處理JPA期間查詢的正確方法是什麼？

Question

我需要在下面生成預期結果。基本上，這是一個按照特定時期彙總值的查詢（每週，每月等）。有一個日期過濾器，開始和結束，我們需要返回所有範圍的值。如果它們不存在，則返回0.

在下面的示例中，開始日期是'2015-08-02'，日期'2015-08-23'且週期爲每週。請注意，對於第2周，我們沒有值，但應該返回零值。

因此，在這種情況下，使用JPA做到這一點的最佳方法是什麼？我們考慮使用臨時表並將結果與​​此表結合以獲得整個範圍的結果，但我不知道是否可以使用JPA，因爲我們需要創建表，加入並銷燬臨時表。

另一種選擇是創建數據庫視圖並將其映射到實體。

在上述情況下，JPQL查詢應該是這樣的：

@Query("select status, sum(totalInvoice), week from Invoice where " + 
     "left join TempTable as tt..." + <-- TEMP TABLE OR VIEW TO GET THE PERIODS 
     "issuer.id = :issuerId and type = :type and (:recipientId is null or recipient.id = :recipientId) and " + 
     "status in ('ISSUED', 'PAID') " + 
     "group by status") 

另一種方法是使用存儲過程，但他們似乎很難與JPA實現，我不認爲他們是必要的。

預期結果：

{ 
    "code":"xxx", 
    "title":"This is the title of the first series" 
    "type":"PERIODIC", 
    "period":"WEEKLY", <-- PERIOD 
    "from":"2015-08-02", 
    "to":"2015-08-29", 
    "labels": ["2015-08-02", "2015-08-09", "2015-08-16", "2015-08-23"], 
    "tabelType": "TEXT", 
    "series":[ 
     { 
     "code":"xxx", 
     "title":"This is the title of the first series" 
     "values":[10, 0, 13, 18] <- in this example, we don't have values for label "2015-08-09" 
     }, 
     { 
     "code":"xxx", 
     "title":"This is the title of the second series" 
     "values":[10, 0, 13, 18] <- in this example, we don't have values for label "2015-08-09" 
     } 
    ] 
} 

Answer 1

@pozs在這裏提供了答案。這隻能用本地查詢（PostgreSQL）完成。結果如下：

/** 
* Returns the counts, totals and averages of the states by their currency, period and status. 
*/ 
@Query(value = "select i.currency, date(p), i.status, count(id), sum(coalesce(i.total, 0)), avg(coalesce(i.total, 0)) " + 
    "from generate_series(date_trunc(:period, cast(:from as timestamp)), date_trunc(:period, cast(:to as timestamp)) + cast('1 ' || :period as interval), cast('1 ' || :period as interval)) p " + 
    "inner join invoice i on i.due_date >= p and i.due_date < p + cast('1 ' || :period as interval) " + 
    "where issuer_id = :issuerId and type = :type and (:recipientId = 0 or recipient_id = :recipientId) and type = :type " + 
    "group by i.currency, date(p), i.status " + 
    "order by i.currency, date(p), i.status", nativeQuery = true) 
List<Object[]> getIssuerStatementTotalsByCurrencyPeriodAndStatus(
    @Param("issuerId") long issuerId, 
    @Param("recipientId") long recipientId, 
    @Param("type") String statementType, 
    @Param("from") String from, 
    @Param("to") String to, 
    @Param("period") String period); 

請注意，這將返回對象數組的列表。另外請注意，我無法將枚舉和複雜參數傳入該方法。我不得不將這些值貶低爲字符串和原語。

我已經把這個結果到一些有意義的事情下面的類：

/** 
    * Contains the result of a single result in an aggregate query. 
    */ 
    public class AggregateResult { 

     private List<String> keys; 
     private List<BigDecimal> values; 

     @SuppressWarnings("unused") 
     public AggregateResult(Object value1, Object value2) { 
      this(new Object[] { value1, value2 }); 
     } 

     @SuppressWarnings("unused") 
     public AggregateResult(Object value1, Object value2, Object value3) { 
      this(new Object[] { value1, value2, value3 }); 
     } 

     @SuppressWarnings("unused") 
     public AggregateResult(Object value1, Object value2, Object value3, Object value4) { 
      this(new Object[] { value1, value2, value3, value4 }); 
     } 

     @SuppressWarnings("unused") 
     public AggregateResult(Object value1, Object value2, Object value3, Object value4, Object value5) { 
      this(new Object[] { value1, value2, value3, value4, value5 }); 
     } 

     public AggregateResult(Object... vals) { 
      values = new ArrayList<>(); 
      while (values.size() < vals.length && vals[vals.length - values.size() - 1] instanceof Number) { 
       Number number = (Number) vals[vals.length - values.size() - 1]; 
       values.add(number instanceof BigDecimal ? (BigDecimal) number : new BigDecimal(number.toString())); 
      } 

      this.keys = Stream.of(ArrayUtils.subarray(vals, 0, vals.length - values.size())).map(Object::toString).collect(toList()); 
     } 

     public List<String> getKeys() { 
      return keys; 
     } 

     public List<BigDecimal> getValues() { 
      return values; 
     } 

     /** 
     * Returns the list of {@link AggregateResult}s for the raw result. The raw result is expected to 
     * have been returned from a native JPA query. 
     */ 
     public static List<AggregateResult> fromNativeResult(List<Object[]> raw) { 
      return raw.stream().map(AggregateResult::new).collect(toList()); 
     } 
    } 

Answer 2

這可能不是直接回答你的問題，而是：你爲什麼需要JPA查詢，而不是直接在Java代碼中做了分組？這種類型的複雜語義分組是使用Java很好完成的，但是SQL數據庫通常不太擅長生成這種類型的結構化數據。如果在數據庫級別上沒有其他原因這樣做（例如，您需要填充可根據期限搜索的數據的視圖），那麼只需加載原始數據並使用Java代碼填充結構 - 它將少得多編碼開銷也可能更高效。