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我怎样才能将这个库包含https://github.com/troydavisson/PHRETS在一个 symfony 项目中？

这是我的composer.json：

{
    "name": "marysalcedo/reactive-pandora",
    "license": "proprietary",
    "type": "project",
    "autoload": {
        "psr-0": {
            "": "src/",
            "SymfonyStandard": "app/"
        },
        "files": []
    },
    "require": {
        "php": ">=5.3.3",
        "symfony/symfony": "2.6.*",
        "doctrine/orm": "~2.2,>=2.2.3,<2.5",
        "doctrine/dbal": "<2.5",
        "twig/extensions": "~1.0",
        "symfony/assetic-bundle": "~2.3",
        "symfony/swiftmailer-bundle": "~2.3",
        "symfony/monolog-bundle": "~2.4",
        "sensio/distribution-bundle": "~3.0,>=3.0.12",
        "sensio/framework-extra-bundle": "~3.0,>=3.0.2",
        "incenteev/composer-parameter-handler": "~2.0",
        "doctrine/mongodb-odm": "1.0.*@dev",
        "doctrine/mongodb-odm-bundle": "3.0.*@dev",
        "maciejczyzewski/bottomline": "*",
        "doctrine/doctrine-fixtures-bundle": "2.2.*",
        "troydavisson/phrets": "2.*"
    },
    "minimum-stability": "dev",
    "require-dev": {
        "sensio/generator-bundle": "~2.3"
    },
    "scripts": {
        "post-root-package-install": [
            "SymfonyStandard\\Composer::hookRootPackageInstall"
        ],
        "post-install-cmd": [
            "Incenteev\\ParameterHandler\\ScriptHandler::buildParameters",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::buildBootstrap",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::clearCache",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::installAssets",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::installRequirementsFile",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::removeSymfonyStandardFiles",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::prepareDeploymentTarget"
        ],
        "post-update-cmd": [
            "Incenteev\\ParameterHandler\\ScriptHandler::buildParameters",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::buildBootstrap",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::clearCache",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::installAssets",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::installRequirementsFile",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::removeSymfonyStandardFiles",
            "Sensio\\Bundle\\DistributionBundle\\Composer\\ScriptHandler::prepareDeploymentTarget"
        ]
    },
    "config": {
        "bin-dir": "bin"
    },
    "extra": {
        "symfony-app-dir": "app",
        "symfony-web-dir": "web",
        "symfony-assets-install": "relative",
        "incenteev-parameters": {
            "file": "app/config/parameters.yml"
        }
    }
}

每当我尝试使用对象 \PHRETS\Configuration 时，都会出现错误：

new \PHRETS\Configuration();

这是我回来的错误：

[Symfony\Component\Debug\Exception\ClassNotFoundException]                   
  Attempted to load class "Configuration" from namespace "PHRETS".             
  Did you forget a "use" statement for e.g.

为什么不包括在内？

4

1 回答 1

1

你确定你真的安装了这个包吗？

1）我使用了自定义的基于作曲家的项目，并尝试创建该类的新实例并且它有效。输出如下。

Symfony22) 然后我安装了( v2.6) 和包的新副本。尝试创建实例，它也工作得很好。

尝试这个：

php composer.phar update

然后在你的控制器中：

$c = new \PHRETS\Configuration();
var_dump($c);

我的设置中的输出：

object(PHRETS\Configuration)[570]
  protected 'username' => null
  protected 'password' => null
  protected 'login_url' => null
  protected 'user_agent' => string 'PHRETS/2.0' (length=10)
  protected 'user_agent_password' => null
  protected 'rets_version' => 
    object(PHRETS\Versions\RETSVersion)[571]
      protected 'number' => string '1.5' (length=3)
      protected 'valid_versions' => 
        array (size=5)
          0 => string '1.5' (length=3)
          1 => string '1.7' (length=3)
          2 => string '1.7.1' (length=5)
          3 => string '1.7.2' (length=5)
          4 => string '1.8' (length=3)
  protected 'http_authentication' => string 'digest' (length=6)
  protected 'strategy' => null
  protected 'options' => 
    array (size=0)
      empty

边注

只是为了消除对use陈述的混淆。

  • 当use您指定类的FQN.

  • 如果你省略了前导反斜杠，PHP 将尝试在你的use语句中找到一个与命名空间匹配的别名。如果没有找到，它将假定当前类的命名空间。以众所周知的Doctrine映射示例：

    use `Doctrine\ORM\Mapping` as ORM
....
@ORM\Column() // This resolves to `Doctrine\ORM\Mapping\Column` `FQN`

  • 如果您只使用类名（例如Configuration）use声明将是强制性的。

于 2015-05-06T22:53:17.397 回答