小编Ris*_*s90的帖子

我正在尝试通过Apache Kafka加载数据并不断收到此错误:

kafka.common.OffsetOutOfRangeException:偏移量1003786超出了kafka.log.Log $ .findRange(Log.scala:46)的范围,位于kafka.log.Log.read(读取:264),位于kafka.server.KafkaRequestHandlers. kafka $ server $ KafkaRequestHandlers $$ readMessageSet(KafkaRequestHandlers.scala:112)at kafka.server.KafkaRequestHandlers $$ anonfun $ 2.apply(KafkaRequestHandlers.scala:101)at kafka.server.KafkaRequestHandlers $$ anonfun $ 2.apply(KafkaRequestHandlers.scala :100)scala.collection.TraversableLike $$ anonfun $ map $ 1.apply(TraversableLike.scala:206)at scala.collection.TraversableLike $$ anonfun $ map $ 1.apply(TraversableLike.scala:206)at scala.collection.在scala.collection.mutable.ArrayOps.foreach(ArrayOps.scala:34)在scala.collection.TraversableLike $ class.map(TraversableLike.scala:206)在scala.collection:IndexedSeqOptimized $ class.foreach(34 IndexedSeqOptimized.scala) .mutable.ArrayOps.map(ArrayOps.scala:34)位于kafka.server.K的kafka.server.KafkaRequestHandlers.handleMultiFetchRequest(KafkaRequestHandlers.scala:100)afkaRequestHandlers $$ anonfun $ handlerFor $ 3.apply(KafkaRequestHandlers.scala:40)在kafka.server.KafkaRequestHandlers $$ anonfun $ handlerFor $ 3.apply(KafkaRequestHandlers.scala:40)在kafka.network.Processor.handle(SocketServer.scala: 296)at kafka.network.Processor.read(SocketServer.scala:319)at kafka.network.Processor.run(SocketServer.scala:214)at java.lang.Thread.run(Thread.java:724)

这个例外意味着什么以及如何解决它？

我有以下代码:

$packed = pack('i',PHP_INT_MAX);
echo unpack('i', $packed)[1];

结果我得到了 -1

我正在使用PHP 5.4.6-1ubuntu1.1 (cli) (built: Nov 15 2012 01:18:34) ,我PHP_INT_MAx的等于9223372036854775807

有没有办法处理pack函数和64位整数？

我正在尝试将我的项目从symfony 2.1更新到2.2.我尝试逐包更新它.当我输入时composer.phar update symfony/symfony,我收到以下错误:

 Problem 1
- Conclusion: don't install symfony/symfony 2.2.x-dev
- Conclusion: don't install symfony/symfony v2.2.0
- Conclusion: don't install symfony/symfony v2.2.0-RC3
- Conclusion: don't install symfony/symfony v2.2.0-RC2
- Installation request for symfony/monolog-bundle == 2.1.9999999.9999999-dev -> satisfiable by symfony/monolog-bundle 2.1.x-dev.
- Conclusion: don't install symfony/symfony v2.2.0-RC1
- Conclusion: don't install symfony/symfony v2.2.0-BETA2
- doctrine/doctrine-bundle v1.2.0-beta1 requires symfony/framework-bundle >=2.2.0-beta2,<2.3-dev -> satisfiable by symfony/symfony 2.2.x-dev, symfony/symfony v2.2.0, symfony/symfony v2.2.0-BETA2, symfony/symfony v2.2.0-RC1, symfony/symfony v2.2.0-RC2, symfony/symfony v2.2.0-RC3, symfony/framework-bundle 2.2.x-dev, symfony/framework-bundle …

有没有办法在php中将显式类型设置为对象字段？像这样的东西

class House{
       private Roof $roof
}

我有mysql表charset的问题.我的数据库中的每个表都有默认的字符集.例如:

CREATE TABLE privacy_settings (
  id_privacy_setting int(11) NOT NULL AUTO_INCREMENT,
  id_account int(11) NOT NULL,
  setting_name varchar(255) NOT NULL DEFAULT '0',
  privacy_level int(11) NOT NULL DEFAULT '0',
  PRIMARY KEY (id_privacy_setting),
  KEY fk_privacy_settings_accounts (id_account),
  CONSTRAINT fk_privacy_settings_accounts FOREIGN KEY (id_account) REFERENCES accounts (id_account) ON DELETE NO ACTION ON UPDATE NO ACTION
) ENGINE=InnoDB DEFAULT CHARSET=utf8  

我想删除DEFAULT CHARSET块,因此表可以使用数据库默认字符集:

CREATE TABLE privacy_settings (
      id_privacy_setting int(11) NOT NULL AUTO_INCREMENT,
      id_account int(11) NOT NULL,
      setting_name varchar(255) NOT NULL DEFAULT '0',
      privacy_level int(11) NOT NULL DEFAULT '0',
      PRIMARY …

我正在尝试在Symfony 2项目中实现更改密码功能.我User在validation.yml文件中有验证规则的实体.在User实体中,我有字段" password",其中包含验证约束validation.yml.
我创建了带有2个字段' password'和' confirmPasswod'的表单.我想对'password'字段使用我的实体验证约束,并检查' passwod'和' confirmPassword'字段之间的相等性.在我的控制者中,我写道

$form = $this->createForm(new SymfonyForm\ChangePasswordType(), new Entity\User());
if ($form->isValid())
    {..............}

在"用户"实体中,我没有"confirmPasswod"字段.所以我得到错误:

Neither property "confirmPassword" nor method "getConfirmPassword()" nor method "isConfirmPassword()" exists in class

有没有办法对某些表单字段使用基于实体的表单验证而不为其他表单字段使用基于实体的验证？提前致谢.

我正在调查symfony 2框架.在我的示例应用程序中,我有Blog实体和BlogEntry实体.它们与一对多关系相关联.这是BlogEntry类:

class BlogEntry
{
    ....
    private $blog;
    ....
    public function getBlog()
    {
        return $this->blog;
    }

    public function setBlog(Blog $blog)
    {
        $this->blog = $blog;
    }
}

我想将方法​​setBlogByBlogId添加到BlogEntry类,我这样看:

public function setBlogByBlogId($blogId)
    {
        if ($blogId && $blog = $this->getDoctrine()->getEntityManager()->getRepository('AppBlogBundle:Blog')->find($blogId))
        {
            $this->setBlog($blog);
        }
        else
        {
            throw \Exception();
        }
    }

这是否可以在模型类中获得学说？从Symfony 2 MVC架构的角度来看这是正确的吗？或者我应该在我的控制器中执行此操作？