FOSRestBundle

Question

我正在使用 FOSRestBundle。我想 return 用户信息，但数据信息较少...该怎么做？这是我目前的回复

{
    "id": 1,
    "user": {
        "id": 1,
        "username": "qqqq",
        "username_canonical": "qqqq",
        "email": "nie@mam.go",
        "email_canonical": "nie@mam.go",
        "enabled": true,
        "salt": "9ds56htrvvokkckgscs0084so8ss40g",
        "password": "ZcmowLmoNVLNfdMLakzp29G3TZKIfJWVWZ7sE3iylKin5dlYDIQfAJvDC6CFAS2PI4KcpJM0A8qPihDU6jH02A==",
        "last_login": "2016-04-03T18:45:06+0200",
        "locked": false,
        "expired": false,
        "roles": [],
        "credentials_expired": false,
        "created_at": "2016-04-03T18:45:06+0200",
        "updated_at": "2016-04-03T18:45:06+0200"
    },
    "text": "asdadadasd",
    "uv": 0,
    "dv": 0,
    "vote_count": 0,
    "public_i_p": "",
    "private_i_p": "",
    "voters": [],
    "created_at": "-0001-11-30T00:00:00+0124",
    "updated_at": "-0001-11-30T00:00:00+0124"
}

我不想显示所有用户信息，只显示用户名、已锁定、已过期

我的回复

public function getEntryAction(Entry $entry)
{
    return View::create()
        ->setStatusCode(200)
        ->setData($entry);
}

Answer 1

我可以推荐JMSSerializerBundle。它无缝集成，您可以定义要序列化的属性。

Bundle documentation

Annotation reference

Answer 2

完成第一个答案：

This bundle needs a serializer to work correctly. In most cases, you'll need to enable a serializer or install one. This bundle tries the following (in the given order) to determine the serializer to use:

The one you configured using fos_rest.service.serializer (if you did).

The JMS serializer, if the JMSSerializerBundle is available (and registered).

The Symfony Serializer if it's enabled (or any service called serializer).

FOSRestBundle 需要使用 Serializer，如 the documentation 中所述。
@riska 的推荐很好，我从未尝试过内置的序列化程序，但 JMSSerializer 将满足您的需求。

永久或动态查看 exclusion strategies 到 expose/exclude 属性。

FOSRestBundle - 响应数据操作

FOSRestBundle - response data manipulation

php

symfony