Question

我有以下数据库结构

mysql> select * from categories limit 30;
+-------------+-------------+-----------------+-------+------+------+------+------+------+------+------+
| category_id | T1          | T2              | T3    | T4   | T5   | T6   | T7   | T8   | T9   | T10  |
+-------------+-------------+-----------------+-------+------+------+------+------+------+------+------+
|           1 | Popcorn     | Regular Pack    | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           1 | Popcorn     | Regular Pack    | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           1 | Popcorn     | Regular Pack    | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           2 | Popcorn     | Bucket          | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           2 | Popcorn     | Bucket          | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           2 | Popcorn     | Bucket          | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           3 | Popcorn     | Jumbo           | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           3 | Popcorn     | Jumbo           | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           4 | Popcorn     | Tub             | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           4 | Popcorn     | Tub             | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           5 | Popcorn     | Combo Cool Bite | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           6 | Popcorn     | Combo Relish    | NULL  | NULL | NULL | NULL | NULL | NULL | NULL | NULL |


|           7 | Soft Drinks | Fountain        | Apple      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           8 | Soft Drinks | Fountain        | Orange     | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           8 | Soft Drinks | Fountain        | Orange     | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           9 | Soft Drinks | Fountain        | Lemon      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|           9 | Soft Drinks | Fountain        | Lemon      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|          10 | Soft Drinks | Fountain        | Clear      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|          10 | Soft Drinks | Fountain        | Clear      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|          11 | Soft Drinks | Tin             | Apple      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|          11 | Soft Drinks | Tin             | Apple      | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|          12 | Soft Drinks | Tin             | Lean Apple | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
|          12 | Soft Drinks | Tin             | Lean Apple | NULL | NULL | NULL | NULL | NULL | NULL | NULL |
+-------------+-------------+-----------------+------------+------+------+------+------+------+------+------+

我正试图以这种格式为上面的内容创建一个Json结构。

例如

以下两个数组。（伪代码）

PopCorn :[Regular Pack, Bucket ,Jumbo,Tub,Combo Cool Bite,Combo Relish]

Soft Drinks:[Fountain,Tin]

我试图获得T1的所有不同元素并组合T2的不同元素

我能够以这种方式做到：

JSONArray T1array=new JSONArray();
JSONArray T2array=new JSONArray();

 ResultSet rs=stmt.executeQuery();
        while(rs.next())
        {
            String T1 = rs.getString("T1");
            String T2 = rs.getString("T2");
            if(!tempvariable.equals(T1))
            {
                T1array.put(T1);
            //  T2array.put(T2);
            }
             tempvariable = T1;

        }

请告诉我如何将T2映射到T1

Answer 1

可能的解决方案可能是

Map<String, Set<String>> tempMap = new HashMap<String, Set<String>>();

ResultSet rs=stmt.executeQuery();
while(rs.next())
{
    String T1 = rs.getString("T1");
    String T2 = rs.getString("T2");
    Set<String> t2 = tempMap.get(T1);
    if(t2 == null)
    {
       t2 = new HashSet<String>();
       tempMap.put(T1, t2);
    }
    t2.add(T2);
}

之后，您在tempMap中将所有不同的T1作为键，并在Set值中为它们设置所有不同的T2。

无法构造此JSON对象

1 个答案: