PHP PHPUnit assertIsNotResource()用法及代碼示例

assertIsNotResource()函數是PHPUnit中的內置函數，用於斷言給定變量是否不是Resource。如果給定變量不是Resource，則此斷言將返回true，否則返回false。如果為真，則通過斷言的測試用例，否則測試用例失敗。

用法:

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

參數：此函數接受上述和以下描述的兩個參數：

• $variable:此參數可以是代表實際數據的任何類型的變量。
• $message:此參數采用字符串值。當測試用例失敗時，此字符串消息將顯示為錯誤消息。

以下示例說明了PHPUnit中的assertIsNotResource()函數：

範例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); 
    } 
   
 }  
?> 

輸出：

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.

範例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"
        ); 
       
    } 
   
 }  
?> 

輸出：

PHPUnit 8.5.8 by Sebastian Bergmann and contributors.

.                                                  1 / 1 (100%)

Time:88 ms, Memory:10.00 MB

OK (1 test, 1 assertion)