php - 将程序代码转换为 Blob 的类

Question

因此，我构建了以下代码，可以从 Azure 存储库中获取一些图像:

$accountName = 'teststorageaccount';
$accountKey = '**';
$containerName = 'users';

$connectionString = "DefaultEndpointsProtocol=http;AccountName={$accountName};AccountKey={$accountKey}";
$blobClient = ServicesBuilder::getInstance()->createBlobService($connectionString);

try {
    $blob_list = $blobClient->listBlobs($containerName);
    $blobs = $blob_list->getBlobs();

    // Grab all the blob links
    foreach ($blobs as $blob) {
        echo $blob->getUrl() . "</br>";
    }
} catch(ServiceException $e) {
    $code = $e->getCode();
    $error_message = $e->getMessage();
    echo $code.": ".$error_message."<br />";
}

然后我得到以下结果:

http://teststorageaccount.blob.core.windows.net/users/ABREUG.jpg
http://teststorageaccount.blob.core.windows.net/users/ABUKHADA.jpg
http://teststorageaccount.blob.core.windows.net/users/ACHANT.jpg
http://teststorageaccount.blob.core.windows.net/users/ACQUISTE.jpg

<小时/>

现在这是我需要一些帮助的地方..我正在练习构建类里面的所有内容，这就是我已经走了多远:

class AzureStorage
{
    private $accountName;
    private $accountKey;
    private $containerName;

    public static function init()
    {
        return new AzureStorage([
            'accountName' => 'teststorageaccount',
            'accountKey' => '***',
            'containerName' => 'users',
        ]);
    }

    /**************************************************************************/

    public function __construct(array $data = [])
    {
        if (count($data) === 0) {
            return;
        }
        $this->load($data);
    }

    public function load(array $data) : void
    {
        if (isset($data['accountName'])) {
            $this->accountName = $data['accountName'];
        }
        if (isset($data['accountKey'])) {
            $this->accountKey = $data['accountKey'];
        }
        if (isset($data['containerName'])) {
            $this->containerName = $data['containerName'];
        }
    }

    public function connect()
    {
        $connectionString = "DefaultEndpointsProtocol=https;AccountName={$this->accountName};AccountKey={$this->accountKey}";
        $blobClient = ServicesBuilder::getInstance()->createBlobService($connectionString);

        return $blobClient;
    }

    public function getContainers() : array
    {
        $containers = $this->connect()->listContainers();
        return $containers->getContainers();
    }

    public function getBlobURLs()
    {
        try {
            $blob_list = $this->connect()->listBlobs($this->containerName);
            $blobs = $blob_list->getBlobs();

            // Grab all the blob links
            foreach ($blobs as $blob) {
                echo $blob->getUrl() . "</br>";
            }
        } catch (ServiceException $e) {
            $code = $e->getCode();
            $error_message = $e->getMessage();
            echo $code.": ".$error_message."<br />";
        }
    }
}

<小时/>

问题:我如何在 getBlobURLs 方法中使用 try 和 catch，然后输出结果？我在类外调用它时无法获得结果。

这就是我正在做的事情:

$test2 = new \AzureStorage\AzureStorage([
    'accountName' => 'teststorageaccount',
    'accountKey' => '***',
    'containerName' => 'users',
]);

现在，如果我调用以下命令(我得到一个容器数组，它工作得很好):

$containers = $test2->getContainers();
//var_dump($containers);

但是如果我执行以下操作(我没有得到任何结果输出):

$blobs = $test2->getBlobURLs();
var_dump($blobs);

有谁知道为什么我可能无法取回网址？

Answer 1

您在 getContainers 函数中有 return 语句，但在 getBlobURLs 中没有。

public function getBlobURLs(){
    try {
        $blob_list = $this->connect()->listBlobs($this->containerName);
        $blobs = $blob_list->getBlobs();

        $blobUrls = [];
        // Grab all the blob links
        foreach ($blobs as $blob) {
            $blobUrls[] = $blob->getUrl();
        }
        return $blobUrls;
    } catch (ServiceException $e) {
        return false; 
    }
}

现在，如果你想显示 blob url 列表，那么

$blobs = $test2->getBlobURLs();

if($blobs === false){
   echo 'Error while getting blob urls.';
}
else{
    foreach($blobs as $blobUrl){
       echo $blobUrl . "</br>";
    }
}