扩展PHPUnit

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如何创建一个扩展PHPUnit_Framework_TestCase的基类，并将其用于子类化实际的测试用例，而无需由PHPUnit测试基类本身?

How can I create a base class that extends PHPUnit_Framework_TestCase and use that for subclassing actual test cases, without having the base class itself tested by PHPUnit?

我有一系列相关的测试用例，为此我创建了一个基类，该基类包含一些将由所有测试用例继承的通用测试:

I have a series of related test cases for which I have created a base class that contains some common tests to be inherited by all test cases:

BaseClass_TestCase.php: class BaseClass_TestCase extends PHPUnit_Framework_TestCase { function test_common() { // Test that should be run for all derived test cases } } MyTestCase1Test.php: include 'BaseClass_TestCase.php'; class MyTestCase1 extends BaseClass_TestCase { function setUp() { // Setting up } function test_this() { // Test particular to MyTestCase1 } } MyTestCase2Test.php: include 'BaseClass_TestCase.php'; class MyTestCase2 extends BaseClass_TestCase { function setUp() { // Setting up } function test_this() { // Test particular to MyTestCase2 } }

我的问题是，当我尝试运行文件夹中的所有测试时，它失败了(没有输出).

My problem is that when I try to run all the tests in the folder, it fails (without output).

尝试调试我发现问题出在基类本身就是PHPUnit_Framework_TestCase的子类，因此PHPUnit也将尝试运行其测试. (直到那时，我天真地认为只有实际测试文件中定义的类(即以Test.php结尾的文件名)才可以被测试.)

Trying to debug I've found that the problem lies with the base class being itself a subclass of PHPUnit_Framework_TestCase and therefore PHPUnit will also try to run its tests. (Until then I naively thought that only classes defined inside actual test files - filenames ending in Test.php - would be tested.)

由于特定实现中的细节，无法在上下文中将基类作为测试用例运行.

Running the base class as a test case out of context doesn't work due to details in my specific implementation.

如何避免测试基类，而仅测试派生类?

How can I avoid the base class being tested, and only test the derived classes?

推荐答案

将其抽象化，PHPUnit应该忽略它.

Make it abstract, PHPUnit should ignore it.