无法创建表“ basque30.newreferral”（错误号：150）

首先，感谢您的帮助和支持。当我从网络上学习MySQL时，我知道我会犯很多错误，在此先感谢您的耐心和理解。

我正在基于Java创建项目，然后创建MySQL数据库和表。通常情况下，一切顺利，直到遇到此问题为止。到目前为止，这给我带来了最大的麻烦。我读了许多关于错误150的文章。我已经在MySQL站点上阅读了错误150，我认为我已经满足了发生错误150的十个原因。

我正在拨打的电话正在创建每个表。以下是一些表格，这些表格涵盖了我所遇到的问题。

这是WorkCalendar表

    private String workCalandar = "CREATE TABLE WorkCalendar ("
        + "WorkCalendarIdNo INT(64) NOT NULL AUTO_INCREMENT,"
        + "WorkCalendarDate Date,"
        + "WorkCalendarDayCount INT(64),"
        + "WorkDayTypeIdNo INT(64),"
        + "PRIMARY KEY(WorkCalendarIdNo),"
        + "FOREIGN KEY(WorkDayTypeIdNo) REFERENCES WorkDayType(WorkDayTypeIdNo)"
        + ")";

这是部门表

    private String department = "CREATE TABLE Department ("
        + "DeptIdNo INT(64) NOT NULL AUTO_INCREMENT,"
        + "DeptName VARCHAR(25),"
        + "PRIMARY KEY(DeptIdNo)"
        + ")";

这是专家表

    private String specialist = "CREATE TABLE Specialist ("
        + "SpecIdNo INT(64) NOT NULL AUTO_INCREMENT,"
        + "PrefixIdNo INT(64),"
        + "SpecFirstName VARCHAR(30),"
        + "SpecSurname VARCHAR(35),"
        + "SpecDisplayName VARCHAR(72),"
        + "DeptIdNo INT(64),"
        + "PRIMARY KEY(SpecIdNo),"
        + "FOREIGN KEY(PrefixIdNo) REFERENCES Prefix(PrefixIdNo),"
        + "FOREIGN KEY(DeptIdNo) REFERENCES Department(DeptIdNo)"
        + ")";

这是NewReferral表

    private String newReferral = "CREATE TABLE NewReferral("
        + "NewReferralIdNo INT(64) NOT NULL AUTO_INCREMENT,"
        + "PatientNumber VARCHAR(12),"
        + "NewReferralDate Date,"
        + "DeptIdNo INT(64),"
        + "SpecIdNo INT(64),"
        + "NewReferralMatched BOOLEAN,"
        + "WorkCalendarIdNo INT(64),"
        + "PRIMARY KEY(NewReferralIdNo),"
        + "FOREIGN KEY(DeptIdNo) REFERENCES Department(DeptIdNo),"
        + "FOREIGN KEY(SpecIdNo) REFERENCES Specialist(SpecIdNo),"
        + "FOREIGN KEY(WorkCalendarIdNo) REFERENCES WorkCalendar(WorkCalendarIdNo)"
        + ")";

错误是无法创建表“ basque30.newreferral”（错误号：150）

我已将错误跟踪到下一行

    + "FOREIGN KEY(SpecIdNo) REFERENCES Specialist(SpecIdNo),"

有人可以指出我的方法中的错误并建议我如何解决它。