三个表之间的关系都是多对一，多个学生对应一个职业，多个职业对应一个公司，然后我查询已经就业的学生姓名，专业，职业，公司名的时候，一个学生对应到了多个职业中。这是为什么呢

你把sql，三张表的数据都发出来呀

USE [毕业学生就业管理系统]
GO
/****** Object: Table [dbo].[manager] Script Date: 2022/6/17 19:41:58 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[manager](
[managerNO] [char](10) NOT NULL,
[Mname] [char](20) NULL,
[Mid] [char](10) NULL,
[Mkey] [char](10) NULL,
[Mtel] [char](20) NULL,
CONSTRAINT [PK_MANAGER] PRIMARY KEY CLUSTERED
(
[managerNO] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
/****** Object: Table [dbo].[公司] Script Date: 2022/6/17 19:41:58 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[公司](
[cno] [char](10) NOT NULL,
[cname] [char](20) NULL,
[require] [smallint] NULL,
[applicant] [smallint] NULL,
CONSTRAINT [PK_公司] PRIMARY KEY CLUSTERED
(
[cno] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
/****** Object: Table [dbo].[学生] Script Date: 2022/6/17 19:41:58 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[学生](
[sno] [char](10) NOT NULL,
[pno] [char](10) NULL,
[sname] [char](20) NULL,
[ssex] [char](2) NULL,
[sbirthday] [date] NULL,
[splace] [char](20) NULL,
[swno] [char](10) NOT NULL,
[scno] [char](10) NULL,
[workflag] [bit] NULL,
CONSTRAINT [PK_学生] PRIMARY KEY CLUSTERED
(
[sno] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON, ALLOW_PAGE_LOCKS = ON) ON [PRIMARY]
) ON [PRIMARY]
GO
/****** Object: Table [dbo].[职业] Script Date: 2022/6/17 19:41:58 ******/
SET ANSI_NULLS ON
GO
SET QUOTED_IDENTIFIER ON
GO
CREATE TABLE [dbo].[职业](
[pno] [char](10) NOT NULL,
[cno] [char](10) NULL,
[pname] [char](20) NULL,
CONSTRAINT [PK_职业] PRIMARY KEY CLUSTERED
(
[pno] ASC
)WITH (PAD_INDEX = OFF, STATISTICS_NORECOMPUTE = OFF, IGNORE_DUP_KEY = OFF, ALLOW_ROW_LOCKS = ON,

1. 创建学生信息表以及完整性约束
create table 学生
(
sno char(10) not null,
pno char(10) null,
sname char(20) null,
ssex char(2) null,
sbirthday date null,
splace char(20) null,
swno char(10) not null,
scno char(10) null,
workflag bit null,
constraint PK_学生 primary key (sno)，
foreign key (pno) references 职业(pno)
);
2. 创建公司信息表以及完整性约束
create table 公司
(
cno char(10) not null,
cname char(20) null,
require smallint null,
applicant smallint null,
constraint PK_公司 primary key (cno)
);
3.创建职业信息表以及完整性约束
create table 职业
(
pno char(10) not null,
cno char(10) null,
pname char(20) null,
constraint PK_职业 primary key (pno)，
foreign key (cno) references 公司(cno)
);

你三个表都没关联，肯定出笛卡尔积了

你查询语句可以发下吗，是不是没有用join关联