©
Dokumen ini menggunakan Manual laman web PHP Cina Lepaskan
(PHP 4, PHP 5, PHP 7)
is_resource — 检测变量是否为资源类型
$var
)
如果给出的参数 var
是 resource 类型, is_resource()
返回 TRUE
,否则返回 FALSE
。
查看 resource 类型文档获取更多的信息。
[#1] CertaiN [2014-01-17 05:05:01]
Try this to know behavior:
<?php
function resource_test($resource, $name) {
echo
'[' . $name. ']',
PHP_EOL,
'(bool)$resource => ',
$resource ? 'TRUE' : 'FALSE',
PHP_EOL,
'get_resource_type($resource) => ',
get_resource_type($resource) ?: 'FALSE',
PHP_EOL,
'is_resoruce($resource) => ',
is_resource($resource) ? 'TRUE' : 'FALSE',
PHP_EOL,
PHP_EOL
;
}
$resource = tmpfile();
resource_test($resource, 'Check Valid Resource');
fclose($resource);
resource_test($resource, 'Check Released Resource');
$resource = null;
resource_test($resource, 'Check NULL');
?>
It will be shown as...
[Check Valid Resource]
(bool)$resource => TRUE
get_resource_type($resource) => stream
is_resoruce($resource) => TRUE
[Check Released Resource]
(bool)$resource => TRUE
get_resource_type($resource) => Unknown
is_resoruce($resource) => FALSE
[Check NULL]
(bool)$resource => FALSE
get_resource_type($resource) => FALSE
Warning: get_resource_type() expects parameter 1 to be resource, null given in ... on line 10
is_resoruce($resource) => FALSE
[#2] btleffler [AT] gmail [DOT] com [2011-05-12 13:55:06]
I was recently trying to loop through some objects and convert them to arrays so that I could encode them to json strings.
I was running into issues when an element of one of my objects was a SoapClient. As it turns out, json_encode() doesn't like any resources to be passed to it. My simple fix was to use is_resource() to determine whether or not the variable I was looking at was a resource.
I quickly realized that is_resource() returns false for two out of the 3 resources that are typically in a SoapClient object. If the resource type is 'Unknown' according to var_dump() and get_resource_type(), is_resource() doesn't think that the variable is a resource!
My work around for this was to use get_resource_type() instead of is_resource(), but that function throws an error if the variable you're checking isn't a resource.
So how are you supposed to know when a variable is a resource if is_resource() is unreliable, and get_resource_type() gives errors if you don't pass it a resource?
I ended up doing something like this:
<?php
function isResource ($possibleResource) { return !is_null(@get_resource_type($possibleResource)); }
?>
The @ operator suppresses the errors thrown by get_resource_type() so it returns null if $possibleResource isn't a resource.
I spent way too long trying to figure this stuff out, so I hope this comment helps someone out if they run into the same problem I did.