Re: [PHP] Strange behaviour with str_ireplace()

2010-01-26 Thread Angus Mann

Grrr...please ignore.

After battling with this for a while I now can't reproduce the problem.

Apologies for an un-necessary post.

- Original Message - 
From: "Angus Mann" 

To: "PHP-General" 
Sent: Wednesday, January 27, 2010 9:02 AM
Subject: [PHP] Strange behaviour with str_ireplace()


Hi all. I found some unexpected behaviour with str_ireplace.

To summarize, if the $replace term is a number, but formatted as currency 
with trailing zero's (34.50) the function drops the trailing zero.


Example :

$price=34.5;
$message="You owe me #amount#.";
$message1 = str_ireplace("#amount#,sprintf("%01.2f",$price),$message);
$message2 = str_ireplace("#amount#,"$".sprintf("%01.2f",$price),$message);


$message1 will yield "You owe me 34.4"
$message2 will yield "You owe me $34.40"

So in other words, if the replace term can be interpreted as a number, it 
will be, including truncation of zero's.
I didn't expect this, because as the function name suggests, it's a *string* 
replace.


Fortunately I can just add a $ to the amount and make it work as a string, 
not a number but this would not always be the case.


Is this by design, or is it a bug?



--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php



[PHP] Strange behaviour with str_ireplace()

2010-01-26 Thread Angus Mann
Hi all. I found some unexpected behaviour with str_ireplace.

To summarize, if the $replace term is a number, but formatted as currency with 
trailing zero's (34.50) the function drops the trailing zero.

Example :

$price=34.5;
$message="You owe me #amount#.";
$message1 = str_ireplace("#amount#,sprintf("%01.2f",$price),$message);
$message2 = str_ireplace("#amount#,"$".sprintf("%01.2f",$price),$message);


$message1 will yield "You owe me 34.4"
$message2 will yield "You owe me $34.40"

So in other words, if the replace term can be interpreted as a number, it will 
be, including truncation of zero's.
I didn't expect this, because as the function name suggests, it's a *string* 
replace.

Fortunately I can just add a $ to the amount and make it work as a string, not 
a number but this would not always be the case.

Is this by design, or is it a bug?