返回声明的问题

我有返回陈述的问题>.<我想将所有杂志名称存储到

ArrayList ListNameMagazine = new ArrayList();

我有一个DB; 在数据库中有一个表name_magazine,并在数据name_magazine的

Magazine1

Magazine2

Magazine3

Magazine4

我的主要是:

    ShowData Show = new ShowData();

    int HowManyMagazine = Show.HowManyMagazine(1); // to make sure there is how many Magazine name in my database

    //System.out.print(HowManyMagazine); //i want to make sure the data is out.

    String nmeMagazine = null; // this variable for get data from return statement 

    // i want to store in ListNameMagazine
    ArrayList ListNameMagazine = new ArrayList();

    for (int numbeer = 0;numbeer <= HowManyMagazine ; numbeer++)
    {
       //Store in 1 variable String, because if arrayList it's error
        nmeMagazine = Show.getResult("Select Name_Magazine from Magazine");
        // Store again in array list
        ListNameMagazine.add(nmeMagazine);
    }

    for (String s : ListNameMagazine)
    {
        System.out.println(s); // show the data
    }

这是我的回复声明:

public String getResult(String sql)
throws SQLException
{
    ResultSet rs = st.executeQuery(sql);
    ResultSetMetaData resultsetmetadata = rs.getMetaData();

    //String just_try = null;
    while (rs.next()) { 
         System.out.println("Result:"+rs.getString(1));
         //just_try = rs.getString(1);
         //return just_try;
    }
    return null;    //return just_try;
}

问题在于返回声明.

当注释(//)我擦除而最后一个返回null; 我删除了.它变得像这里:

public String getResult(String sql)
throws SQLException
{         
     ResultSet rs = st.executeQuery(sql);
     ResultSetMetaData resultsetmetadata = rs.getMetaData();
     String just_try = null;

     while (rs.next()) { 
         //System.out.println("Result:"+rs.getString(1));
        just_try = rs.getString(1);
        return just_try;
     }

     return just_try;
 }

当我使用此语句显示数据时.

for (String s : ListNameMagazine)
    {
        System.out.println(s); // show the data
    }

结果而已

Magazine4

Magazine4

Magazine4

Magazine4

@.@我迷惑@ @的地方.@

但是当我在这样的return语句中显示数据时

public String getResult(String sql)
throws SQLException
    {         
    ResultSet rs = st.executeQuery(sql);
    ResultSetMetaData resultsetmetadata = rs.getMetaData();
    String just_try = null;

    while (rs.next()) {     
        System.out.println("Result:"+rs.getString(1));
        //just_try = rs.getString(1);
        //return just_try;
    }

    return null;
}

数据显示了我想要的东西.我知道我只想念某个地方,但我不知道@.@.我希望你们能找到它.THX

你的问题是return只返回一件事,它会立即返回,函数将退出!你正在重新调整杂志的名称just_try.

while (rs.next()) { 

    //System.out.println("Result:"+rs.getString(1));

   just_try = rs.getString(1);

  return just_try;

}

所以,你开始迭代rs,你得到的名字:

just_try = rs.getString(1);

然后你告诉代码返回just_try.

return just_try;

此时将返回just_try,函数将退出!我觉得你的问题是,你期待继续下去,并回头率值来调用它的代码的功能,但这不是它的工作方式.

我怀疑你想做的是这样的:

ArrayList ListNameMagazine;
ListNameMagazine = Show.getResult("Select Name_Magazine from Magazine");

然后在函数getResult中:

public ArrayList getResult(String sql) throws SQLException {

    ResultSet rs = st.executeQuery(sql);

    ResultSetMetaData resultsetmetadata = rs.getMetaData();

    ArrayList returnArrayList = new ArrayList();

    while (rs.next()) { 

        returnArrayList.add(rs.getString(1));

    }
    return returnArrayList;

}