Question

我遇到了django的问题，我试图在我的应用程序上为django创建一个数据库我已经设置了这样的连接

DATABASES = {
    'default': {
        'ENGINE': 'django.db.backends.postgresql_psycopg2',
        'NAME': 'Proyecto_OSC',
        'USER': 'posgresql',
        'PASSWORD': 'aa121292',
        'HOST': '',
        'PORT': '5432'
    }
}

然后我尝试创建这样的模型：

class TipoOrganizacion(models.Model):
    nombre = models.CharField(max_length=140)

    def __str__(self):
        return self.nombre

    class Meta:
        verbose_name_plural = "Tipos de Organizacion"


class Actividades(models.Model):
    nombre = models.CharField(max_length=255)

    def __str__(self):
        return self.nombre

    class Meta:
        verbose_name_plural = "Actividades"


class AreasInteres(models.Model):
    nombre = models.CharField(max_length=255)

    def __str__(self):
        return self.nombre

    class Meta:
        verbose_name_plural = "Areas de interes"
        verbose_name = "Area de interes"

class TiposRedesSociales(models.Model):
    tipo = models.CharField(max_length=255)

    def __str__(self):
        return self.tipo

    class Meta:
        verbose_name_plural = "Tipos de redes sociales"
        verbose_name = "tipo de red social"

class RedSocial(models.Model):
    nombre = models.CharField(max_length=255)
    tipo = models.ManyToManyField(TiposRedesSociales)

    def __str__(self):
        return self.nombre

    class Meta:
        verbose_name_plural = "Redes sociales"
        verbose_name = "Red social"

class Organizacion(models.Model):
    nombre = models.CharField(max_length=150)
    poblacion = models.CharField(max_length=100)
    direccion = models.CharField(max_length=200)
    fecha = models.DateField()
    telefono = models.CharField(max_length=15)
    dias = models.CharField(max_length=200)
    iniciohora = models.TimeField()
    finhora = models.TimeField()
    nombrecontacto = models.CharField(max_length=150)
    numeropersonas = models.CharField(max_length=3)
    recursos = models.CharField(max_length=255)
    tipo = models.ForeignKey(TipoOrganizacion)
    actividades = models.ManyToManyField(Actividades)
    areas = models.ManyToManyField(AreasInteres)
    red = models.ForeignKey(RedSocial)

    def __str__(self):
        return self.nombre

    class Meta:
        verbose_name_plural = "organizaciones"
        verbose_name = "organizacion"

我已经尝试过manage.py验证以验证我的模型上的语法错误，并说出0错误，然后在执行syncdb后发生这种情况：

C:\Users\Abdul Hamid\PycharmProjects\Proyecto_OSC>python manage.py syncdb 
Traceback (most recent call last):
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 127, in ensure_connection
    self.connect()
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 115, in connect
    self.connection = self.get_new_connection(conn_params)
  File "C:\Python34\lib\site-packages\django\db\backends\postgresql_psycopg2\base.py", line 114, in get_new_connection
    return Database.connect(**conn_params)
  File "C:\Python34\lib\site-packages\psycopg2\__init__.py", line 164, in connect
    conn = _connect(dsn, connection_factory=connection_factory, async=async)
psycopg2.OperationalError

The above exception was the direct cause of the following exception:

Traceback (most recent call last):
  File "manage.py", line 10, in <module>
    execute_from_command_line(sys.argv)
  File "C:\Python34\lib\site-packages\django\core\management\__init__.py", line 399, in execute_from_command_line
    utility.execute()
  File "C:\Python34\lib\site-packages\django\core\management\__init__.py", line 392, in execute
    self.fetch_command(subcommand).run_from_argv(self.argv)
  File "C:\Python34\lib\site-packages\django\core\management\base.py", line 242, in run_from_argv
    self.execute(*args, **options.__dict__)
  File "C:\Python34\lib\site-packages\django\core\management\base.py", line 285, in execute
    output = self.handle(*args, **options)
  File "C:\Python34\lib\site-packages\django\core\management\base.py", line 415, in handle
    return self.handle_noargs(**options)
  File "C:\Python34\lib\site-packages\django\core\management\commands\syncdb.py", line 57, in handle_noargs
    cursor = connection.cursor()
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 160, in cursor
    cursor = self.make_debug_cursor(self._cursor())
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 132, in _cursor
    self.ensure_connection()
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 127, in ensure_connection
    self.connect()
  File "C:\Python34\lib\site-packages\django\db\utils.py", line 99, in __exit__
    six.reraise(dj_exc_type, dj_exc_value, traceback)
  File "C:\Python34\lib\site-packages\django\utils\six.py", line 549, in reraise
    raise value.with_traceback(tb)
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 127, in ensure_connection
    self.connect()
  File "C:\Python34\lib\site-packages\django\db\backends\__init__.py", line 115, in connect
    self.connection = self.get_new_connection(conn_params)
  File "C:\Python34\lib\site-packages\django\db\backends\postgresql_psycopg2\base.py", line 114, in get_new_connection
    return Database.connect(**conn_params)
  File "C:\Python34\lib\site-packages\psycopg2\__init__.py", line 164, in connect
    conn = _connect(dsn, connection_factory=connection_factory, async=async)
django.db.utils.OperationalError

所以我已经尝试使用pgadminIII在postgres中创建数据库而不创建表并且它一直在破坏这个错误，我想知道我是否必须用所有表创建整个数据库，或者django是否只是编写模型来创建整个数据库？难道我做错了什么？我相信我，但我不确定

我不是一个专业的想法我还在学习

对不起我的英语，这不是我的第一语言

顺便提前谢谢

Answer 1

如果您与其他人共享文件，则他们的postgres密码很可能与您的密码不同，因此请务必正确设置“PASSWORD”：以便您使用自己的密码。

postgresql中的Django操作错误

1 个答案: