Question

我有Multi-Dimesional我想重建这个数组，以便所有第一级数组（array[]）具有另一个数组所缺少的数据（基本上替换丢失的数据）。我正在用这个数组构建一个图形，所以我需要用0填充缺少的区域，否则我的图形中会有很大的漏洞。

[numberof]是此给定情节的金额。 [date]是它发生的月份。如你所见，我从今天的月份开始订购（3）ASC：

当前输出

Array
 (
[First] => Array
    (
        [0] => Array
            (
                [numberof] => 1
                [date] => 12
            )

        [1] => Array
            (
                [numberof] => 2
                [date] => 1
            )

        [2] => Array
            (
                [numberof] => 2
                [date] => 2
            )

        [3] => Array
            (
                [numberof] => 25
                [date] => 3
            )

    )

[Second] => Array
    (
        [0] => Array
            (
                [numberof] => 1
                [date] => 1
            )

        [1] => Array
            (
                [numberof] => 2
                [date] => 2
            )

        [2] => Array
            (
                [numberof] => 1
                [date] => 3
            )

    )

[Third] => Array
    (
        [0] => Array
            (
                [numberof] => 1
                [date] => 2
            )

    )
)

期望的结果

Array
 (
[First] => Array
    (
        [0] => Array
            (
                [numberof] => 1
                [date] => 12
            )

        [1] => Array
            (
                [numberof] => 2
                [date] => 1
            )

        [2] => Array
            (
                [numberof] => 2
                [date] => 2
            )

        [3] => Array
            (
                [numberof] => 25
                [date] => 3
            )

    )

[Second] => Array
    (
        [0] => Array //ADDED and given a value of 0
            (
                [numberof] => 0
                [date] => 12
            )

        [1] => Array
            (
                [numberof] => 1
                [date] => 1
            )

        [2] => Array
            (
                [numberof] => 2
                [date] => 2
            )

        [3] => Array
            (
                [numberof] => 1
                [date] => 3
            )

    )

[Third] => Array
    (

        [0] => Array //ADDED and given a value of 0
            (
                [numberof] => 0
                [date] => 12
            )

        [1] => Array  //ADDED and given a value of 0
            (
                [numberof] => 0
                [date] => 1
            )
        [2] => Array
            (
                [numberof] => 1
                [date] => 2
            )

        [3] => Array //ADDED and given a value of 0
            (
                [numberof] => 0
                [date] => 3
            )
    )
)

在我看来，这是不可能的。但也许我没有掌握技能，到目前为止我已经尝试过了：

n.b这是我大声说话 ....我无法理解如何做到这一点，循环遍历数组，检查它是否有值numbr，如果它没有添加它。但是，如果最后一行是月份最多且第一个月只有1个月呢？

修改

MySQL：

$sql = sprintf("SELECT count(*) as numberOf, MONTH(s.sextime) as date FROM sex s, users u WHERE %s AND s.uid = u.uid AND s.sextime >= DATE_SUB(CURRENT_DATE, INTERVAL %d MONTH) GROUP BY MONTH(s.sextime) ORDER BY s.sextime ASC;", $type, $months);

这是PHP生成上述数组并调用sql：

    $info['info']['First'] = $this->dataStore->grabSexesByMonth('AND s.sexnumber=1');
    $info['info']['Second'] = $this->dataStore->grabSexesByMonth('AND s.sexnumber=2');
    $info['info']['Third'] = $this->dataStore->grabSexesByMonth('AND s.sexnumber=3');

Answer 1

三步流程可能有效：

$fillTo = array('key'=>'','count'=>0);
foreach($array as $k => $data) {
    if (count($data) > $fillTo['count']) {
        $fillTo['key'] = $k;
        $fillTo['count'] = count($data);
    }
}

$fillData = $array[$fillTo['key']];
foreach($fillData as $k => &$data) {
    $data['numberof'] => 0;
}

foreach($array as $k => &$data) {
    if ($k === $fillTo['key'])
        continue;

    $data = $data + $fillData;
}

var_dump($array);

Answer 2

不幸的是MySQL没有设置返回功能，例如PostgreSQL generate_series。这使事情变得复杂，因为您需要以编程方式生成所有日期的集合。另外，您应该使用MONTH(sextime)代替EXTRACT(YEAR_MONTH FROM sextime)，否则您将获得不同年份的混合金额。

$months = 4;

# emulate GENERATE_SERIES
for ($i = $months-1; $i >= 0; $i--) {
  $gen_series[] = "SELECT EXTRACT(YEAR_MONTH FROM CURRENT_DATE - INTERVAL $i MONTH) extr";
}
$gen_series = join(' UNION ', $gen_series);


$sql = sprintf("SELECT
                  fll.sexnumber,
                  fll.extr as date,
                  COUNT(u.uid) as numberOf
                FROM
                  (SELECT
                    dts.extr,
                    sxn.sexnumber
                  FROM
                    (%s) dts,
                    (SELECT DISTINCT sexnumber FROM sex) sxn) fll LEFT JOIN sex s
                ON
                  fll.extr = EXTRACT(YEAR_MONTH FROM s.sextime) AND fll.sexnumber=s.sexnumber LEFT JOIN users u
                ON
                  s.uid = u.uid
                GROUP BY
                  fll.sexnumber,
                  fll.extr
                ORDER BY
                  fll.sexnumber,
                  fll.extr", $gen_series);

echo $sql;

上面的PHP代码将输出以下SQL查询。

SELECT
  fll.sexnumber,
  fll.extr as date,
  COUNT(u.uid) as numberOf
FROM
  (SELECT
    dts.extr,
    sxn.sexnumber
  FROM
    (SELECT EXTRACT(YEAR_MONTH FROM CURRENT_DATE - INTERVAL 3 MONTH) extr UNION SELECT EXTRACT(YEAR_MONTH FROM CURRENT_DATE - INTERVAL 2 MONTH) extr UNION SELECT EXTRACT(YEAR_MONTH FROM CURRENT_DATE - INTERVAL 1 MONTH) extr UNION SELECT EXTRACT(YEAR_MONTH FROM CURRENT_DATE - INTERVAL 0 MONTH) extr) dts,
    (SELECT DISTINCT sexnumber FROM sex) sxn) fll LEFT JOIN sex s
ON
  fll.extr = EXTRACT(YEAR_MONTH FROM s.sextime) AND fll.sexnumber=s.sexnumber LEFT JOIN users u
ON
  s.uid = u.uid
GROUP BY
  fll.sexnumber,
  fll.extr
ORDER BY
  fll.sexnumber,
  fll.extr;

在同一数据集上运行它，您将从MySQL获得以下输出。

+-----------+--------+----------+
| sexnumber | date   | numberOf |
+-----------+--------+----------+
|         1 | 201312 |        1 |
|         1 | 201401 |        2 |
|         1 | 201402 |        2 |
|         1 | 201403 |       25 |
|         2 | 201312 |        0 |
|         2 | 201401 |        1 |
|         2 | 201402 |        2 |
|         2 | 201403 |        1 |
|         3 | 201312 |        0 |
|         3 | 201401 |        0 |
|         3 | 201402 |        1 |
|         3 | 201403 |        0 |
+-----------+--------+----------+
12 rows in set (0.01 sec)

所以只需1个查询（而不是3个）就可以得到你需要的东西，而且没有多余的数组玩杂耍。

如果缺少数组元素，PHP将重建多维数组

2 个答案: