "Y" && in_array($CheckArr, $type))
{
PRINT "$approvalcode";
PRINT " ";
PRINT "$type";
}
Th
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