get_resource_type

(PHP 4 >= 4.0.2, PHP 5)

get_resource_type — 자원형을 반환

설명

string get_resource_type ( resource $handle )

주어진 자원의 자료형을 얻습니다.

인수

handle: 평가할 자원 핸들.

반환값

주어진 handle이 자원이면, 이 함수는 그 자료형의 문자열 표현을 반환합니다. 자료형을 식별할 수 없으면, 반환값은 문자열 Unknown입니다.

handle이 resource가 아닐 경우, FALSE를 반환하고 오류를 발생합니다.

예제

Example #1 get_resource_type() 예제


<?php
// 출력: mysql link
$c = mysql_connect();
echo get_resource_type($c) . "\n";

// 출력: file
$fp = fopen("foo", "w");
echo get_resource_type($fp) . "\n";

// 출력: domxml document
$doc = new_xmldoc("1.0");
echo get_resource_type($doc->doc) . "\n";
?>

gettype

get_defined_vars

[edit] Last updated: Sat, 07 Jan 2012

add a note User Contributed Notes get_resource_type

avp200681 at gmail dot com 12-Jun-2011 09:11


This function will generate an error and return NULL instead FALSE in the next example:



<?php



$var = 3;

var_dump(get_resource_type($var));



?>

hessodreamy at gmail dot com 28-May-2007 10:07


Furthermore, this function returns 'stream' on php 4.3 on windows & linux.

You if you want to differentiate between a file resource and other stream resources, as stated before stream_get_meta_data() gives you an array which, on my system, contains the following elements which might suggest a file handle:

[wrapper_type] => plainfile

[stream_type] => STDIO



However I've only compared this to a stream from a url, so I include this only as a suggestion.

jdhawk _at_ gmail 05-Oct-2005 07:18


In PHP5 (on Linux), the above example identifies the "file" handler as a 'stream', not 'file'.



This can be particularly worrisome if you're trying to determine the type of stream using this function.



The alternative is to use <?php stream_get_meta_data() ?>, which returns an array containing "stream_type (string) - a label describing the underlying implementation of the stream."

add a note