Python postgresql中的Django操作错误
我对django有一个问题,我试图为我的应用程序在django上创建一个数据库,我已经这样设置了连接Python postgresql中的Django操作错误,python,django,postgresql,Python,Django,Postgresql,我对django有一个问题,我试图为我的应用程序在django上创建一个数据库,我已经这样设置了连接 DATABASES = { 'default': { 'ENGINE': 'django.db.backends.postgresql_psycopg2', 'NAME': 'Proyecto_OSC', 'USER': 'posgresql', 'PASSWORD': 'aa121292', 'HOST':
DATABASES = {
'default': {
'ENGINE': 'django.db.backends.postgresql_psycopg2',
'NAME': 'Proyecto_OSC',
'USER': 'posgresql',
'PASSWORD': 'aa121292',
'HOST': '',
'PORT': '5432'
}
}
然后我尝试创建如下模型:
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"
我已经尝试使用manage.py validate验证模型上的语法错误,并显示0个错误,然后在执行syncdb后发生以下情况:
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
C:\Users\Abdul Hamid\PycharmProjects\Proyecto\u OSC>python manage.py syncdb
回溯(最近一次呼叫最后一次):
文件“C:\Python34\lib\site packages\django\db\backends\\ uuuu init\uuuuu.py”,第127行,在确保连接中
self.connect()
文件“C:\Python34\lib\site packages\django\db\backends\\uuuu init\uuuuu.py”,第115行,在connect中
self.connection=self.get_new_connection(连接参数)
文件“C:\Python34\lib\site packages\django\db\backends\postgresql\u psycopg2\base.py”,第114行,在get\u new\u连接中
返回数据库。连接(**连接参数)
文件“C:\Python34\lib\site packages\psycopg2\\uuuu init\uuuu.py”,第164行,在connect中
连接=连接(dsn,连接工厂=连接工厂,异步=异步)
psycopg2.0错误
上述异常是以下异常的直接原因:
回溯(最近一次呼叫最后一次):
文件“manage.py”,第10行,在
从命令行(sys.argv)执行命令
文件“C:\Python34\lib\site packages\django\core\management\\ uuuu init\uuuu.py”,第399行,从命令行执行
utility.execute()
文件“C:\Python34\lib\site packages\django\core\management\\uuuu init\uuuuu.py”,第392行,在execute中
self.fetch_命令(子命令)。从_argv(self.argv)运行_
文件“C:\Python34\lib\site packages\django\core\management\base.py”,第242行,运行于\u argv
self.execute(*args,**选项._dict__;
文件“C:\Python34\lib\site packages\django\core\management\base.py”,第285行,执行
输出=self.handle(*args,**选项)
文件“C:\Python34\lib\site packages\django\core\management\base.py”,第415行,位于句柄中
返回self.handle\u noargs(**选项)
文件“C:\Python34\lib\site packages\django\core\management\commands\syncdb.py”,第57行,在handle\u noargs中
cursor=connection.cursor()
光标中第160行的文件“C:\Python34\lib\site packages\django\db\backends\\ uuuu init\uuuu.py”
cursor=self.make\u debug\u cursor(self.\u cursor())
文件“C:\Python34\lib\site packages\django\db\backends\\uuuu init\uuuuu.py”,第132行,在光标处
self.sure_连接()
文件“C:\Python34\lib\site packages\django\db\backends\\ uuuu init\uuuuu.py”,第127行,在确保连接中
self.connect()
文件“C:\Python34\lib\site packages\django\db\utils.py”,第99行,在退出时__
6.重新播放(dj_exc_类型、dj_exc_值、回溯)
文件“C:\Python34\lib\site packages\django\utils\six.py”,第549行,在reraise中
通过_回溯(tb)提升值
文件“C:\Python34\lib\site packages\django\db\backends\\ uuuu init\uuuuu.py”,第127行,在确保连接中
self.connect()
文件“C:\Python34\lib\site packages\django\db\backends\\uuuu init\uuuuu.py”,第115行,在connect中
self.connection=self.get_new_connection(连接参数)
文件“C:\Python34\lib\site packages\django\db\backends\postgresql\u psycopg2\base.py”,第114行,在get\u new\u连接中
返回数据库。连接(**连接参数)
文件“C:\Python34\lib\site packages\psycopg2\\uuuu init\uuuu.py”,第164行,在connect中
连接=连接(dsn,连接工厂=连接工厂,异步=异步)
django.db.utils.error
因此,我已经尝试在postgres中使用pgadminIII创建数据库,而不创建表
它不断地出现这个错误,我想知道是我必须用所有的表来创建整个数据库,还是django仅仅通过编写模型来创建整个数据库?
我做错什么了吗?我相信我是,但我不确定
我不是专业人士
我还在学习
对不起,我的英语不是我的第一语言
顺便说一句,提前谢谢您是否“pip安装了psycopg2”?是的,当然,很抱歉没有在说明中指出这一点。请检查用户名和密码是否有拼写错误?我注意到用户很多次都是“postgresql”用户是“postgresql”可能你缺少了“t”?好吧,我把它改成了postgresql,因为我的用户就是我在postgresql中使用的用户,但它会给我带来相同的操作错误异常,我必须指出,我试图使用djangook在postgresql上创建表,我意识到用户名是postgres而不是postgresql,并且postgres服务器没有监听localhost