Question

我正在尝试填充数组，以便我可以使用chart.js

绘制图表

我有一个看起来像这样的数组：

[0] => Array
    (
        [id] => 1
        [date] => 09-04-2018
        [length] => 10
    )

[1] => Array
    (
        [id] => 2
        [date] => 09-04-2018
        [length] => 20
    )

[2] => Array
    (
        [id] => 1
        [date] => 10-04-2018
        [length] => 11
    )

[3] => Array
    (
        [id] => 2
        [date] => 10-04-2018
        [length] => 21
    )

[4] => Array
    (
        [id] => 1
        [date] => 11-04-2018
        [length] => 12
    )

[5] => Array
    (
        [id] => 1
        [date] => 12-04-2018
        [length] => 13
    )

[6] => Array
    (
        [id] => 2
        [date] => 12-04-2018
        [length] => 23
    )

[7] => Array
    (
        [id] => 1
        [date] => 13-04-2018
        [length] => 14
    )

我需要将这个数组拆分为2个数据集（id 1和id 2）但是我不能在这些数据集中留有空白，因为chartJS不喜欢它。

我需要数组看起来像：

[1] => Array
(
    [09-04-2018] => 10
    [10-04-2018] => 11
    [11-04-2018] => 12
    [12-04-2018] => 13
    [13-04-2018] => 14
)

[2] => Array
(
    [09-04-2018] => 20
    [10-04-2018] => 21
    [11-04-2018] => 0
    [12-04-2018] => 23
    [13-04-2018] => 0
)

使用以下代码： -

foreach ($array as $item)
{
    $id = $item['id'];
    $date = $item['date'];
    $length = $item['length'];

    $output[$id][$date] = $length;
}

我可以按照我喜欢的格式生成一个数组，但由于初始数组缺少数据集2的2个日期，所以我没有添加这些数据，结果如下： -

[1] => Array
(
    [09-04-2018] => 10
    [10-04-2018] => 11
    [11-04-2018] => 12
    [12-04-2018] => 13
    [13-04-2018] => 14
)

[2] => Array
(
    [09-04-2018] => 20
    [10-04-2018] => 21
    [12-04-2018] => 23
)

有没有办法以我想要的方式填充数组，所以如果其中一个数据集使用该日期，两者（所有可能是3或4个数据集）都必须使用该日期，只需添加0？

Answer 1

<强> INPUT

$array = array(
    array('id' => 1,'date' => '09-04-2018','length' => 10),
    array('id' => 2,'date' => '09-04-2018','length' => 20),
    array('id' => 1,'date' => '10-04-2018','length' => 11),
    array('id' => 2,'date' => '10-04-2018','length' => 21),
    array('id' => 1,'date' => '11-04-2018','length' => 12),
    array('id' => 1,'date' => '12-04-2018','length' => 13),
    array('id' => 2,'date' => '12-04-2018','length' => 23),
    array('id' => 1,'date' => '13-04-2018','length' => 14)
);

<强>解

    $id = array();
foreach($array as $r){
    $data[$r['id']][$r['date']] = $r['length'];
    if(!in_array($r['id'],$id)) $id[] = $r['id'];
}
for($i=0;$i<count($id);$i++){
    $cid = $id[$i];
    $nid = $id;
    unset($nid[$i]);
    foreach($data[$cid] as $k => $r){
        foreach($nid as $n){
            if(!isset($data[$n][$k]))$data[$n][$k] = 0;
        }
    }
}
echo "<pre>";print_r($data);

<强>输出

Array
(
    [1] => Array
        (
            [09-04-2018] => 10
            [10-04-2018] => 11
            [11-04-2018] => 12
            [12-04-2018] => 13
            [13-04-2018] => 14
        )

    [2] => Array
        (
            [09-04-2018] => 20
            [10-04-2018] => 21
            [11-04-2018] => 0
            [12-04-2018] => 23
            [13-04-2018] => 0
        )

)

Answer 2

此解决方案中，如果id

中没有任何日期

然后会添加date 0值

$a = array(
        array('id' => '1','date' => '09-04-2018','length' => '10'),
        array('id' => '2','date' => '09-04-2018','length' => '20'),
        array('id' => '1','date' => '10-04-2018','length' => '11'),
        array('id' => '2','date' => '10-04-2018','length' => '21'),
        array('id' => '1','date' => '11-04-2018','length' => '12'),
        array('id' => '1','date' => '12-04-2018','length' => '13'),
        array('id' => '2','date' => '12-04-2018','length' => '23'),
        array('id' => '1','date' => '13-04-2018','length' => '14')
    );

$b = array_unique(array_column($a,'date'));
$a1 = $a2 = [];

foreach ($b as $key => $date) {
    $a1[1][$date] = 0;
    $a1[2][$date] = 0;
    foreach ($a as $key2 => $value2) {
        if($value2['id'] == 1){
            if($date == $value2['date']){
                $a1[1][$date] = $value2['length'];
            }
        }elseif($value2['id'] == 2){
            if($date == $value2['date']){
                $a1[2][$date] = $value2['length'];
            }
        }
    }
}

适用于2个以上的数据集

$b = array_unique(array_column($a,'date'));
$sets = array_unique(array_column($a,'id'));
$a1 = $a2 = [];

foreach ($sets as $s) {
    foreach ($b as $key => $date) {
        $a1[$s][$date] = 0;
        foreach ($a as $key2 => $value2) {
            if($value2['id'] == $s){
                if($date == $value2['date']){
                    $a1[$s][$date] = $value2['length'];
                }
            }
        }
    }   
}

echo "<pre>";
print_r($a1);
echo "</pre>";

<强>输出

Array
(
    [1] => Array
        (
            [09-04-2018] => 10
            [10-04-2018] => 11
            [11-04-2018] => 12
            [12-04-2018] => 13
            [13-04-2018] => 14
        )

    [2] => Array
        (
            [09-04-2018] => 20
            [10-04-2018] => 21
            [11-04-2018] => 0
            [12-04-2018] => 23
            [13-04-2018] => 0
        )

)

Answer 3

只需添加一个条件即可获得

$data_set = array();
    foreach ($array as $item)
    { 
        $data_set[$item['id']][$item['date']] = (($item['length'] == 0)?0:$item['length']);
    }
echo '<pre>'; print_r($data_set); echo '</pre>';exit();

Answer 4

<?php
$dataset=array();
for ($i=0;$i<=10;$i++){
    for ($k=0;$k<=100;$k++){
$int= mt_rand(time()-1000000,time());
$dataset[$i][date("Y-m-d",$int)] = mt_rand(0,30) ;
}
}

print_r($dataset);

你会得到像

这样的东西

Php将虚拟数据添加到数组

4 个答案: