php中具有相同名称的对象的总值

Question

我遇到了一个问题：我试图在json中将所有值相同的“id”加起来示例：

[{id: "100002475575341", name: "N1", point: "2"},
 {id: "100002993363229", name: "N2", point: "2"},
 {id: "100002993363229", name: "N2", point: "2"},
 {id: "100002475575341", name: "N1", point: "2"}]

和预期结果：

[{id: "100002475575341", name: "N1", point: "4"},
 {id: "100002993363229", name: "N2", point: "4"}]

Answer 1

注意：我通过在索引名称周围添加引号，将JSON字符串更改为有效

将JSON解码为数组，然后循环它。将它添加到由$results索引的name数组中。如果元素存在，请将这些点一起添加。如果它不存在，请创建它。

$json = '[{"id": "100002475575341", "name": "N1", "point": "2"},
          {"id": "100002993363229", "name": "N2", "point": "2"},
          {"id": "100002993363229", "name": "N2", "point": "2"},
          {"id": "100002475575341", "name": "N1", "point": "2"}]';

$array = json_decode($json);
$result = [];

foreach ($array as $a) {
    if (isset($result[$a->name])) {
        $result[$a->name]->point += $a->point;
    } else {
        $result[$a->name] = $a;
    }
}

$output = json_encode(array_values($result));

• 在https://3v4l.org/CVOmt现场演示

Answer 2

我以你的代码为例，你可以试试这个

注意：我更改了您的JSON，因为它是无效的JSON格式。

<?php   

  $jsonString = '[{
      "id": "100002475575341",
      "name": "N1",
      "point": "2"
    },
    {
      "id": "100002993363229",
      "name": "N2",
      "point": "2"
    },
    {
      "id": "100002993363229",
      "name": "N2",
      "point": "2"
    },
    {
      "id": "100002475575341",
      "name": "N1",
      "point": "2"
    }
  ]';

  $jsonDecoded = json_decode($jsonString, true);

  $res = [];
  foreach ($jsonDecoded as $value) {
    if(array_key_exists($value['id'], $res)){
        $res[$value['id']] += $value['point'];
    }else{
        $res[$value['id']] = $value['point'];
    }
  }
  print_r($res);

?>