[PHP] Newbie continued..wrong datatype

2002-07-11 Thread Rw 

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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Re: [PHP] Newbie continued..wrong datatype

2002-07-11 Thread Martin Clifford 

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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Re: [PHP] Newbie continued..wrong datatype

2002-07-11 Thread Rw 

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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Re: [PHP] Newbie continued..wrong datatype

2002-07-11 Thread Chris Hewitt 

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





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