blocks|key|3595230|text|你也许应该用一个正常的顺序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3595231|但是，如果必须将日历信息用于键，则可能需要将其存储为“string”类型，然后使用php的DateTime格式：|3595232|+$currentData->setDate(new+\DateTime($dateStr)->format('yyyy/mm/dd');|code-block|syntax|javascript|3595233|如果您尝试使用datetime对象作为键，您可能会陷入困境。|3595234|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|O|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

You probably should just use a normal sequence. 

However, if you must use calendar information to key, you might want to store it as a 'string' type and then use php's DateTime Format:

<pre><code> $currentData-&gt;setDate(new \DateTime($dateStr)-&gt;format('yyyy/mm/dd');
</code></pre>

You are probably headed for misery if you try to use a datetime object for key.

blocks|key|3396480|text|我在这里也有同样的问题。我用这个来解决这个问题：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3396481|/**
+*+@var+string
+*
+*+@ORM\Id
+*+@ORM\Column(type="string")
+*/
private+$date;

/**
+*+@return+\DateTime
+*/
public+function+getDate()
{
++++return+\DateTime::createFromFormat('Y-m-d%7C',+$this->date);
}

/**
+*+@param+\DateTime+$date
+*/
public+function+__construct(\DateTime+$date)
{
++++$this->date+=+$date->format('Y-m-d');
}|code-block|syntax|javascript|3396482|如果要使用datetime，则应该使用不同的格式，如\DateTime：：if+8601。在使用时区保存东西时要小心。|3396483|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I had the same problem here. I worked around it by using this:

<pre><code>/**
 * @var string
 *
 * @ORM\Id
 * @ORM\Column(type="string")
 */
private $date;

/**
 * @return \DateTime
 */
public function getDate()
{
 return \DateTime::createFromFormat('Y-m-d|', $this-&gt;date);
}

/**
 * @param \DateTime $date
 */
public function __construct(\DateTime $date)
{
 $this-&gt;date = $date-&gt;format('Y-m-d');
}
</code></pre>

if you want to use datetime, you should use a different format like \DateTime::ISO8601. Be careful at saving stuff with timezones.

blocks|key|2559030|text|这方面的一个roubust解决方案是使用实现了DateTime+()的__toString()子代来实现自己的DBAL类型：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2559031|<?php
class+DateKey+extends+\DateTime{
++++function+__toString()+{
++++++++return+$this->format('c');
++++}

++++static+function+fromDateTime(\DateTime+$dateTime)+{
++++++++return+new+static($dateTime->format('c'));
++++}
}

class+DateKeyType+extends+\Doctrine\DBAL\Types\DateType{
++++public+function+convertToPHPValue($value,+\Doctrine\DBAL\Platforms\AbstractPlatform+$platform)+{
++++++++$value+=+parent::convertToPHPValue($value,+$platform);
++++++++if+($value+!==+NULL)+{
++++++++++++$value+=+DateKey::fromDateTime($value);
++++++++}
++++++++return+$value;
++++}
++++public+function+getName()
++++{
++++++++return+'DateKey';
++++}
}

\Doctrine\DBAL\Types\Type::addType('datekey',+'DateKeyType');
//edit:+do+not+forget+this+after+creating+entity+manager.
//otherwise,+you+will+get+into+problems+with+doctrine+database+diff+/+migrations.
$platform+=+$entityManager->getConnection()->getDatabasePlatform();
$platform->registerDoctrineTypeMapping('datekey',+'datekey');
$platform->markDoctrineTypeCommented(\Doctrine\DBAL\Types\Type::getType('datekey'));|code-block|syntax|javascript|2559032|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

A roubust solution to this is to implement your own DBAL type, using a DateTime descendant with __toString() implemented:

<pre><code>&lt;?php
class DateKey extends \DateTime{
 function __toString() {
 return $this-&gt;format('c');
 }

 static function fromDateTime(\DateTime $dateTime) {
 return new static($dateTime-&gt;format('c'));
 }
}

class DateKeyType extends \Doctrine\DBAL\Types\DateType{
 public function convertToPHPValue($value, \Doctrine\DBAL\Platforms\AbstractPlatform $platform) {
 $value = parent::convertToPHPValue($value, $platform);
 if ($value !== NULL) {
 $value = DateKey::fromDateTime($value);
 }
 return $value;
 }
 public function getName()
 {
 return 'DateKey';
 }
}

\Doctrine\DBAL\Types\Type::addType('datekey', 'DateKeyType');
//edit: do not forget this after creating entity manager.
//otherwise, you will get into problems with doctrine database diff / migrations.
$platform = $entityManager-&gt;getConnection()-&gt;getDatabasePlatform();
$platform-&gt;registerDoctrineTypeMapping('datekey', 'datekey');
$platform-&gt;markDoctrineTypeCommented(\Doctrine\DBAL\Types\Type::getType('datekey'));
</code></pre>

I am trying to make an Entity using a date as a primary key. The problem is that Symfony can't convert the DateTime I'm using into a string to introduce it in the IdentityMap. I get the following error during the persist of the entity:

<pre><code>Catchable Fatal Error: Object of class DateTime could not be converted to string in..
</code></pre>

I'm using this code in the entity:

<pre><code>/**
 * @ORM\Id
 * @ORM\Column(type="datetime")
 */
protected $date;
</code></pre>

The error appears in the entity repository:

<pre><code>$em = $this-&gt;getEntityManager();
$currentData = new CurrentData();
...
$currentData-&gt;setDate(new \DateTime($dateStr));
...
$em-&gt;persist($currentData);
$em-&gt;flush();
</code></pre>

How can I solve this problem? Thank you.

Symfony/Doctrine: DateTime as primary key

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋 

腾讯云代码助手

CODING DevOps

Cloud Studio

SDK中心

API中心

命令行工具

我试图使用日期作为主键来创建一个实体。问题是，Symfony不能将我使用的DateTime转换为字符串，以便在IdentityMap中引入它。在实体持久化过程中，我得到以下错误：Catchable Fatal Error: Object of class DateTime could not be converted ...

问Symfony/Doctrine: DateTime作为主键
EN

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问Symfony/Doctrine: DateTime作为主键EN