Question

以下是2个查询：

SELECT COUNT(BUG.BG_BUG_ID) AS BUG_ID_SIT_CNT 
FROM BUG 
WHERE BUG.BG_USER_02 = 'SIT'

SELECT COUNT(BUG.BG_BUG_ID) AS BUG_ID_UAT_CNT 
FROM BUG 
WHERE BUG.BG_USER_02 = 'UAT'

如何编写查询以使用上述查询的结果并执行以下操作：

缺陷去除效率：BUG_ID_SIT_CNT/(BUG_ID_SIT_CNT + BUG_ID_UAT_CNT)

Answer 1

SELECT BUG_ID_SIT_CNT * 1. / BUG_ID_ALL
FROM (
    SELECT
        COUNT(CASE WHEN BG_USER_02 = 'SIT' THEN BG_BUG_ID END) AS BUG_ID_SIT_CNT,
        COUNT(BG_BUG_ID) AS BUG_ID_ALL
    FROM BUG
    WHERE BG_USER_02 IN ('SIT', 'UAT')
) t

Answer 2

您可以在

之类的外部查询中获取它

SELECT BUG_ID_SIT_CNT/(BUG_ID_SIT_CNT + BUG_ID_UAT_CNT) as testcol
FROM (
SELECT SUM(CASE WHEN BUG.BG_USER_02 = 'SIT' THEN 1 ELSE 0 END) AS BUG_ID_SIT_CNT,
SUM(CASE WHEN BUG.BG_USER_02 = 'UAT' THEN 1 ELSE 0 END) AS BUG_ID_UAT_CNT 
FROM BUG ) xx;

Answer 3

您可以将两个查询包装在WITH语句中，然后在SELECT中执行计算。

Answer 4

WITH sit
AS
(
    SELECT COUNT(BUG.BG_BUG_ID) AS BUG_ID_SIT_CNT 
    FROM BUG 
    WHERE BUG.BG_USER_02 = 'SIT'
),
uat
AS
(
    SELECT COUNT(BUG.BG_BUG_ID) AS BUG_ID_UAT_CNT 
    FROM BUG 
    WHERE BUG.BG_USER_02 = 'UAT'
)
SELECT sit.BUG_ID_SIT_CNT/(sit.BUG_ID_SIT_CNT + uat.BUG_ID_UAT_CNT) 
FROM sit, uat

Answer 5

我会用参数来做，因为我发现它更具可读性：

<template>
    <form method="post" action="" v-on:submit.prevent="search">
        <input type="text" placeholder="Person name" required v-model="name" v-el="nameInput" />
        <input type="text" placeholder="Company" required v-model="company" v-el="domainInput" />
        <input type="submit" value="Search" class="btn show-m" />
    </form>
</template>

<script>
export default {
    data () {
        return {
            contacts: [],
            name: null,
            company: null
        }
    },
    ready: {
        // I tried the following :
        this.$$.nameInput.focus();
        this.$els.nameInput.focus();
        // None of them worked !
    }
    methods: {
        search: function (event) {
            // ...

            // I also would like to give the focus here, once the form has been submitted.
            // And here also, this.$$ and this.$els doesn't work
        },
    }
}
</script>

Answer 6

SELECT cast(BUG_ID_SIT_CNT as decimal) / cast((BUG_ID_SIT_CNT + BUG_ID_UAT_CNT)as decimal) 
FROM (
 SELECT
    COUNT(CASE WHEN BG_USER_02 = 'SIT' THEN BG_BUG_ID END) AS BUG_ID_SIT_CNT
    COUNT(CASE WHEN BG_USER_02 = 'UAT' THEN BG_BUG_ID END) AS BUG_ID_UAT_CNT
FROM BUG
WHERE BG_USER_02 IN ('SIT', 'UAT')
) t

如何对2个查询的输出执行算术运算

6 个答案: