Question

我一直收到以下错误。

com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException: 
You have an error in your SQL syntax; check
the manual that corresponds to your MySQL server version for 
the right syntax to use near '= 3' at line 1

我浏览了整个代码，此时只有2个SQL函数被调用，能够导致此错误：

读

public void Read (String userName) /*have checked validity*/
try {
    con = getConnection();

    PreparedStatement pstmt = con.prepareStatement("SELECT * FROM " + 
                              tableName + 
                             " WHERE userName=?");

        pstmt.setString(1,userName);

    ResultSet rs = pstmt.executeQuery();

    User user;
    PrintQuery(rs);
    rs.close();
    pstmt.close();
    releaseConnection(con);
    return user;

}

的GetItem

 public void getItem(int userid) /*have checked validity*/
 try {
    con = getConnection();

    Statement stmt = con.createStatement();

    PreparedStatement pstmt = con.prepareStatement("SELECT * FROM " 
                              + tableName +"WHERE UserId=?");
    pstmt.setInt(1,userid);
    rs = pstmt.executeQuery();
    PrintQuery(rs);
    stmt.close();
    releaseConnection(con);

}

我现在被困在这几个小时，我无法解决它。我已仔细检查以查看这些参数是否为空等等。它们不是。事实上，它们都很好地存在于数据库中。我无法弄清楚我哪里出错了。任何帮助将不胜感激。

Answer 1

错误在GetName上，因为在tableName和WHERE子句的连接期间缺少额外的空间

+ tableName +" WHERE UserId=?");
           // ^ add extra space here

无法在以下函数中发现SQL语法错误：导致异常

1 个答案: