Question

此脚本需要永远运行，有时甚至会使整个服务器崩溃

shop_designs有3000多个条目，shop_tshirts有超过50,000个条目

如何优化它以加快执行速度？

$query = "SELECT * FROM shop_designs";
$res = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($res)) {
$design = $row["design_id"];
$rating = $row["rating"];
$hits = $row["hits"];
$retours = $row["returns"];
$sales = $row["sales"];
$plussales = $row["sales_plus"];
$featured = $row["featured"];
$created_timestamp = $row["created_timestamp"];
$returns = $row["returns"];
$wishlisted = $row["wishlisted"];
$vector = $row["vector"];
$sales = $sales + $plussales - $returns;

    if ($sales > 10) {
    $bonus = $sales * 100;
    } else {
        $bonus = $sales * 50;
    }

$unan = 60 * 60 * 24 * 365;
$unjour = 60 * 60 * 24;
$age_du_design = time() - $created_timestamp;
$age_du_design_en_jours = $age_du_design / $unjour;
$age_du_design_an = $age_du_design / $unan;
    if ($age_du_design < $unan) {
    $age_du_design = $unan;
    $age_du_design_an = "1";
    }

if ($sales == "1") {
$sales_per_year = $sales / $age_du_design_an;
} else {
$sales_per_year = $sales / $age_du_design_an;
}
    if ($sales < 1) {
    $sales_per_year = "0";
    }

$indicehits = ($hits / 1000) / 3;
$calculsales = $sales + $wishlisted - $retours;
$cote = $calculsales + $indicehits;
$cote = round($cote, 2);
if ($vector == "1") {
$cote * 1.25;
}


echo "Update D#$design.......";
$yquery = "UPDATE shop_designs SET rating='$cote', sales_per_year='$sales_per_year' WHERE design_id='$design'";
mysql_query($yquery) or die(mysql_error());

$rating_hits = $hits + $bonus;
$rating_hits = $rating_hits / 2;
$rating_hits = round($rating_hits);
$zyquery = "UPDATE shop_tshirts SET wishlisted='$wishlisted', rating='$cote', hits='$hits', rating_hits='$rating_hits',featured='$featured',tshirt_sales='$sales',sales_per_year='$sales_per_year' WHERE design_id='$design'";
mysql_query($zyquery) or die(mysql_error());

}

Answer 1

通过痛苦的行（RBAR）处理每一行，每行都会变慢。该算法的设计完全忽略了SQL处理集数据的能力。

如何重新设计此项以获得更好的性能：重写此项以运行更少的SQL UPDATE语句。

在我们开始之前......代码中有一些奇怪的逻辑。例如，结果是什么：

 if ($vector == "1") {
 $cote * 1.25;
 }

我们看到$cote乘以1.25，但该操作的返回不会存储在任何位置。结果被丢弃了。

为什么我们需要在这里进行条件测试：

 if ($sales == "1") {
 $sales_per_year = $sales / $age_du_design_an;
 } else {
 $sales_per_year = $sales / $age_du_design_an;
 }

因此，如果某些条件为真，我们会为$sales_per_year指定一个值。否则，我们将完全相同的值分配给$sales_per_year。为什么我们需要条件测试？

整个过程中，“年龄”以秒为单位，年龄以小时为单位，年龄以年为单位......基本上可以归结为a）中最大的一年，或b）计算出的年龄。

从$calculsales计算$retours减去$returns（又名$sales）。这不是无效的，但有点好奇，因为先前已从$returns中减去$sales。

除了这些问题之外，我在这里看不到任何在SQL语句中无法在SQL表达式中执行的操作。

例如：

 SELECT n.*
   FROM ( SELECT v.design_id
               , v.wishlisted
               , v.rating
               , v.hits
               , ROUND(v.hits + (v.sales * IF(v.sales > 10, 100, 50)) AS rating_hits
               , v.featured
               , v.sales
               , IF(v.sales < 1, 0, v.sales / v.age_du_design_an) AS sales_per_year
               , ROUND(v.sales + v.wishlisted - v.returns + v.hits/3000,2)
                 -- * CASE WHEN v.vector = 1 THEN 1.25 ELSE 1.00 END
                 AS cote
            FROM (
                   SELECT d.design_id
                        , d.rating
                        , d.hits
                        , d.featured
                        , (d.sales + d.plussales - d.returns) AS sales
                        , GREATEST((UNIX_TIMESTAMP(NOW())-s.created_timestamp)/(60*60*24*365),1) AS age_du_design_an
                        , d.returns
                        , d.wishlisted
                        , d.vector
                     FROM shop_designs d
                 ) v
        ) n

但是，不是获取单个行，然后发出大量的单个更新语句，每个design_id一个，我们可以在上面的查询和目标表之间执行JOIN操作更新。

我们可以先将它作为SELECT语句编写，以进行测试。然后将其转换为UPDATE语句，如下所示：

 UPDATE shop_tshirts t
   JOIN ( SELECT v.design_id
               , v.wishlisted
               , v.rating
               , v.hits
               , ROUND(v.hits + (v.sales * IF(v.sales > 10, 100, 50)) AS rating_hits
               , v.featured
               , v.sales
               , IF(v.sales < 1, 0, v.sales / v.age_du_design_an) AS sales_per_year
               , ROUND(v.sales + v.wishlisted - v.returns + v.hits/3000,2)
                 -- * CASE WHEN v.vector = 1 THEN 1.25 ELSE 1.00 END
                 AS cote
            FROM (
                   SELECT d.design_id
                        , d.rating
                        , d.hits
                        , d.featured
                        , (d.sales + d.plussales - d.returns) AS sales
                        , GREATEST((UNIX_TIMESTAMP(NOW())-s.created_timestamp)/(60*60*24*365),1) AS age_du_design_an
                        , d.returns
                        , d.wishlisted
                        , d.vector
                     FROM shop_designs d
                 ) v
        ) n
     ON n.design_id = t.design_id
    SET t.wishlisted     = n.wishlisted
      , t.rating         = n.cote
      , t.hits           = n.hits
      , t.rating_hits    = n.rating_hits
      , t.featured       = n.featured
      , t.tshirt_sales   = n.sales
      , t.sales_per_year = n.sales_per_year

只需执行该语句的一次执行，它将一举更新shop_tshirts中的所有行。我们也可以为另一个表做类似的操作。

这就是我们改善表现的方式。

<强>后续

如果您不将UPDATE作为集处理，而是通过痛苦行（RBAR）处理行，那么请确保在shop_tshirts上定义了合适的索引， shop_designs个表格，其中design_id为前导列。

Answer 2

我的意思是这样的

<?php 
$query = "SELECT * FROM shop_designs";
$res = mysql_query($query) or die(mysql_error());
while($row = mysql_fetch_array($res)) {
.
.
.
echo "Update D#$design.......";
$query += "UPDATE shop_designs SET rating='$cote', sales_per_year='$sales_per_year' WHERE design_id='$design';";
.
.
.

$query += "UPDATE shop_tshirts SET wishlisted='$wishlisted', rating='$cote', hits='$hits', rating_hits='$rating_hits',featured='$featured',tshirt_sales='$sales',sales_per_year='$sales_per_year' WHERE design_id='$design';";
}

mysql_query($query) or die(mysql_error());

在你之前的实现中，php将访问数据库连接超过6000次，但现在php只需要访问它两次...首先选择查询，然后再批量更新查询。

<强> [编辑]
此外，$unan = 60 * 60 * 24 * 365; $unjour = 60 * 60 * 24;和time()也可以放在循环之外。它不会产生明显的内存差异，但这种方式更为正确。

优化重复的mysql查询

2 个答案: