Question

免责声明：我知道我的代码中已经弃用了mysql函数。那是在我的 todo 列表中。

我有一个MySql选择，在房屋预订系统中为我提供不同项目的季节。

种类：

Low season: 2010-01-01, 2010-03-01, 100 //meaning start,end,price

这是我的第一个sql：

while($season_row=mysql_fetch_assoc($season_res)){
    $seasonsArray[$season_row['id_item']][] = array(
        $season_row['season_start'], 
        $season_row['season_end'],
        $season_row['daily_price']
    );
}

这里定义的日期（到达YYYY-mm-dd的功能）：

function seasonPrice($from,$to,$requested_item){

    $start = round(strtotime($from)/86400)*86400; // like 2008-01-01
    $end = round(strtotime($to)/86400)*86400;     // to 2015-01-01

    $formattedStart = date('Y-m-d', $start);
    $formattedEnd   = date('Y-m-d', $end);

现在我需要在$seasonsArray的项目之间循环2个日期，然后检查该特定日期内该项目的价格。

我是这样做的：

foreach($seasonsArray as $item=>$value){            
    for( $thisDay = $start; $thisDay < $end; $thisDay = $thisDay + 86400){

        foreach($value as $innerValue){     
            $season_start = roundToSeconds($innerValue[0]);
            $season_end = roundToSeconds($innerValue[1]);
            if($thisDay >= $season_start && $thisDay <= $season_end) {  
                $foundPrice[] = round($innerValue[2]);
            }
        }

        $thisSerie[] = array($thisDay * 1000, isset($foundPrice) ? $foundPrice[0] : 0);

        // security check to avoid double assigned seasons to same day
        if(count($foundPrice) > 1){ die('There is double bookings in item: '.$item);}   

        unset($foundPrice);
    }
    $seasonPrices[] = array(
        'data'=> $thisSerie,
        'label'=> 'House ID: '.$item, 
    );  
}

但我明白了：Fatal error: Allowed memory size of 100663296 bytes exhausted

有关我的代码可以改进到何处不需要这么多内存的任何建议吗？或者是否有错误，我看不到它？

Answer 1

我会生成一系列天数并加入您的季节表，并使用单个查询来获得所需的结果集，例如：

SELECT dates.Date,
       coalesce(s.price, 0) AS price
FROM
  (SELECT a.Date
   FROM
     ( SELECT curdate() - INTERVAL (a.a + (10 * b.a) + (100 * c.a)) DAY AS Date, '0' AS price
      FROM
        (SELECT 0 AS a
         UNION ALL SELECT 1
         UNION ALL SELECT 2
         UNION ALL SELECT 3
         UNION ALL SELECT 4
         UNION ALL SELECT 5
         UNION ALL SELECT 6
         UNION ALL SELECT 7
         UNION ALL SELECT 8
         UNION ALL SELECT 9) AS a
      CROSS JOIN
        (SELECT 0 AS a
         UNION ALL SELECT 1
         UNION ALL SELECT 2
         UNION ALL SELECT 3
         UNION ALL SELECT 4
         UNION ALL SELECT 5
         UNION ALL SELECT 6
         UNION ALL SELECT 7
         UNION ALL SELECT 8
         UNION ALL SELECT 9) AS b
      CROSS JOIN
        (SELECT 0 AS a
         UNION ALL SELECT 1
         UNION ALL SELECT 2
         UNION ALL SELECT 3
         UNION ALL SELECT 4
         UNION ALL SELECT 5
         UNION ALL SELECT 6
         UNION ALL SELECT 7
         UNION ALL SELECT 8
         UNION ALL SELECT 9) AS c) a
   WHERE a.Date BETWEEN '$from' AND '$to'
   ORDER BY a.Date) dates
LEFT JOIN seasons s ON dates.Date BETWEEN s.start AND s.END

复杂的内部查询避免创建临时表（取自generate days from date range）并且最多可以工作1000天，但是创建临时表就可以了。

Answer 2

看起来好像你永远不会离开for循环。 $ start和$ end的设置在哪里。通过打印出来验证它们的值。

就优化而言，没有必要每天循环。跳过计算天数的for循环，并使用$ season_start和$ season_end计算第二个foreach循环中的日期。

事实上，现在有一个bug，除非$ session_start和$ session_end总是超过一天，因为有时候事件会发生在你循环的24小时之间。

内部循环内部循环达到内存限制

2 个答案: