Question

在这个方法中，我在哪里可以使用return语句？里面终于？还是试试？

我对在try catch语句中返回字符串感到有点困惑。

这是我的代码。

   public List<String> getEmailAddr(String strAccountnbr, String strCode) throws Exception {

        String strQuery2 = null;
        ResultSet rs = null;
        PreparedStatement ps = null;

     try

     {
        List<String> emailAddress= new ArrayList<String>();
        strQuery2 =  "SELECT c.EmailAddress AS EmailAddress" +

            " FROM customeremailid c " +
            "WHERE c.AccountNbr = ? " +
            "AND c.Code = ? ";

          logMsg("strQuery2: "+strQuery2);

          ps = getDBConn().prepareStatement(strQuery2);
          ps.setString(1, strAccountnbr);   
          ps.setString(2, strCode); 
          rs = ps.executeQuery();

        while(rs.next())
        {   
            emailAddress.add((rs.getString("EmailAddress")));   
        } 


     }

     catch(Exception e)
     {
         e.printStackTrace();
     }  

     return emailAddress;
     }

我收到错误emailAddress cannot resolved to a variable. 有什么帮助吗？

Answer 1

我收到错误，因为emailAddress无法解析为变量。任何帮助

List<String> emailAddress= new ArrayList<String>();//Declare it outside the block
try{...

OR 在外部声明，但在try内初始化，以便在catch块中访问它。

List<String> emailAddress=null;
try{...
emailAddress= new ArrayList<String>();//Initialize it inside the block

现在您的emailAddress只能在try{//Block}中访问，但不能在catch{//Block}中访问

Answer 2

将您的代码更改为

List<String> emailAddress= new ArrayList<String>();
try
{
....
}

return emailAddress;

以前，您尝试返回的变量仅限于try块范围。

Answer 3

return可以在try块内部或try - catch块之外使用，finally块基本上用于放弃不用于写入返回语句的资源。

Answer 4

public List<String> getEmailAddr(String strAccountnbr, String strCode) throws Exception {

        String strQuery2 = null;
        ResultSet rs = null;
        PreparedStatement ps = null;
        List<String> emailAddress= new ArrayList<String>();

     try{

         strQuery2 =  "SELECT c.EmailAddress AS EmailAddress" +
                      " FROM customeremailid c " +
                      " WHERE c.AccountNbr = ? " +
                      " AND c.Code = ? ";

          logMsg("strQuery2: "+strQuery2);
          ps = getDBConn().prepareStatement(strQuery2);
          ps.setString(1, strAccountnbr);   
          ps.setString(2, strCode); 
          rs = ps.executeQuery();

        while(rs.next())
        {   
            emailAddress.add((rs.getString("EmailAddress")));   
        } 


     }

     catch(Exception e)
     {
         //some error especify your exception or uses a generic..
         throw new EmailAdressException("BLA BLA BLA...")
     }  
     //if everything is fine return  email
     return emailAddress;
     }

Answer 5

您需要考虑范围。在您的代码中，当您在块（try-block）中声明您的emailAddress时，emailAddress仅在该块内可用。 try-catch仅适用于在执行查询时处理可能的异常。尝试将try-catch限制为仅引发异常的代码（在这种情况下，您可以/应该重构代码，以便查询在其自己的方法中完成）。

如果在try-catch块之外声明emailAddress，则在catch块之后返回它。

public List<String> getEmailAddr(String strAccountnbr, String strCode) throws Exception {
    List<String> emailAddress= new ArrayList<String>();

     try {
         // Handle query here
     } catch(Exception e) {
         // You should not catch Exception, but a more fine-grained exception.
         // In this case, SQLException
     } 

     return emailAddress;
 }

在java中的try catch语句中返回

5 个答案: