标题是我收到的错误,当我单击“加载”时,我的程序卡住了。我认为这是因为我在声明中执行声明,但从我看来,这是解决我的问题的唯一解决方案。通过加载,我只想重新填充患者列表,但要做到这一点,我还需要了解他们的条件。代码有效,底部方法是我试图修复的。我认为问题是我有 2 份声明未结,但我不确定。 负载:
public void DatabaseLoad()
{
try
{
String Name = "Wayne";
String Pass= "Wayne";
String Host = "jdbc:derby://localhost:1527/Patients";
Connection con = DriverManager.getConnection( Host,Name, Pass);
PatientList.clear();
Statement stmt8 = con.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE,
ResultSet.CONCUR_UPDATABLE);
String SQL8 = "SELECT * FROM PATIENTS";
ResultSet rs8 = stmt8.executeQuery( SQL8 );
ArrayList<PatientCondition> PatientConditions1 = new ArrayList();
while(rs8.next())
{
PatientConditions1 = LoadPatientConditions();
}
Statement stmt = con.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE,
ResultSet.CONCUR_UPDATABLE);
String SQL = "SELECT * FROM PATIENTS";
ResultSet rs = stmt.executeQuery( SQL );
while(rs.next())
{
int id = (rs.getInt("ID"));
String name = (rs.getString("NAME"));
int age = (rs.getInt("AGE"));
String address = (rs.getString("ADDRESS"));
String sex = (rs.getString("SEX"));
String phone = (rs.getString("PHONE"));
Patient p = new Patient(id, name, age, address, sex, phone,
PatientConditions1);
PatientList.add(p);
}
UpdateTable();
UpdateAllViews();
DefaultListModel PatientListModel = new DefaultListModel();
for (Patient s : PatientList) {
PatientListModel.addElement(s.getAccountNumber() + "-" + s.getName());
}
PatientJList.setModel(PatientListModel);
}
catch(SQLException err)
{
System.out.println(err.getMessage());
}
}
这是返回患者状况的 ArrayList
的方法
public ArrayList LoadPatientConditions()
{
ArrayList<PatientCondition> PatientConditionsTemp = new ArrayList();
try
{
String Name = "Wayne";
String Pass= "Wayne";
String Host = "jdbc:derby://localhost:1527/Patients";
Connection con = DriverManager.getConnection( Host,Name, Pass);
Statement stmt = con.createStatement(ResultSet.TYPE_SCROLL_INSENSITIVE,
ResultSet.CONCUR_UPDATABLE);
String SQL = "SELECT * FROM PATIENTCONDITIONS";
ResultSet rs5 = stmt.executeQuery( SQL );
int e = 0;
while(rs5.next())
{
e++;
String ConName = (rs5.getString("CONDITION"));
PatientCondition k = new PatientCondition(e,ConName);
PatientConditionsTemp.add(k);
}
}
catch(SQLException err)
{
System.out.println(err.getMessage());
}
return PatientConditionsTemp;
}
最佳答案
我也遇到了类似的问题。 我正在连接到本地服务器上托管的 derby 数据库。 我创建了 2 个同时连接:
- 与松鼠
- 使用ij工具
错误 40XL1:无法在请求的时间内获得锁定
要解决此问题,修改表的连接必须提交其事务。
希望这可以帮助 !
关于java - 无法在请求的时间内获得锁定问题,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/15582682/