我很难理解为什么这不起作用,如果我直接在 MySQL 控制台中输入完全相同的内容,它会接受它,但当我尝试运行它时,它会报告语法错误。
com.mysql.jdbc.exceptions.jdbc4.MySQLSyntaxErrorException:您的 SQL 语法有错误;检查与您的 MySQL 服务器版本相对应的手册,了解在第 1 行 '= '6'' 附近使用的正确语法
我想做的就是接收行中的数据,其中包含用户输入的member_id 值。出于测试目的,该值始终为 6,我尝试将其解析为 int 而不是字符串,这给出了相同的错误,并且我尝试将 ID 变量添加到字符串的末尾而不是使用占位符,但它没有也不那么喜欢。
这是代码:
public class MemberDAO {
public PreparedStatement ps = null;
public Connection dbConnection = null;
public List<Member> getMembersDetails(String ID) throws SQLException{
List<Member> membersDetails = new ArrayList();
String getMembershipDetails = "SELECT first_name, last_name, phone_number, email, over_18, date_joined, date_expire, fines FROM members"
+ "WHERE member_id = ?";
try {
DBConnection mc = new DBConnection();
dbConnection = mc.getConnection();
ps = dbConnection.prepareStatement(getMembershipDetails);
ps.setString(1, ID);
ps.executeQuery();
ResultSet rs = ps.executeQuery(getMembershipDetails);
String firstName = rs.getString("first_name");
String lastName = rs.getString("last_name");
String phoneNumber = rs.getString("phone_number");
String email = rs.getString("email");
String over18 = rs.getString("over_18");
String dateJoined = rs.getString("date_joined");
String dateExpired = rs.getString("date_expire");
String fines = rs.getString("fines");
Member m;
m = new Member(firstName, lastName, phoneNumber, email, over18, dateJoined, dateExpired, fines);
membersDetails.add(m);
} catch (SQLException ex){
System.err.println(ex);
System.out.println("Failed to get Membership Details.");
return null;
} finally{
if (ps != null){
ps.close();
}
if (dbConnection != null){
dbConnection.close();
}
} return membersDetails;
}
这就是它的名字:
private void btnChangeCustomerActionPerformed(java.awt.event.ActionEvent evt) {
customerID = JOptionPane.showInputDialog(null, "Enter Customer ID.");
MemberDAO member = new MemberDAO();
try {
List membersDetails = member.getMembersDetails(customerID);
txtFullName.setText(membersDetails.get(0) + " " + membersDetails.get(1));
} catch (SQLException ex) {
System.err.println(ex);
System.out.println("Failed to get Details.");
JOptionPane.showMessageDialog(null, "Failed to retrieve data.");
}
}
欢迎任何意见。
最佳答案
您的查询缺少空格:
...fines FROM members"
+ "WHERE...
将会导致
FROM membersWHERE
哪个 SQL 无效
更改为
+ " WHERE....
关于java - JDBC 如何在带有 where 子句的准备好的语句中使用占位符?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/24840038/