2012-08-15 85 views
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我正在写图像下载服务写在PHP测试用例。我们使用phpunit。如何检查检索到的二进制数据是否为图像?Phpunit图像下载测试

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,以确定是否一个URL是在图像[最好的方式重复PHP的](http://stackoverflow.com/questions/676949/best-way-to-determine-if-a-url-is-an-image-in-php)和http://stackoverflow.com/questions/ 10662915 /检查文件是否是图像或不是和http://stackoverflow.com/questions/6391916/is-it-important-to-verify-that-the-uploaded-file- is-an-an-image-file – 2012-08-15 12:25:51

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'getimagesize()'是某些图像格式的家喻户晓的名称。你需要支持哪些? – 2012-08-15 12:26:11

回答

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使用exif_imagetype(请参阅manual)很好,但确实需要您必须将文件放在本地磁盘上。如果你不介意硬编码一些神奇的数字,你可以检查图像类型直接看到testFetchWithoutSaving在下面的例子:

class ImageTest extends PHPUnit_Framework_TestCase 
{ 

/** 
* @see http://stackoverflow.com/a/676975/841830 
*/ 
public function testFetchWithoutSaving(){ 
    $s=file_get_contents("https://www.google.com/images/srpr/logo3w.png"); 
    $this->assertEquals("\x89PNG\x0d\x0a\x1a\x0a",substr($s,0,8)); 

    $s=file_get_contents("https://www.google.com/"); 
    $this->assertEquals("\x89PNG\x0d\x0a\x1a\x0a",substr($s,0,8),"Fails: first 8 bytes are actually '<!doctyp'"); 
    } 

/** 
* @see http://php.net/manual/en/function.exif-imagetype.php 
*/ 
public function testFetchWithTempFile(){ 
    $s=file_get_contents("https://www.google.com/images/srpr/logo3w.png"); 
    $tempFilename="/tmp/phpunit.testImage.testFetchWithTempFile"; 
    file_put_contents($tempFilename,$s); 
    $type=exif_imagetype($tempFilename); 
    unlink($tempFilename); 
    $this->assertTrue($type!==false); //Any recognized image type 
    $this->assertEquals(IMAGETYPE_PNG,$type); //A specific image type 
    } 

}