Question

我将以下字符串转换为json数组。我要做的所有事情都是将具有相同值的id_number关键字组合在一起，并将amount加到每个关键字中，并在循环外打印结果值。

$response = '{
               "nhif":[
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T20:18:50"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T19:59:14"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T09:07:34"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T09:04:27"
                  },
                  {
                     "id_number":"12345678",
                     "amount":"5",
                     "date":"2018-09-14T09:04:21"
                  }
               ]
            }';
$json_response = json_decode($response, true);

//Start NHIF
foreach ($json_response['nhif'] as $nhif) {
    echo ''.$nhif{"id_number"}.' '.$nhif{"amount"}.' '.$nhif{"date"}.'<br/>';
}
//END NHIF

我的预期输出是：

#id_number #amount
AA112233    20
12345678    5

有任何解决方法吗？

Answer 1

您可以使用array_reduce通过amount对id_number值求和：

$sums = array_reduce($json_response['nhif'], function ($c, $i) { 
     $c[$i['id_number']] = (isset($c[$i['id_number']]) ? $c[$i['id_number']] : 0) + $i['amount']; 
     return $c; 
}, array());
print_r($sums);

输出：

AA112233    20
12345678    5

Demo on rextester

如果您使用的是PHP7，则此

isset($c[$i['id_number']]) ? $c[$i['id_number']] : 0

可以简化为

$c[$i['id_number']] ?? 0

Answer 2

您可以使用数组解决此问题

$response = '{
               "nhif":[
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T20:18:50"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T19:59:14"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T09:07:34"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T09:04:27"
                  },
                  {
                     "id_number":"12345678",
                     "amount":"5",
                     "date":"2018-09-14T09:04:21"
                  }
               ]
            }';
$json_response = json_decode($response, true);

$results = array();  // will map the unique keys with it's sum
//Start NHIF
foreach ($json_response['nhif'] as $nhif) {
    if(!isset($results[$nhif['id_number']]))  // new key initialized
        $results[$nhif['id_number']] = $nhif['amount'];
    else
        $results[$nhif['id_number']] += $nhif['amount'];
}
//END NHIF

foreach($results as $key => $value) {
    echo $key .' '.$value;  // printing the result
}

Answer 3

对数组排序[如果未对数据进行排序，则不需要排序]。
执行for循环和求和。

尝试使用此代码示例，根据解决问题的技巧，还有许多其他方法可以解决此问题。

$response = '{
               "nhif":[
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T20:18:50"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T19:59:14"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T09:07:34"
                  },
                  {
                     "id_number":"AA112233",
                     "amount":"5",
                     "date":"2018-09-14T09:04:27"
                  },
                  {
                     "id_number":"12345678",
                     "amount":"5",
                     "date":"2018-09-14T09:04:21"
                  }
               ]
            }';
$data = json_decode($response, true);
usort($data['nhif'], function($a, $b) {
    return $b['id_number'] <=> $a['id_number'];
});

//Start NHIF
$amount = 0;
$id_number = null;
foreach($data['nhif'] as $nhif) {
    if(($nhif['id_number'] != $id_number) && ($id_number != null)){
        echo $id_number. "\t". $amount . "<br />";
        $id_number = $nhif['id_number'];
        $amount = $nhif['amount'];
    }
    else{
        $id_number = $nhif['id_number'];
        $amount = $amount + $nhif['amount'];
    }
    // echo ''.$nhif{"id_number"}.' '.$nhif{"amount"}.' '.$nhif{"date"}.'<br />';
}
echo $id_number. "\t". $amount . "<br />";
//END NHIF

Answer 4

您可以创建一个新数组，例如$r，并检查$nhif['id_number']的密钥是否已存在。

如果有，请使用$nhif['amount']为+=添加值。如果不是，请使用=

设置值

此示例使用ternary ?:运算符：

$r = [];
foreach ($json_response['nhif'] as $nhif) {
    isset($r[$nhif['id_number']]) ? $r[$nhif['id_number']] += $nhif['amount'] : $r[$nhif['id_number']] = $nhif['amount'];
}

Demo

如何获取不同键的总和-> PHP数组中的值

4 个答案: