Question

我必须将数据库的数据放在这种格式中（显然值会有所不同），我认为它被称为关联数组（我对术语很糟糕）。

$values=array(
            "Jan" => 110,
            "Feb" => 130,
            "Mar" => 215,
            "Apr" => 81,
            "May" => 310,
            "Jun" => 110,
            "Jul" => 190,
            "Aug" => 175,
            "Sep" => 390,
            "Oct" => 286,
            "Nov" => 150,
            "Dec" => 196
    );

这是我开发的内容：

    $sql = "SELECT MONTH(AddDate) AS Date, column_name FROM table ORDER BY AddDate ASC";
    $res = mysql_query($sql) or die(mysql_error());
    $prev_date = null;

$values=array();

while ( $row = mysql_fetch_assoc($res) ) {
    if ( $row['Date'] != $prev_date) {
        $month = $row['Date'];
        $sql = "SELECT count(MONTH(AddDate)) AS EntryAmount FROM `table` WHERE MONTH(AddDate)=$month ";
        $countResults = mysql_query($sql) or die(mysql_error());
        if( $entryAmount = mysql_fetch_array($countResults) ) {
            $values[$row['Date']] =  $entryAmount['EntryAmount'];
        }
        $prev_date = $row['Date'];
    }
}

输出：

Array ( [9] => 999 [10] => 986 [11] => 264 )

Answer 1

您可以在一个查询中完成所有这些操作：

select month(AddDate) as theMonth,
       count(*) as numberOfRows
from   `table`
group by theMonth
order by theMonth

然后外环消失，内环变为：

if( $row = mysql_fetch_array($results) ) 
{
    $values[$row['theMonth']] = $row['numberOfRows'];
}

对于中等大小的数据集，它应该表现得更好。但请注意，通过使用month函数，您将失去您可能能够使用的索引的任何好处。

Answer 2

使用GROUP BY（未经测试）：

$sql = "SELECT MONTH(AddDate) AS Date, count(AddDate) AS EntryAmount FROM table ORDER BY AddDate ASC GROUP BY MONTH(AddDate)";
$res = mysql_query($sql) or die(mysql_error());
$values=array();

while ( $row = mysql_fetch_assoc($res) ) {
    $values[$row['Date']] =  $row['EntryAmount'];
}

用于组织数据库中的值的代码的效率

2 个答案: