小编Roh*_*war的帖子

我尝试运行命令: app/console cache:clear

这回复了我的错误:

执行"'cache:clear --no-warmup'"命令时
发生错误:PHP致命错误:未捕获的Symfony\Component\Debug\Exception\FatalThrowable
错误:致命错误:在/ var/www中找不到类'DOMDocument'/html/vendor/s
ymfony/symfony/src/Symfony/Component/Config/Util/XmlUtils.php:52

Ubuntu 14.04 LTS我使用PHP 7并安装php5-dom没有解决我的问题: sudo apt-get install php5-dom

JSON 规范：

"responses": {\n          "200": {\n            "description": "\xd0\xa3\xd1\x81\xd0\xbf\xd0\xb5\xd1\x88\xd0\xbd\xd1\x8b\xd0\xb9 \xd0\xbe\xd1\x82\xd0\xb2\xd0\xb5\xd1\x82 \xd1\x81\xd0\xb5\xd1\x80\xd0\xb2\xd0\xb8\xd1\x81\xd0\xb0",\n            "schema": {\n              "$ref": "#/definitions/BaseResponse"\n            },\n            "examples": {\n              "application/json": {\n                "status": true,\n                "response": {\n                  "$ref": "#/definitions/Product"\n                },\n                "errors": null\n              }\n            }\n          }\n}\n

结果：\n

但是我需要：

{\n  "status": true,\n  "response": {\n      "ProductNumber": "number",\n      "Barcode": "number",\n      "Length": 12,\n      "Width": 34,\n      "Height": 423,\n      "Volume": 1232\n    }\n  },\n  "errors": null\n}\n

我如何使用 $refs 到示例数组中进行自定义格式响应？\n这是一个典型案例，但我找不到它的文档。感谢您的反馈。

我的实体FosUserBundle

namespace AppBundle\Entity;

use JMS\Serializer\Annotation\Expose;
use JMS\Serializer\Annotation\Exclude;
use JMS\Serializer\Annotation\ExclusionPolicy;

use Doctrine\ORM\Mapping as ORM;
use FOS\UserBundle\Model\User as BaseUser;
use FOS\UserBundle\Model\Group;

/**
 * User
 *
 * @ORM\Table(name="user")
 * @ORM\Entity(repositoryClass="AppBundle\Repository\UserRepository")
 * @ExclusionPolicy("all")
 */
class User extends BaseUser
{
    /**
     * @ORM\Id
     * @ORM\Column(type="integer")
     * @ORM\GeneratedValue(strategy="AUTO")
     * @Exclude
     */
    protected $id;

    /**
     * @ORM\Column(type="integer")
     */
    private $balance = 0;

但是,如果我尝试序列化App\Entity\User对象:

$this->get('jms_serializer')->serialize($user, 'json');

它归还给我ID字段!

 {
  "id": 1,
  "username": "admin",
  "username_canonical": "admin",
  "email": "admin",
  "email_canonical": "admin",
  "enabled": true,
  "salt": "o12yxgxp3vkk0w4sck80408w8s8o84s",
  "password": "$2y$13$o12yxgxp3vkk0w4sck804uSVjSMSB1W0qwEjunGTHomBqqoGvkW9G",
  "last_login": "2016-02-28T17:28:19+0300",
  "locked": false, …

当我尝试创建它时,我在测试类中看到了警告.我不想为测试类创建PHPDoc和更多,但是PhpStorm向我显示警告,因为PhpStorm不能将测试类标记为phpcs的排除.

我如何运行我的phpcs来验证我的类的编码标准:

php bin/phpcs ./src -p --encoding=utf-8 --extensions=php --ignore=Tests --standard=./vendor/escapestudios/symfony2-coding-standard/Symfony2

如何为PhpStorm 2016.3.2实现它？我尝试创建配置:

<target name="phpcs-ci"
        description="Find coding standard violations using PHP_CodeSniffer creating a log file for the continuous integration server">
    <exec executable="phpcs" output="/dev/null">
        <arg value="--encoding=utf-8" />
        <arg value="--extensions=php" />
        <arg value="--ignore=Tests" />
        <arg value="--standard=./vendor/escapestudios/symfony2-coding-standard/Symfony2" />
        <arg path="${basedir}" />
    </exec>
</target>

并在PhpStorm中导入此配置:

但我的测试再次向我显示警告: