ohn W. Holmes [mailto:[EMAIL PROTECTED]
-> Verzonden: vrijdag 12 september 2003 19:06
-> Aan: Wouter van Vliet; Chris Boget; Thomas Panzarella;
-> [EMAIL PROTECTED]
-> Onderwerp: Re: [PHP] Simple (?) var_dump question
->
->
-> From: "Wouter van Vliet" <[EMAIL P
From: "CPT John W. Holmes" <[EMAIL PROTECTED]>
> The only difference between print_r() and var_export() produces valid PHP
> code.
That should say var_export() produces valid PHP code, while print_r() does
not.
---John Holmes...
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From: "Wouter van Vliet" <[EMAIL PROTECTED]>
> aiaiaia ... you are all pretty right, for another function was forged.
> var_export has a "return" flag as second argument where print_r and
var_dump
> both have other variables as second (and third, and fourth, and sixth) ..
>
> So you can do: $logg
will log the
var to your log.
-> -Oorspronkelijk bericht-
-> Van: Chris Boget [mailto:[EMAIL PROTECTED]
-> Verzonden: vrijdag 12 september 2003 18:34
-> Aan: CPT John W. Holmes; Thomas Panzarella; [EMAIL PROTECTED]
-> Onderwerp: Re: [PHP] Simple (?) var_dump question
> > So, the above line would get a "var_dump
> > stringified" version of $foo and pass it to my
> > logger's debug method.
> You need to use output buffering to capture the output with var_dump.
> Or you can use print_r(), which will give you a similar output and can be
> returned to a varible.
No
From: "Thomas Panzarella" <[EMAIL PROTECTED]>
> Hi. I'm new to the list so sorry if this has already
> been asked before ...
>
> I want to find a way to capture the var_dump
> output as a string so I can write it to a log file ...
> (for example):
>
> $logger->debug(var_dump($foo));
>
> So, the a
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