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, 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 supply the third argument as TRUE, then the search becomes case
sensitive, I believe.
HTH!
>>> "Rw" <[EMAIL PROTECTED]> 07/11/02 02:25PM >>>
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, $type))
{
PRINT "<BR>$approvalcode";
PRINT " ";
PRINT "$type";
}
This line: IF (SUBSTR($approvalcode,0,1) == "Y" && in_array($CheckArr,
$type))
Causes this error:
Warning: Wrong datatype for second argument in call to in_array in
/home/www/globalspacesolutions/php3/billingtrx.php on line 47
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