PHP 7.1.5 Released


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is_resource 检测变量是否为资源类型


bool is_resource ( mixed $var )

如果给出的参数 varresource 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。

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User Contributed Notes 2 notes

btleffler [AT] gmail [DOT] com
6 years ago
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:


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.
3 years ago
Try this to know behavior:

function resource_test($resource, $name) {
'[' . $name. ']',
'(bool)$resource => ',
$resource ? 'TRUE' : 'FALSE',
'get_resource_type($resource) => ',
get_resource_type($resource) ?: 'FALSE',
'is_resoruce($resource) => ',
is_resource($resource) ? 'TRUE' : 'FALSE',

$resource = tmpfile();
resource_test($resource, 'Check Valid 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
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