MySQL查詢能不能做到更好

Question

我目前在做後續的過去6個月，但它似乎是它可以做更好的方式（較少的代碼行，可能更優化）

$monthones = mktime(0, 0, 0, date("n") - 1, 1); 
$monthonef = mktime(0, 0, 0-1, date("n"), 1); 

$query = "SELECT COUNT(*) FROM users WHERE type = '1' AND status = '1' AND (date >= '".$monthones."' AND date <= '".$monthonef."');"; 
$result = $pdo->query($query); 
$monthone = $result->fetchColumn(); 

$query = "SELECT COUNT(*) FROM users WHERE type = '3' AND status = '1' AND (date >= '".$monthones."' AND date <= '".$monthonef."');"; 
$result = $pdo->query($query); 
$monthone2 = $result->fetchColumn(); 

$monthtwos = mktime(0, 0, 0, date("n") - 2, 1); 
$monthtwof = mktime(0, 0, 0-1, date("n") - 1, 1); 

$query = "SELECT COUNT(*) FROM users WHERE type = '1' AND status = '1' AND (date >= '".$monthtwos."' AND date <= '".$monthtwof."');"; 
$result = $pdo->query($query); 
$monthtwo = $result->fetchColumn(); 

$query = "SELECT COUNT(*) FROM users WHERE type = '3' AND status = '1' AND (date >= '".$monthtwos."' AND date <= '".$monthtwof."');"; 
$result = $pdo->query($query); 
$monthtwo2 = $result->fetchColumn(); 

我還有4個月的休息時間。任何建議如何我可以改善這一點？我正在考慮結合查詢並在循環結果時開始計數？或者會更糟？

Answer 1

似乎運行一個對數據庫的查詢會更有效率，一次通過數據。

對於日期範圍的條件下，而不是減去第二，使用

<= last second of month 

我更喜歡使用使用

< first second of following month 

圖案如果在原始代碼中的邏輯爲正確生成「開始」和「結束」邊界，我們只需要7個邊界值，每月的第一秒：

$month00 = mktime(0, 0, 0, date("n") - 0, 1); 
$month01 = mktime(0, 0, 0, date("n") - 1, 1); 
$month02 = mktime(0, 0, 0, date("n") - 2, 1); 
$month03 = mktime(0, 0, 0, date("n") - 3, 1); 
$month04 = mktime(0, 0, 0, date("n") - 4, 1); 
$month05 = mktime(0, 0, 0, date("n") - 5, 1); 
$month06 = mktime(0, 0, 0, date("n") - 6, 1); 

然後查詢可以是這樣的：

$query = " 
SELECT SUM(u.type = '1' AND u.date >= '$month01' AND u.date < '$month00') AS monthone 
    , SUM(u.type = '3' AND u.date >= '$month01' AND u.date < '$month00') AS monthone2 

    , SUM(u.type = '1' AND u.date >= '$month02' AND u.date < '$month01') AS monthtwo 
    , SUM(u.type = '3' AND u.date >= '$month02' AND u.date < '$month01') AS monthtwo2 

    , ... 
    , ... 

    , SUM(u.type = '1' AND u.date >= '$month06' AND u.date < '$month05') AS monthsix 
    , SUM(u.type = '3' AND u.date >= '$month06' AND u.date < '$month05') AS monthsix2 

    FROM users u 
WHERE u.status = '1' 
    AND u.type IN ('1','3') 
    AND u.date >= '$month06' 
    AND u.date < '$month00' 
"; 

的...將通過重複表達模式返回列monththree/monththree2更換，monthfour/monthfour2，monthfive/monthfive2

注意，這使用MySQL簡寫，SUM（）聚合內的表達式評估爲1（如果條件爲TRUE）或0（如果條件爲FALSE）或NULL。

我們必須抓取整行來獲取所有計數。我們不得不消除fetchColumn()並改用fetch()。

$result = $pdo->query($query); 
$row = $result->fetch(PDO::FETCH_ASSOC); 

如果我們需要在標變量的值，所以我們不必更改後續代碼：

$monthone = $row['monthone']; 
$monthone2 = $row['monthone2']; 
$monthtwo = $row['monthtwo']; 
$monthtwo2 = $row['monthtwo2']; 
... 
... 
$monthsix = $row['monthsix']; 
$monthsix2 = $row['monthsix2']; 

Answer 2

你可以返回所有的計數使用conditional aggregation一個查詢：

select count(case when type = '1' AND 
        date >= '".$monthones."' AND date <= '".$monthonef."' then 1 end) cnt1, 
     count(case when type = '3' AND 
        date >= '".$monthones."' AND date <= '".$monthonef."' then 1 end) cnt2, 
     count(case when type = '1' AND 
        date >= '".$monthtwos."' AND date <= '".$monthtwof."' then 1 end) cnt3, 
     count(case when type = '3' AND 
        date >= '".$monthtwos."' AND date <= '".$monthtwof."' then 1 end) cnt4 
from users 
where status = '1' 

Answer 3

你可以只使用一個SQL查詢解決。 請試試這個：

select count(case when type= '1' then 1 end) type_one 
    , count(case when type= '3' then 1 end) type_three 
    , DATE_FORMAT(FROM_UNIXTIME(date),'%Y-%m') month 
from users 
where status = '1' 
    and date between UNIX_TIMESTAMP(DATE_ADD(CURDATE(), INTERVAL -6 MONTH)) and UNIX_TIMESTAMP(CURDATE()) 
group by DATE_FORMAT(FROM_UNIXTIME(date),'%Y-%m') 

您可以用參數替換CURDATE（）。