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:
>
> $CheckArr = array("Periodic", "Sale", "Return");
> IF (SUBSTR($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
>
>
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
