Question

我是mysql的新手，我必须减少以下更新查询的执行时间

UPDATE temp_countcalculations,
(
    SELECT count(*) as insuffcounts,CRP_RefNo as ref
    FROM testsymphony7.p_education 
    WHERE testsymphony7.p_education.EducationStatusId=6 
    GROUP BY CRP_RefNo
) as icounts
SET Edu_pending=icounts.insuffcounts
WHERE temp_countcalculations.crp_refno=icounts.ref;

Answer 1

根据您提供的有限信息，我会尽可能地扩展您的可能性（如果您还没有这样做）

首先，让我们重写您的查询：

UPDATE 
    temp_countcalculations t
        JOIN (
            SELECT 
                COUNT(*) as insuffcounts,
                CRP_RefNo as ref
            FROM 
                testsymphony7.p_education p
            WHERE 
                p.EducationStatusId = 6 
            GROUP BY 
                CRP_RefNo
        ) i ON t.crp_refno = icounts.ref
SET 
    t.Edu_pending = i.insuffcounts;

好。

因此，您根据他们的参考，使用t.Edu_pending更新所有i.insuffcounts。这里有2个要优化的查询。

（1）：

SELECT 
    COUNT(*) AS insuffcounts,
    CRP_RefNo as ref
FROM 
    testsymphony7.p_education p
WHERE 
    p.EducationStatusId = 6 
GROUP BY 
    CRP_RefNo

和（2）：

SELECT 1 
FROM
    temp_countcalculations t
        JOIN ((1)) i ON t.crp_refno = icounts.ref

优化（1）：

列的理想索引：CRP_RefNo, EducationStatusId（复合）
专栏testsymphony7.p_education.crp_refno NOT NULL，如果可能，UNIQUE。

优化（2）：

列temp_countcalculations.crp_refno
专栏temp_countcalculations.crp_refno NOT NULL，如果可能，UNIQUE。

基于此，我们可能会更进一步了解您的结果：

EXPLAIN 
SELECT 1 
FROM 
    temp_countcalculations t
        JOIN (
            SELECT 
                COUNT(*) as insuffcounts,
                CRP_RefNo as ref
            FROM 
                testsymphony7.p_education p
            WHERE 
                p.EducationStatusId = 6 
            GROUP BY 
                CRP_RefNo
        ) i ON t.crp_refno = icounts.ref

在MySQL查询下优化：

1 个答案: