0

クエリから動的にテーブルを構築しようとしています

ここにクエリがあります

SELECT MONTH(c.pat_added_date) AS month、COUNT(a.ana_id) as total、YEAR(c.pat_added_date) AS year FROM lab_patient_analysis AS b

                      LEFT JOIN lab_analysis AS a
                      ON a.ana_id = b.ana_id

                      LEFT JOIN lab_patients AS c
                      ON c.pat_id = b.pat_id

                      WHERE 
                      a.ana_id = '3' AND 
                      YEAR(c.pat_added_date) BETWEEN '2012' AND '2013'
                      GROUP BY MONTH(c.pat_added_date)
                      ORDER BY YEAR(c.pat_added_date), MONTH(c.pat_added_date)

そして、ここにクエリが返すものがあります

month | total | year
1         13     2012
7         9      2012
8         33     2012
3         21     2013
6         8      2013
....

これは、配列に使用したphp関数です

function array_set(&$a, $path, $value) {
    if(!is_array($path))
        $path = explode($path[0], substr($path, 1));
    $key = array_pop($path);
    foreach($path as $k) {
        if(!isset($a[$k]))
            $a[$k] = array();
        $a = &$a[$k];
    }
    $a[$key ? $key : count($a)] = $value;
}

そして私が立ち往生しているポイント

$yr = null;
while($row ... ))
{
 if($yr != $row['year']) {
   $yr = $row['year'];
   array_set($array, array($row['year'], $row['month']), $row['total']);
   echo '</tr></tr>';
   echo '<th scope="row">'.$row['year'].'</th>';
 }
  
 echo '<td>'.$row['total'];
}

返された配列

Array
(
    [2012] => Array
        (
            [1] => 13
            [7] => 9
            [8] => 33
        )

    [2013] => Array
        (
            [3] => 21
            [6] => 8
        )

)

そして、次の表を作成したいと思います

        Jan| Feb| Mar | Apr | May | Jun | Jul | Sep | Oct | Nov | Dec
  2012   13   0    0     0     0     0     9    33     0     0     0
  2013   0    0    21    0     0     8     0     0     0     0     0

それが理にかなっていることを願っています

4

2 に答える 2

1

私はそれからこれを得ることができました：

<?php

// Build source array.. You should already have this layout by the looks of it.

$array = array('2012' => array("1" => 13, "7" => 9, "8" => 33),
                '2013' => array("3" => 21, "6" => 8));

// Loop over every year.

foreach ($array as $year => $values) {

        for ($i = 1; $i <= 12; $i++) { // Loop through months 1 .. 12

                if (!$values[$i]) { // If there is no value for this month, pad it with 0 (Note string "0", else it leaves it empty)
                        $array[$year][$i] = "0";
                }
        }

        // Sort it out, probably won't need this actually, but it doesn't hurt to have it.
        ksort($array[$year], SORT_NATURAL);
}

// Print out the array 
var_dump($array);

私はこの出力を得る：

array(2) {
  [2012]=>
  array(12) {
    [1]=>
    int(13)
    [2]=>
    string(1) "0"
    [3]=>
    string(1) "0"
    [4]=>
    string(1) "0"
    [5]=>
    string(1) "0"
    [6]=>
    string(1) "0"
    [7]=>
    int(9)
    [8]=>
    int(33)
    [9]=>
    string(1) "0"
    [10]=>
    string(1) "0"
    [11]=>
    string(1) "0"
    [12]=>
    string(1) "0"
  }
  [2013]=>
  array(12) {
    [1]=>
    string(1) "0"
    [2]=>
    string(1) "0"
    [3]=>
    int(21)
    [4]=>
    string(1) "0"
    [5]=>
    string(1) "0"
    [6]=>
    int(8)
    [7]=>
    string(1) "0"
    [8]=>
    string(1) "0"
    [9]=>
    string(1) "0"
    [10]=>
    string(1) "0"
    [11]=>
    string(1) "0"
    [12]=>
    string(1) "0"
  }
}
于 2013-05-30T22:48:10.407 に答える
0

これが私が行う方法です：

  1. 配列を作成し、ゼロで埋めます (例では 2*12 配列になります)
  2. クエリ データベース
  3. クエリ結果をウォークし、配列に入力します ($myarray[$data['month']][$data['$year']] = data['total'])
  4. 配列内の値を含むテーブルを表示します
于 2013-05-30T22:47:18.677 に答える