Re: json_encode()

2008/12/15 mike <mike503@gmail.com>: > On Mon, Dec 15, 2008 at 9:50 AM, Rasmus Lerdorf <rasmus@lerdorf.com> wrote: > >> 1. Document the fact that if you want to strictly conform to the JSON >> spec and be sure your json_encode output will work in various JSON >> parsers, you have to pass it a PHP array or object. > > Instead of json_encode(34) the suggestion would be > json_encode(array(34)) ? Seems kind of like a lame thing to require. > IMHO the language should "do the right thing" as far as the consumer > (javascript/JSON parser) is concerned. However, Douglas is infinitely > more educated than I am here. This is just my $0.02 as a PHP user. I was one of those that read the PHP dox and not the RFC/standard (http://bugs.php.net/bug.php?id=38680). I would say that having PHP "correct" my mistake is wrong. If I say... json_encode(34); I am saying that I expect the result to be ... var i_SomeInt = 34; And from there I would expect to be able to say ... var i_SomeOtherInt = i_SomeInt + 10; This is not going to work if PHP "corrects" my mistake. I'd be perfectly happy for the standard to be enforced and an E_STRICT warning to be raised. If I want to shoot myself then I have to at least take the safety off first - turning off E_STRICT that is. Essentially, I don't like computers guessing my intent. If I don't state it clearly enough then it may be that I don't know what I'm doing. GIMGO (Garbage In, Maybe Garbage Out) isn't a good way to go. Richard. -- ----- Richard Quadling Zend Certified Engineer : http://zend.com/zce.php?c=ZEND002498&r=213474731 "Standing on the shoulders of some very clever giants!"