Question

我有两个包含一些值的数组。

$type = array("first", "second", "third");

$date = array(
          0 => "2019-04-30",
          1 => "2019-05-01",
          2 => "2019-05-02",
          3 => "2019-05-03"
        );

我需要输出如下内容：

[
    type :first, 
    date: [
        "2019-04-30": 1.2,
        .....
    ]

]

但是由于某种原因，我没有采用这种格式。

这是我尝试过的代码。

$newArr = array();

foreach($type as $tt) {
    $newArr[]['type'] = $tt;
    $newDate = array();
    foreach ($date as $d) {
        $newDate[$d] = 1.2;
    }
    $newArr[]['date'] = $newDate;
}

任何人都可以证明我做错了什么。

谢谢。

Answer 1

它只涉及构建数组，然后以正确的顺序添加它，这将构建所有数据并在循环结束时一口气添加它...

$newArr = array();

foreach($type as $tt) {
    $newDate = array();
    foreach ($date as $d) {
        $newDate[$d] = 1.2;
    }
    $newArr[] = [ 'type' => $tt, 'date' => $newDate];
}

您可以将其缩短为这个长度，但这并没有太大的不同...

foreach($type as $tt) {
    $newArr[] = [ 'type' => $tt, 'date' => array_fill_keys($date, 1.2)];
}

Answer 2

让它变得更简单

$newArr = array();
$newdates = array();
  foreach($dates as $date){
  $newdates[$date] = 1.2;
}

    foreach($type as $tt) {
        $newArry[] = array("type"=>$tt,"date"=>$newdates);
    }

Answer 3

您可以使用array_map和array_fill_keys以获得所需的结果

 $type = ["first", "second", "third"];
 $date = [
      0 => "2019-04-30",
      1 => "2019-05-01",
      2 => "2019-05-02",
      3 => "2019-05-03"
    ];
$newArr = [];
array_map(function($t, $d) use ($date, &$newArr){
  $newArr[] = [
    'type' => $t,
    'date' => array_fill_keys($date, 1.2)
  ];
}, $type, $date);

将第二个数组推入第一个数组

3 个答案: