Question

我有一个名为CREATED_AT的表我写了一个查询并收到如下响应： Score D 27 01 17 02 80 03 55 06

这是我的查询：

SELECT SUM(AMOUNT) as scores,to_char(CREATED_AT, 'MONTH') as  d  
FROM C_SCORES 
WHERE USER_ID =201 and to_char(CREATED_AT, 'YYYY') ='1398' 
GROUP BY to_char(CREATED_AT, 'MONTH');

实体代码为：

    public class ScoreCategorizeWithMonth {

        private int Scores;
        private int month;
    private String monthfa;


 public String getMonthfa() {
        switch (month) {
            case 1:
                monthfa = "فروردین";
                break;
            case 2:
                monthfa = "اردیبهشت";
                break;
            case 3:
                monthfa = "خرداد";
                break;
            case 4:
                monthfa = "تیر";
                break;
            case 5:
                monthfa = "مرداد";
                break;
            case 6:
                monthfa = "شهریور";
                break;
            case 7:
                monthfa = "مهر";
                break;
            case 8:
                monthfa = "آبان";
                break;
            case 9:
                monthfa = "آذر";
                break;
            case 10:
                monthfa = "دی";
                break;
            case 11:
                monthfa = "بهمن";
                break;
            case 12:
                monthfa = "اسفند";
            default:
                monthfa = "";
        }



        return monthfa;
    }

    }

此代码在控制器中：

 @GetMapping(value = "score-Categorize-months")
    public List<ScoreCategorizeWithMonth> scoreCategorizedWithMonths()
    {
        return   iScoresSrv.scoreCategorizeWithMonths(userInfo.getUserId());
    }

我的问题是：

2-我的回答是数据库中存在的月份。如何显示月份甚至不在数据库中

 {
    "scores": 27,
    "month": 1,
    "monthfa": "فروردین"
  },
  {
    "scores": 17,
    "month": 2,
    "monthfa": "اردیبهشت"
  },
  {
    "scores": 84,
    "month": 3,
    "monthfa": "خرداد"
  },
  {
    "scores": 55,
    "month": 6,
    "monthfa": "شهریور"
  }

我还要显示分数== 0的4..5月份

谢谢

Answer 1

第2步和第3步：您可以通过以下查询来实现：

select sum(AMOUNT) as scores, month from  
(select AMOUNT ,to_char(CREATED_AT, 'MONTH') as month from C_SCORES WHERE USER_ID 
 =201 and to_char(CREATED_AT, 'YYYY') ='1398'
 union
 select 0 as AMOUNT, Month from
 (SELECT 1 AS MONTH
 UNION SELECT 2 AS MONTH
 UNION SELECT 3 AS MONTH
 UNION SELECT 4 AS MONTH
 UNION SELECT 5 AS MONTH
 UNION SELECT 6 AS MONTH
 UNION SELECT 7 AS MONTH
 UNION SELECT 8 AS MONTH
 UNION SELECT 9 AS MONTH
 UNION SELECT 10 AS MONTH
 UNION SELECT 11 AS MONTH
 UNION SELECT 12 AS MONTH) TBL1) TBL2
 group by month;

只需将1-12替换为您的月份值

Answer 2

您可以使用属性文件...

此链接提供了许多示例： http://zetcode.com/java/resourcebundle/

在该教程中，'words.properties'是您的默认语言和后备语言文件，而words_de.properties和words_sk.properties是针对自定义翻译语言的

在这一行：source.setBasenames("messages/words");中，他试图将翻译文件传递给算法。

并使用messageSource.getMessage("w1", null, Locale.GERMAN)获取所需的语言。...

请注意，sk中的words_sk.properties是语言iso代码，波斯语为fa，英语为en ...

某些系统也使用本地化，它们定义了自定义字符和格式，我不确定该算法是否支持这种命名方式……您可以传递额外的参数：fa-IR或{{ 1}}

说实话，我看上去并不深，但是自从几年前使用它以来，这就是原理。

Answer 3

我只是在Java中添加sume代码来填充无效的索引。解决起来很容易。但是我有点困惑，没有答案没有帮助我

 public List<ScoreCategorizedWithMonth> scoreCategorizeWithMonths(Long userId) {
        List<ScoreCategorizedWithMonth> monthCategory = scoreDao.scoreCategorizeWithMonths(userId);
        List<ScoreCategorizedWithMonth> result = new ArrayList<>();
        for (int i=1; i<=12;i++)
        {
            ScoreCategorizedWithMonth scoreCategorizedWithMonth = checkMonthCatIsExist(i, monthCategory);
            result.add(scoreCategorizedWithMonth);
        }
        return result;
    }

    private ScoreCategorizedWithMonth checkMonthCatIsExist(int month, List<ScoreCategorizedWithMonth> monthCategory) {
        for (ScoreCategorizedWithMonth cat:monthCategory) {
            if(cat.getMonth() == month){
                return cat;
            }
        }
        ScoreCategorizedWithMonth temp= new  ScoreCategorizedWithMonth();
        temp.setMonth(month);
        temp.setScores(0);
        return temp;
    }

翻译月份名称回复

3 个答案: