Question

我有两个数组，我试图根据它们的日期合并。这是数组的样子：

$a[0][0] = '11/15/08';
$a[0][1] = '50';
$a[1][0] = '11/20/08';
$a[1][1] = '75';
$a[2][0] = '01/04/09';
$a[2][1] = '23';
$a[3][0] = '01/09/09';
$a[3][1] = '92';

和

$b[0][0] = '01/04/09';
$b[0][1] = '30';
$b[1][0] = '01/05/09';
$b[1][1] = '54';
$b[2][0] = '01/08/09';
$b[2][1] = '89';
$b[3][0] = '01/09/09';
$a[3][1] = '62';

在合并结束时，我希望：

$n[0][0] = '11/15/08';
$n[0][1] = '50';
$n[0][2] = '0';
$n[1][0] = '11/20/08';
$n[1][1] = '75';
$n[1][2] = '0';
$n[2][0] = '01/04/09';
$n[2][1] = '23';
$n[2][2] = '30';
$n[3][0] = '01/05/09';
$n[3][1] = '0';
$n[3][2] = '54';
$n[4][0] = '01/08/09';
$n[4][1] = '0';
$n[4][2] = '89';
$n[4][0] = '01/09/09';
$n[4][1] = '92';
$n[4][2] = '62';

这可能吗？

Answer 1

我首先将数组排序为一个以时间戳为键，数组为数值的数组。这将很容易排序。例如：

$result = array();
foreach (array_merge($a, $b) as $item) {
    $date = $item[0];
    $dateParts = explode('/', $date);
    $timestamp = mktime(0,0,0,$dateParts[0], $dateParts[1], $dateParts[2]);
    if (!isset($result[$timestamp])) {
        $result[$timestamp] = array();
    }   

    $result[$timestamp][] = $item[1];
}

因为键是日期时间戳，所以我们可以按日期轻松地对数组进行排序：

ksort($result);

然后我们可以将数组转换为您想要的格式

$result2 = array();
foreach ($result as $timestamp => $item) {
    $tmp = array();

    $tmp[] = date('m/d/y', $timestamp);
    $tmp = array_merge($tmp, $item);
    $result2[] = $tmp;
}

print_r($result2);

Answer 2

最简单的方法是将日期用作索引

    #Initialization

    $n = array();
    $n[$date] = array();
    $n[$date][0] = "";
    $n[$date][1] = "";

    #and then

    $n["10/01/09"][0] = "10";
    $n["10/01/09"][1] = "50";
    $n["11/02/09"][1] = "70";
    $n["01/05/09"][0] = "90";

--- EDIT

你可以在“array_merge”，“array_combine”中看到这个函数，但是在这个例子中，这可以解决了

Answer 3

这是我的解决方案：

$a[0][0] = '11/15/08';
$a[0][1] = '50';
$a[1][0] = '11/20/08';
$a[1][1] = '75';
$a[2][0] = '01/04/09';
$a[2][1] = '23';
$a[3][0] = '01/09/09';
$a[3][1] = '92';

$b[0][0] = '01/04/09';
$b[0][1] = '30';
$b[1][0] = '01/05/09';
$b[1][1] = '54';
$b[2][0] = '01/08/09';
$b[2][1] = '89';
$b[3][0] = '01/09/09';
$b[3][1] = '62';

function dates($arr, &$result)
{
    foreach ($arr as $dates)
    {
        $result[$dates[0]][] = $dates[1];
    }
}

$result = array();
dates($a, $result);
dates($b, $result);

$dates = array();
$i = 0;
foreach ($result as $date => $res)
{
    $dates[$i][0] = $date;
    $dates[$i][1] = $res[0];
    if (count($res) == 1)
    {
        $dates[$i][2] = 0;
    } else {
        $dates[$i][2] = $res[1];
    }
    $i++;
}
print_r($dates);

输出：

Array
(
    [0] => Array
        (
            [0] => 11/15/08
            [1] => 50
            [2] => 0
        )

    [1] => Array
        (
            [0] => 11/20/08
            [1] => 75
            [2] => 0
        )

    [2] => Array
        (
            [0] => 01/04/09
            [1] => 23
            [2] => 30
        )

    [3] => Array
        (
            [0] => 01/09/09
            [1] => 92
            [2] => 62
        )

    [4] => Array
        (
            [0] => 01/05/09
            [1] => 54
            [2] => 0
        )

    [5] => Array
        (
            [0] => 01/08/09
            [1] => 89
            [2] => 0
        )

)

Php在日期合并阵列

3 个答案: