如何通过比较删除特定索引

Question

我想通过比较删除一些索引，但这并不容易第一个我们的 php 版本是 5，例如我的结果是这样的，它是 json_encode 的回显，它是数组，数据是动态的[        {            "idx": "1",            "mom_member_idx": "1",            "teacher_member_idx": "2",            "care_start_datetime": "2019-09-09 08:30:00",         },        {            "idx": "2",            "mom_member_idx": "1",            "teacher_member_idx": "2",            "care_start_datetime": "2019-09-10 08:30:00",         },         {            "idx": "3",            "mom_member_idx": "2",            "teacher_member_idx": "2",            "care_start_datetime": "2019-09-09 08:30:00",         }]我想要这样的结果与最新的比较和取消设置[        {            "idx": "1",            "mom_member_idx": "1",            "teacher_member_idx": "2",            "care_start_datetime": "2019-09-09 08:30:00",         },         {            "idx": "3",            "mom_member_idx": "2",            "teacher_member_idx": "2",            "care_start_datetime": "2019-09-09 08:30:00",         }]我试过这样，但它不工作while ($r = mysqli_fetch_assoc($chat_list)) {    $dbdata[] = $r;}for($r=0;$r<sizeof($dbdata);$r++){    if ($dbdata[$r]['mom_member_idx']+$dbdata[$r]['teacher_member_idx']==$dbdata[$r+1]['mom_member_idx']+$dbdata[$r+1]['teacher_member_idx'])    {        if($dbdata[$r]['care_start_datetime']<$dbdata[$r+1]['care_start_datetime']){            unset($dbdata[$r]);        }else {            unset($dbdata[$r+1]);        }    }}

Answer 1

首先，将您的 JSON 数组转换为 PHP 数组。然后您可以使用数组索引删除数组元素

$someArray = json_decode($someJSON, true);

Answer 2

您可以care_start_datetime使用usort()和unset对数据进行排序的第一个索引形式按降序对数据进行排序。

代码示例：

$data = json_decode('[

    {"idx": "1","mom_member_idx": "1","teacher_member_idx": "2","care_start_datetime": "2019-09-09 08:30:00"},

    {"idx": "2","mom_member_idx": "1","teacher_member_idx": "2","care_start_datetime": "2019-09-10 08:30:00"},

    {"idx": "3","mom_member_idx": "2","teacher_member_idx": "2","care_start_datetime": "2019-09-09 08:30:00"}

]', true);

usort($data, function($a, $b) { 

    return strtotime($a['care_start_datetime']) > strtotime($b['care_start_datetime']) ? -1 : 1; 

});

unset($data[0]);

print_r($data);

Answer 3

您对列求和的变体并不完全正确，因为某些整数的总和会给您错误的结果，您可能会意外删除错误的结果。喜欢2 + 3 == 1 + 4不正确。

也许该代码可以帮助您：

$groupedData = array();

while ($r = mysqli_fetch_assoc($chat_list)) {

    // We create unique key for group same rows

    // (with the same 'mom_member_idx' and 'teacher_member_idx')

    $uniqueKey = implode('_', [$r['mom_member_idx'], $r['teacher_member_idx']]);

    if (!array_key_exists($uniqueKey, $groupedData)) {

        $groupedData[$uniqueKey] = array();

    }

    // Add row to grouped array by our unique key

    $groupedData[$uniqueKey][] = $row;

}

$result = array();

foreach ($groupedData as $rows) {

    // If in grouped array with that unique key only one row - just add it

    if (count($rows) <= 1) {

        $result[] = $rows[0];

        continue;

    }

    // Go for all same rows and check date

    $neededRow = null;

    foreach ($rows as $row) {

        // Here I don't understand what kind of date do you want - minimum or maximum

        // You just can reverse the sign

        if (

            is_null($neededRow) ||

            strtotime($row['care_start_datetime']) > strtotime($neededRow['care_start_datetime'])

        ) {

            $neededRow = $row

        }

    }

    if ($neededRow) {

        $result[] = $neededRow;

    }

}