Question

我正在使用带有java的select2，我需要将resultset放入名为 resutls 的数组中。

这是我的代码：

的java

public static String convertResultSetToJson(ResultSet resultSet) throws SQLException {
        JSONArray json = new JSONArray();
        ResultSetMetaData metadata = resultSet.getMetaData();
        int numColumns = metadata.getColumnCount();

        //iterate rows
        while (resultSet.next())  {
            JSONObject obj = new JSONObject();      //extends HashMap
            //iterate columns
            for (int i = 1; i <= numColumns; ++i) {
                String column_name = metadata.getColumnName(i);
                obj.put(column_name, resultSet.getObject(column_name).toString());
            }
            json.add(obj);
        }

        return json.toJSONString();
    }

我称之为：

ResultSet list = connection.executeWithResult("select * from test where id::text like '%" + term + "%'");
out.print(functions.convertResultSetToJson(list));

输出：

[{date: "1996-12-11", systemId: "3", creator: "3", boss: "3", creatorChart: "1", subSystemId: "4",…}]

我的问题是，如何生成这样的输出？

 {results:[date: "1996-12-11", systemId: "3", creator: "3", boss: "3", creatorChart: "1", subSystemId: "4",…]}

Answer 1

您当前的输出是有效的JSON。这是一个对象数组。

[{
   date: "1996-12-11",
   systemId: "3",
   creator: "3",
   boss: "3",
   creatorChart: "1",
   subSystemId: "4"
}]

但是你想要实现的不是有效的JSON。

{results:[date: "1996-12-11", systemId: "3", creator: "3", boss: "3", creatorChart: "1", subSystemId: "4",…]}

正如@Harald在评论中提到的，如果你想将数组包装在一个包装器对象中，那么你可以这样做

public static String convertResultSetToJson(ResultSet resultSet) throws SQLException {
    JSONArray json = new JSONArray();
    ResultSetMetaData metadata = resultSet.getMetaData();
    int numColumns = metadata.getColumnCount();

    //iterate rows
    while (resultSet.next())  {
        JSONObject obj = new JSONObject();      //extends HashMap
        //iterate columns
        for (int i = 1; i <= numColumns; ++i) {
            String column_name = metadata.getColumnName(i);
            obj.put(column_name, resultSet.getObject(column_name).toString());
        }
        json.add(obj);
    }
    JSONObject resultsObject = new JSONObject();
    resultsObject.put('results', json); 
    return resultsObject.toJSONString();
}

这将生成以下JSON。

{
   results: [{
       date: "1996-12-11",
       systemId: "3",
       creator: "3",
       boss: "3",
       creatorChart: "1",
       subSystemId: "4"
   }]
}

将结果集放入java中的数组中

1 个答案: