在 grails 3.3.9 中提供默认的辅助排序列

Question

我遇到一个问题，当使用 createCriteria().list(params) 以 10 records/page 为单位翻阅一组 12 条记录时，第 2 页上显示的 2 条记录出现在第 1 页上。默认 'sort' 在域映射中指定，但指定的列包含重复项，因此我假设排序列中包含重复项的子集的顺序是不确定的。

这也使用可排序列 headers，当您使用 generate-all 时，grails 也是如此。这是否意味着任何时候选定的排序列包含重复项，顺序可能是不确定的？有没有办法（除了蛮力）传递一个默认的辅助排序列（可能是 ID），除了选定的列之外，还要使用它来保证顺序？

这是在 SQL 服务器上，这些是生成的查询：

第 1 页：

select TOP(?) 
    this_.procedure_occurrence_id as procedur1_22_1_, 
    this_.provenance_id as provenan2_22_1_, 
    this_.procedure_date as procedur3_22_1_, 
    this_.procedure_type_concept_id as procedur4_22_1_, 
    this_.procedure_source_value as procedur5_22_1_, 
    this_.associated_provider_id as associat6_22_1_, 
    this_.relevant_condition_concept_id as relevant7_22_1_, 
    this_.visit_occurrence_id as visit_oc8_22_1_, 
    this_.person_id as person_i9_22_1_, 
    this_.procedure_concept_id as procedu10_22_1_, 
    person_ali1_.person_id as person_i1_20_0_, 
    person_ali1_.person_source_value as person_s2_20_0_, 
    person_ali1_.provenance_id as provenan3_20_0_, 
    person_ali1_.location_id as location4_20_0_, 
    person_ali1_.gender_concept_id as gender_c5_20_0_, 
    person_ali1_.care_site_id as care_sit6_20_0_, 
    person_ali1_.year_of_birth as year_of_7_20_0_, 
    person_ali1_.provider_id as provider8_20_0_, 
    person_ali1_.race_concept_id as race_con9_20_0_, 
    person_ali1_.month_of_birth as month_o10_20_0_, 
    person_ali1_.ethnicity_concept_id as ethnici11_20_0_, 
    person_ali1_.ethnicity_source_value as ethnici12_20_0_, 
    person_ali1_.race_source_value as race_so13_20_0_, 
    person_ali1_.gender_source_value as gender_14_20_0_, 
    person_ali1_.day_of_birth as day_of_15_20_0_ 
from procedure_occurrence this_ 
inner join person person_ali1_ on this_.person_id=person_ali1_.person_id 
where (person_ali1_.person_id=?) 
and this_.provenance_id=? 
and this_.procedure_concept_id in (?, ?, ?, ?) 
order by this_.procedure_date desc

第 2 页：

WITH query AS (
    SELECT inner_query.*, ROW_NUMBER() 
    OVER (ORDER BY CURRENT_TIMESTAMP) as __hibernate_row_nr__ 
    FROM ( 
        select TOP(?) this_.procedure_occurrence_id as procedur1_22_1_, 
            this_.provenance_id as provenan2_22_1_, 
            this_.procedure_date as procedur3_22_1_, 
            this_.procedure_type_concept_id as procedur4_22_1_, 
            this_.procedure_source_value as procedur5_22_1_, 
            this_.associated_provider_id as associat6_22_1_, 
            this_.relevant_condition_concept_id as relevant7_22_1_, 
            this_.visit_occurrence_id as visit_oc8_22_1_, 
            this_.person_id as person_i9_22_1_, 
            this_.procedure_concept_id as procedu10_22_1_, 
            person_ali1_.person_id as person_i1_20_0_, 
            person_ali1_.person_source_value as person_s2_20_0_, 
            person_ali1_.provenance_id as provenan3_20_0_, 
            person_ali1_.location_id as location4_20_0_, 
            person_ali1_.gender_concept_id as gender_c5_20_0_, 
            person_ali1_.care_site_id as care_sit6_20_0_, 
            person_ali1_.year_of_birth as year_of_7_20_0_, 
            person_ali1_.provider_id as provider8_20_0_, 
            person_ali1_.race_concept_id as race_con9_20_0_, 
            person_ali1_.month_of_birth as month_o10_20_0_, 
            person_ali1_.ethnicity_concept_id as ethnici11_20_0_, 
            person_ali1_.ethnicity_source_value as ethnici12_20_0_, 
            person_ali1_.race_source_value as race_so13_20_0_, 
            person_ali1_.gender_source_value as gender_14_20_0_, 
            person_ali1_.day_of_birth as day_of_15_20_0_ 
        from procedure_occurrence this_ 
        inner join person person_ali1_ on this_.person_id=person_ali1_.person_id 
        where (person_ali1_.person_id=?) 
        and this_.provenance_id=? 
        and this_.procedure_concept_id in (?, ?, ?, ?) 
        order by this_.procedure_date desc ) inner_query ) 
    SELECT 
        procedur1_22_1_, 
        provenan2_22_1_, 
        procedur3_22_1_, 
        procedur4_22_1_, 
        procedur5_22_1_, 
        associat6_22_1_, 
        relevant7_22_1_, 
        visit_oc8_22_1_, 
        person_i9_22_1_, 
        procedu10_22_1_, 
        person_i1_20_0_, 
        person_s2_20_0_, 
        provenan3_20_0_, 
        location4_20_0_, 
        gender_c5_20_0_, 
        care_sit6_20_0_, 
        year_of_7_20_0_, 
        provider8_20_0_, 
        race_con9_20_0_, 
        month_o10_20_0_, 
        ethnici11_20_0_, 
        ethnici12_20_0_, 
        race_so13_20_0_, 
        gender_14_20_0_, 
        day_of_15_20_0_ 
    FROM query 
    WHERE __hibernate_row_nr__ >= ? 
    AND __hibernate_row_nr__ < ?

Answer 1

我认为可以在域中定义多个排序列 class:

    static mapping = { 
          sort(["property1" : "asc",
                "property2" : "desc"] ) 
    }

在 grails 3.3.9 中提供默认的辅助排序列

Provide a default secondary sort column in grails 3.3.9

sql-server

grails

hibernate