Question

我们有一个DB，它有几个模式，每个模式都有一个Fact表。我需要准备一个结构集，其中包含来自Fact的模式名称max（MTH_DT）以及来自每个Fact表的不同MTH_DT计数。

SCHEMA_NAME  MAX(MTH_DT)    DISTINCT_COUNT(MTH_DT)
SCHM_1       11/30/2015     24
SCHM_2       10/31/2015     24
SCHM_3       11/30/2015     36
SCHM_4       10/31/2015     24
SCHM_5       11/30/2015     24

如何以这种方式获取结果集？

Answer 1

这就是

当您确定查询并运行查询时，只需取消注释EXECUTE语句：

<?php
//require('includes/html_table.class.php');
require_once "php/mysql.php";
require('includes/fx.php');
$debug=FALSE;
//////////////////
// DEBUG TOGGLE //
$debug = TRUE;//
//////////////////

    //////////////////
// RESULT PRINT TOGGLE //
$resultOutput = TRUE;//
//////////////////

//variables
$starttime = microtime(true); //stopwatch
//all important tea off date is set below

function makeReport(){
$youngest='2015-9-21 23:59:59';
//$youngest = $startDate;
$youngest = strtotime($youngest);
$oldest=$youngest-2505600; //to figure out what plans the customer is using in the past 30 days


$arrayCompPlans = array();
$seqint=0; //for navigating arrays
$regBusHours=TRUE;

//if we want to start only looking at m-f use this code
//will be put in a function for production
 $count=0; //total weekdays polled
    for($x=0; $x <= 29; $x++)
    {
        if($regBusHours){
            $day1 = strtotime("-$x day",$youngest);
            $day2 = strtotime('-1 day',$day1);
            if (!isweekend($day1))
            {               
                $count=1+$count;
            } 
        }
        else $count=30;
    } 
echo $count;
$SQL = "SELECT *
        ,(PlanMinUsage/$count) AS AvgDailyUsage
        ,(PlanMinLimit-PlanMinUsage) AS RemainingMins
        FROM (
        SELECT b.company Company
        , a.subid CompanyID
        , c.description Plan
        , a.planid PlanID
        , c.monthlymins PlanMinLimit
        , count(a.planid) PlanCalls
        , sum(a.minsused) PlanMinUsage
         FROM transactions a 
         INNER JOIN subscribers b 
         ON a.subid=b.subid 
         INNER JOIN plans c 
         ON a.planid=c.id 
         WHERE (a.timestamp>=1440388799 AND a.timestamp<=1442894399) 
         AND planid!='0' 
         GROUP BY a.subid,a.planid) as subQ";   
debugOutStr($SQL,$debug);
$theRES = mysql_query($SQL, $db);
$seqint=0;

while($row=mysql_fetch_assoc($theRES))
{
    $arrayCompPlans[]=$row;
}
print_r($arrayCompPlans);
return $arrayCompPlans;
}


$array = makeReport();
print_r($array);
$endtime = microtime(true);
$tottime=number_format(($endtime-$starttime),2,'.',',');

if($resultOutput){
    debugOutStr(("<br>Total Run Time: ".$tottime." seconds<br>"),$resultOutput);
    debugOutArray($array, $resultOutput);
}

Answer 2

这是一种比使用INFORMATION_SCHEMA.TABLES更安全的构建动态sql的方法。我离开了小组，因为它是一个常数，所以你实际上并不需要一个小组。

declare @SQL nvarchar(MAX) = ''

select @SQL = @SQL + 'select ''' + name + ''' as SchemaName, MAX(MTH_DT) as MaxMTH_DT, COUNT(distinct MTH_DT) as DISTINCT_COUNT_MTH_DT from ' + name + '.Fact union all ' 
from sys.schemas
where SCHEMA_ID > 4
    and SCHEMA_ID < 16384

Select @SQL = LEFT(@SQL, LEN(@SQL) - 9) + ' order by SchemaName'

select @SQL

来自不同模式的事实的结果集

2 个答案: