如何使用 PHP while 语句构建多维 json 数组？

Question

正如您在下面的代码中看到的，2020-01-13 上有两个约会，但它们是作为单独的数组生成的。因为它们在同一天，我怎样才能把它们放在同一个阵列上？我正在使用 mysqli 从数据库中获取所有信息，但为了这段代码，我只打印必要的信息。它有效，但我需要知道如何在同一数组中收集具有相同日期的约会。

php

//declare variable
$result = array();

//iterate
while($schedule = $result_data->fetch_assoc()){

//start time conversion into american time
// 24-hour time to 12-hour time 
$start_time  =date("g:i a", strtotime($schedule['start_time']));

//end time conversion into american time
$end_time  =date("g:i a", strtotime($schedule['end_time']));

$my_schedule_start .= $start_time;
$my_schedule_end .= $end_time;

$schedule_date= $schedule['cleaning_date'];


$result[] = array(
"$schedule_date" => [
    "number"=>'1',
    "badgeClass"=>'',
    "url"=>'url',

    "dayEvents"=>[
    "title"=>$schedule['first_name'].' '.$schedule['last_name'],
    "status"=>$schedule['cleaning_status'],
    "time"=>$start_time .' - '.$end_time
    ]
]
);  

}

//send response back to jquery
header('Content-Type: application/json');
echo json_encode($result);

这个 php 打印这个

[{"2020-01-15":{
    "number":"1",
    "badgeClass":"",
    "url":"url",
    "dayEvents":{
        "title":"Jen Doe",
        "status":"Booked",
        "time":"11:00 am - 12:00 pm"}
        }
        },
        {"2020-01-13":{
            "number":"1",
            "badgeClass":"",
            "url":"url",
            "dayEvents":{
                "title":"John Doe",
                "status":"Booked",
                "time":"2:00 pm - 5:00 pm"}
                }
                },
                {"2020-01-13":{
                    "number":"1",
                    "badgeClass":"",
                    "url":"url",
                    "dayEvents":{
                        "title":"Alfred Doe",
                        "status":"Booked",
                        "time":"11:00 am - 12:00 pm"}
                        }
                        }
                     ]

但我想要的结果是这样

[{
        "2020-01-15": {
            "number": "1",
            "badgeClass": "",
            "url": "url",
            "dayEvents": {
                "title": "Jen Doe",
                "status": "Booked",
                "time": "11:00 am - 12:00 pm"
            }
        }
    },
    {
        "2020-01-13": {
            "number": "2",
            "badgeClass": "",
            "url": "url",
            "dayEvents": [{
                    "title": "John Doe",
                    "status": "Booked",
                    "time": "2:00 pm - 5:00 pm"
                },
                {
                    "title": "Alfred Doe",
                    "status": "Booked",
                    "time": "11:00 am - 12:00 pm"
                }
            ]
        }
    }
]

Answer 1

这段代码在创建日程表时，会创建由日期本身索引的每一天事件，然后当发现新事件时，它将检查是否已设置此日期。如果是这样，它将增加计数并添加新事件。唯一复杂的是它需要检查它是否只有 1 个现有条目——这需要先转换为数组。要删除索引，json_encode() 首先使用 array_values()...

while($schedule = $result_data->fetch_assoc()){
    //start time conversion into american time
    // 24-hour time to 12-hour time
    $start_time  =date("g:i a", strtotime($schedule['start_time']));

    //end time conversion into american time
    $end_time  =date("g:i a", strtotime($schedule['end_time']));

    $my_schedule_start .= $start_time;
    $my_schedule_end .= $end_time;

    $schedule_date= $schedule['cleaning_date'];

    if ( isset($result[$schedule_date])){
        // Increment count
        $result[$schedule_date][$schedule_date]["number"]++;
        // Convert single item to array
        if ( !isset($result[$schedule_date][$schedule_date]["dayEvents"][0]))    {
            $result[$schedule_date][$schedule_date]["dayEvents"] = 
                [$result[$schedule_date][$schedule_date]["dayEvents"]];
        }
        // Add in new event to dayEvents
        $result[$schedule_date][$schedule_date]["dayEvents"][] = [
            "title"=>$schedule['first_name'].' '.$schedule['last_name'],
            "status"=>$schedule['cleaning_status'],
            "time"=>$start_time .' - '.$end_time
        ];

    }
    else    {
        // Create new days schedule
        $result[$schedule_date] = [
            "$schedule_date" => [
                "number"=>'1',
                "badgeClass"=>'',
                "url"=>'url',

                "dayEvents"=>[
                    "title"=>$schedule['first_name'].' '.$schedule['last_name'],
                    "status"=>$schedule['cleaning_status'],
                    "time"=>$start_time .' - '.$end_time
                ]
            ]
        ];
    }

}

header('Content-Type: application/json');
echo json_encode(array_values($result), JSON_PRETTY_PRINT);

Answer 2

根据我对上面提供的代码和所需输出的观察，您可以这样做：

$scheduledDate = date('Y-m-d');
$my_schedule_start = "";
$my_schedule_end = "";
$finalResult = [];

while($schedule = $result_data->fetch_assoc()){
    $start_time  =date("g:i a", strtotime($schedule['start_time']));
    $end_time  =date("g:i a", strtotime($schedule['end_time']));

    $my_schedule_start .= $start_time;
    $my_schedule_end .= $end_time;

    $schedule_date= $schedule['cleaning_date'];

    $res = new stdClass;

    $res->number = 1;
    $res->badgeClass = "";
    $res->url = 'url';

    $dayEvents = new stdClass;
    $dayEvents->title = $schedule['first_name'].' '.$schedule['last_name'];
    $dayEvents->status = $schedule['cleaning_status'];
    $dayEvents->time = $start_time .' - '.$end_time;

    $res->dayEvents = $dayEvents;
    $finalResult[] = $res;
}

echo json_encode($finalResult);

这里需要注意的事项：

1.) 我已经根据提供的输出中的第一个条目给出了代码库。如果你想更深入的话，而不是在$dayEvents上使用stdClass，你可以直接将$dayEvents声明为一个数组并在里面使用单独的stdClass并将它推到dayEvents 就像我为 $finalResult 数组所做的那样。

2.) 您可能会收到 $my_schedule_start 和 $my_schedule_end 之前未声明的通知，因此我冒昧地在 While 循环之外用空值声明了它们。但请随时根据您的需要更改它们。

3.) 如果 $res = new stdClass; 不起作用，那么您可以尝试 $res = new \stdClass。

4.) 如果您发现任何语法错误，请原谅我，因为我也是人类:)

总结：

stdClass 是 PHP 的内部功能，用于在 PHP 中声明对象。您还可以使用 $dayEvents = {}; 之类的短语法代替 $dayEvents = new stdClass;.

您可以在 stdClass 上找到更多信息 Here

希望这对您有所帮助。

如何使用 PHP while 语句构建多维 json 数组？

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