PHP PHPUnit assertIsNotResource()用法及代码示例

assertIsNotResource()函数是PHPUnit中的内置函数，用于断言给定变量是否不是Resource。如果给定变量不是Resource，则此断言将返回true，否则返回false。如果为真，则通过断言的测试用例，否则测试用例失败。

用法:

assertIsNotResource($actual[, $message = ''])

参数：此函数接受上述和以下描述的两个参数：

• $variable:此参数可以是代表实际数据的任何类型的变量。
• $message:此参数采用字符串值。当测试用例失败时，此字符串消息将显示为错误消息。

以下示例说明了PHPUnit中的assertIsNotResource()函数：

范例1：

<?php  
use PHPUnit\Framework\TestCase;  
    
class GeeksPhpunitTestCase extends TestCase  
{  
    public function testNegativeTestcaseForassertIsNotResource() 
    {  
       $variable =  fopen('http://www.google.com','r'); 
        // Assert function to test whether assert 
        //  variable is not resource 
        $this->assertIsNotResource( 
            $variable, 
            "assert variable is not resource"
        ); 
       fclose($variable); 
    } 
   
 }  
?> 

输出：

PHPUnit 8.5.8 by Sebastian Bergmann and contributors.

F                                                  1 / 1 (100%)

Time:759 ms, Memory:10.00 MB

There was 1 failure:

1) GeeksPhpunitTestCase::testNegativeTestcaseForassertIsNotResource
assert variable is not resource
Failed asserting that resource(1227) of type (stream) is not of type "resource".

/home/lovely/Documents/php/test.php:14

FAILURES!
Tests:1, Assertions:1, Failures:1.

范例2：

<?php  
use PHPUnit\Framework\TestCase;  
    
class GeeksPhpunitTestCase extends TestCase  
{  
    public function testNegativeTestcaseForassertIsNotResource() 
    {  
       $variable =  555; 
        // Assert function to test whether assert 
        //  variable is not resource 
        $this->assertIsNotResource( 
            $variable, 
            "assert variable is not resource"
        ); 
       
    } 
   
 }  
?> 

输出：

PHPUnit 8.5.8 by Sebastian Bergmann and contributors.

.                                                  1 / 1 (100%)

Time:88 ms, Memory:10.00 MB

OK (1 test, 1 assertion)