PHPMYADMIN- SQL创建表错误＃1064

Question

我正在尝试创建一些表，它们之间有一对一的关系。我遇到了这个错误，我已经阅读了其他线程，并怀疑这与FK有关。任何帮助都很棒！：D谢谢

我在XAMPP服务器上使用phpmyadmin。我创建了数据库，单击SQL选项卡，然后将其放置。我在下面有错误。

错误：

Error
Static analysis:
2 errors were found during analysis.
Unexpected beginning of statement. (near "userID" at position 307)
Unrecognized statement type. (near "INT" at position 314)
SQL query:
CREATE TABLE reportTable ( reportID INT AUTO_INCREMENT, submitDate DATETIME, submitBy VARCHAR(255), reportInformation VARCHAR(255), reviewedBy VARCHAR(255), PRIMARY KEY(reportID), FOREIGN KEY(userID) REFERENCES userTable(userID) ) CREATE TABLE userTable ( userID INT AUTO_INCREMENT, username VARCHAR(255), PRIMARY KEY(userID) )
MySQL said: Documentation
#1064 - You have an error in your SQL syntax; check the manual that corresponds to your MariaDB server version for the right syntax to use near 'CREATE TABLE userTable (
    userID INT AUTO_INCREMENT,
    username VARCHAR(2' at line 11

代码：

CREATE TABLE reportTable (
    reportID INT AUTO_INCREMENT,
    userID INT,
    submitDate DATE,
    submitBy VARCHAR(255),
    reportInformation VARCHAR(255),
    reviewedBy VARCHAR(255),
    PRIMARY KEY(reportID),
    FOREIGN KEY(userID) REFERENCES userTable(userID)
);
CREATE TABLE userTable (
    userID INT AUTO_INCREMENT,
    username VARCHAR(255),
    PRIMARY KEY(userID)
);

Answer 1

正如我在评论中提到的那样，半彩色通常用于结束语句。否则，MySQL不知道您何时要停止。在phpmyadmin中进行一个查询时，它将正确解释它，但不能做出一个以上的语句。这将修复您的运行语句。

CREATE TABLE reportTable (
    reportID INT AUTO_INCREMENT,
    submitDate DATE,
    submitBy VARCHAR(255),
    reportInformation VARCHAR(255),
    reviewedBy VARCHAR(255),
    PRIMARY KEY(reportID),
    FOREIGN KEY(userID) REFERENCES userTable(userID)
);
CREATE TABLE userTable (
    userID INT AUTO_INCREMENT,
    username VARCHAR(255),
    PRIMARY KEY(userID)
);