Php 如何构建对象/数组？

我对PHP相当陌生，所以我不知道如何使用这些数据集。我做了一个MySQL选择，然后得到一个如下的对象：

{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 1,
  "dailyworkhourssum": "7.70"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 3,
  "dailyworkhourssum": "1.2"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-05",
  "categoryid": 3,
  "dailyworkhourssum": "7.70"
},

{
      "membername": "NAME",
      "bookingdate": "2020-02-03",
      "categoryid1": true,
      "categorid3": true,
      "dailyworkhourssum1": "7.70",
      "dailyworkhourssum3": "1.2"
    },
    {
      "membername": "NAME",
      "bookingdate": "2020-02-05",
      "categoryid": 3,
      "dailyworkhourssum": "7.70"
    },

    $books = [
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 1,
       "dailyworkhourssum" => "7.70"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 3,
       "dailyworkhourssum" => "1.2"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-05",
       "categoryid" => 3,
       "dailyworkhourssum" => "7.70"
     ]
   ];

$merged = [];
foreach ($books as $book) {
    $date = $book['bookingdate'];
    if (isset($merged[$date])) {
        $merged[$date] = $merged[$date] + $book;
    } else {
        $merged[$date] = $book;
    }
}

我想重复一下，最后应该是这样的：

{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 1,
  "dailyworkhourssum": "7.70"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 3,
  "dailyworkhourssum": "1.2"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-05",
  "categoryid": 3,
  "dailyworkhourssum": "7.70"
},

{
      "membername": "NAME",
      "bookingdate": "2020-02-03",
      "categoryid1": true,
      "categorid3": true,
      "dailyworkhourssum1": "7.70",
      "dailyworkhourssum3": "1.2"
    },
    {
      "membername": "NAME",
      "bookingdate": "2020-02-05",
      "categoryid": 3,
      "dailyworkhourssum": "7.70"
    },

    $books = [
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 1,
       "dailyworkhourssum" => "7.70"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 3,
       "dailyworkhourssum" => "1.2"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-05",
       "categoryid" => 3,
       "dailyworkhourssum" => "7.70"
     ]
   ];

$merged = [];
foreach ($books as $book) {
    $date = $book['bookingdate'];
    if (isset($merged[$date])) {
        $merged[$date] = $merged[$date] + $book;
    } else {
        $merged[$date] = $book;
    }
}

这样做的目的是将两个字段合并在一起（如果它们具有相同的

bookingdate

），这样我就可以在表中显示它，而不会重复日期。 我的问题是：

我不知道这类数据叫什么

我不知道如何创造这样的东西

---

我可以使用

$data->newField=example

向此类数据添加字段，因此我认为这是一个对象。

在JS中称为对象，但在PHP中，您将使用关联数组。 在你的例子中，我认为，你有一个关联数组。看起来是这样的：

{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 1,
  "dailyworkhourssum": "7.70"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 3,
  "dailyworkhourssum": "1.2"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-05",
  "categoryid": 3,
  "dailyworkhourssum": "7.70"
},

{
      "membername": "NAME",
      "bookingdate": "2020-02-03",
      "categoryid1": true,
      "categorid3": true,
      "dailyworkhourssum1": "7.70",
      "dailyworkhourssum3": "1.2"
    },
    {
      "membername": "NAME",
      "bookingdate": "2020-02-05",
      "categoryid": 3,
      "dailyworkhourssum": "7.70"
    },

    $books = [
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 1,
       "dailyworkhourssum" => "7.70"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 3,
       "dailyworkhourssum" => "1.2"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-05",
       "categoryid" => 3,
       "dailyworkhourssum" => "7.70"
     ]
   ];

$merged = [];
foreach ($books as $book) {
    $date = $book['bookingdate'];
    if (isset($merged[$date])) {
        $merged[$date] = $merged[$date] + $book;
    } else {
        $merged[$date] = $book;
    }
}

如果要合并具有相同“bookingdate”的数组，我建议您循环此数组，并将其元素添加到另一个具有bookingdate作为键的关联数组中，然后检查是否已经存在此类键，然后合并数组，如下所示：

{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 1,
  "dailyworkhourssum": "7.70"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-03",
  "categoryid": 3,
  "dailyworkhourssum": "1.2"
},
{
  "membername": "NAME",
  "bookingdate": "2020-02-05",
  "categoryid": 3,
  "dailyworkhourssum": "7.70"
},

{
      "membername": "NAME",
      "bookingdate": "2020-02-03",
      "categoryid1": true,
      "categorid3": true,
      "dailyworkhourssum1": "7.70",
      "dailyworkhourssum3": "1.2"
    },
    {
      "membername": "NAME",
      "bookingdate": "2020-02-05",
      "categoryid": 3,
      "dailyworkhourssum": "7.70"
    },

    $books = [
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 1,
       "dailyworkhourssum" => "7.70"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-03",
       "categoryid" => 3,
       "dailyworkhourssum" => "1.2"
     ],
     [
       "membername" => "NAME",
       "bookingdate" => "2020-02-05",
       "categoryid" => 3,
       "dailyworkhourssum" => "7.70"
     ]
   ];

$merged = [];
foreach ($books as $book) {
    $date = $book['bookingdate'];
    if (isset($merged[$date])) {
        $merged[$date] = $merged[$date] + $book;
    } else {
        $merged[$date] = $book;
    }
}

我认为这不是一个有效的代码（没有时间，对不起），但我希望你能接受这个想法

如果希望使用“列表”而不是关联数组，则可以执行以下操作：

 $mergedList = array_values($merged);

---

因此，您将摆脱字符串键。

如果我理解正确，您将获得一个具有4列和可变行数的表，并希望将其转换为具有可变列数的表。因此，使用一个数据结构，其中每个项目都不同于前一个项目，这会使一切变得比它需要的更困难。我建议您使用固定结构：

// I'm assuming you have a PHP array as starting point
$input = [
    [
        'membername' => 'NAME',
        'bookingdate' => '2020-02-03',
        'categoryid' => 1,
        'dailyworkhourssum' => '7.70',
    ],
    [
        'membername' => 'NAME',
        'bookingdate' => '2020-02-03',
        'categoryid' => 3,
        'dailyworkhourssum' => '1.2',
    ],
    [
       'membername' => 'NAME',
       'bookingdate' => '2020-02-05',
       'categoryid' => 3,
       'dailyworkhourssum' => '7.70',
    ],
];

$output = [];
foreach ($input as $data) {
    // We'll group by booking date
    if (!isset($output[$data['bookingdate']])) {
        $output[$data['bookingdate']] = [
            'membername' => $data['membername'],
            'bookingdate' => $data['bookingdate'],
            'categoryid' => $data['categoryid'],
            'dailyworkhourssum' => [],
        ];
    }
    // A single date may have several daily work hours
    $output[$data['bookingdate']]['dailyworkhourssum'][] = $data['dailyworkhourssum'];
}

// We discard array keys (we only needed them to group)
echo json_encode(array_values($output));

---

无论在何处使用此JSON，都只需循环

dailyworkhourssum

数组。在打印表格之前，您可能还希望循环整个结构，并保留一个计数器，以确定最大列数，以便在需要时绘制空单元格（表格为矩形）。

您的最终目标是什么？您建议的数据结构将非常难以使用。SQL结果集基本上已经是表了，目标是将此结构返回到ajax调用，并将其显示在表中。每个日期只能打印一次，因此我必须将重复出现的日期合并在一起。通过select语句删除重复记录似乎更有意义。否则，您选择的内容超出了需要，只能将其删除。加倍努力。到目前为止你试过什么？你被困在哪里？