如何在PHPUnit中实现If / Else的代码覆盖率？

Question

Disclaimer: I'm new to unit testing and just getting my feet wet. That being said, I understand 100% is not always possible, but I'm curious if there is a solution to code coverage with if/else statements such as this or if they are just untestable.

The function:

public function getAll() 
{

    // Query table for all rows sorted by name
    $select = $this->select();
    $select->order('name ASC');
    $rowset = $this->fetchAll($select);

    // Validate and return row
    if($rowset->current())
    {   
        // Return rowset
        return $rowset;
    }
    else { return false; }

}

The Unit Test

public function testCanGetAll()
{
    // Try to get all states
    $result = $this->model->getAll();
    $this->assertNotNull($result);
    $this->assertNotEquals(false,$result);
}

The Result

As you can see, I can't cover the return false; line. The only way I can think of to test it is to rename my database table or something drastic like that.

Is there another way to write functions with if/else statements (which are a fairly common practice) which allows for easier code coverage?

[EDIT] Perhaps it's better to not check results in the function, and just return it? Whatever calls this function has to do it's own checking?

public function getAll() 
{

    // Query table for all rows sorted by name
    $select = $this->select();
    $select->order('name ASC');
    $rowset = $this->fetchAll($select);

    // Return rowset whether valid or not
    return $rowset; 

}

Answer 1

Create a db context where the result would return no rows, then in your second test method create a row in that db context.

Its usually not a good idea to test against your actual application's DB

you should be cloning it for your tests so that you can manipulate the data to test how the application will respond to varying data contexts.

--EDIT----

Following this guide may help: http://framework.zend.com/manual/ru/zend.test.phpunit.db.html