Php 无法将数组数组转换为laravel中没有索引的对象数组

我从一个API得到了以下JSON响应

[
  [
   {"companyID":"U72200TG2005PTC045191","companyName":"OPEN TEXT TECHNOLOGIES INDIA PRIVATE LIMITED"},
   {"companyID":"U51909HR2002PTC034956","companyName":"OPEN WAYS MARKETING PRIVATE LIMITED"},
   {"companyID":"U00892HR2005PTC035863","companyName":"OPEN HOUSE DIGITAL MEDIA PRIVATE LIMITED"},
   {"companyID":"U00804KA1999PTC024813","companyName":"OPEN LEAF STOCKS AND SHARES PRIVATE LIMITED"},
   {"companyID":"U72300KA2000PTC026648","companyName":"OPEN STREAM TECHNOLOGIES (INDIA) PRIVATE LIMITED"},
   {"companyID":"U72200KA2003PTC032219","companyName":"OPENCLOVIS SOLUTIONS PRIVATE LIMITED"},
   {"companyID":"U72200KA2003PTC032083","companyName":"OPEN-SILICON RESEARCH PRIVATE LIMITED"},
   {"companyID":"U72900KA2003PTC033042","companyName":"OPENGEAR NETWORKS PRIVATE LIMITED"},
   {"companyID":"U72200DL2005PTC170412","companyName":"OPEN SOLUTIONS SOFTWARE SERVICES PRIVATE LIMITED"},
   {"companyID":"U72112MP1985PTC002837","companyName":"OPEN END SPINNERS PVT LTD"}
  ]
]

我想把它转换成一个合适的对象数组

[
{"companyID":"U72200TG2005PTC045191","companyName":"OPEN TEXT TECHNOLOGIES INDIA PRIVATE LIMITED"},
{"companyID":"U51909HR2002PTC034956","companyName":"OPEN WAYS MARKETING PRIVATE LIMITED"},
{"companyID":"U00892HR2005PTC035863","companyName":"OPEN HOUSE DIGITAL MEDIA PRIVATE LIMITED"}
]

但我无法做到这一点

这是我尝试过的代码

$data = json_decode($result,true);

$new_array = array();
foreach ($data[0] as $to_obj)
{
  new_array[] = (object)$to_obj;
}

return ($new_array);

$result是来自API的JSON。当我这样做时，$new_数组返回

array:10 [
  0 => {#546
    +"companyID": "U72200TG2005PTC045191"
    +"companyName": "OPEN TEXT TECHNOLOGIES INDIA PRIVATE LIMITED"
  }
  1 => {#576
    +"companyID": "U51909HR2002PTC034956"
    +"companyName": "OPEN WAYS MARKETING PRIVATE LIMITED"
  }
  2 => {#574
    +"companyID": "U00892HR2005PTC035863"
    +"companyName": "OPEN HOUSE DIGITAL MEDIA PRIVATE LIMITED"
  }
  3 => {#552
    +"companyID": "U00804KA1999PTC024813"
    +"companyName": "OPEN LEAF STOCKS AND SHARES PRIVATE LIMITED"
  }
  4 => {#548
    +"companyID": "U72300KA2000PTC026648"
    +"companyName": "OPEN STREAM TECHNOLOGIES (INDIA) PRIVATE LIMITED"
  }
  5 => {#572
    +"companyID": "U72200KA2003PTC032219"
    +"companyName": "OPENCLOVIS SOLUTIONS PRIVATE LIMITED"
  }
  6 => {#571
    +"companyID": "U72200KA2003PTC032083"
    +"companyName": "OPEN-SILICON RESEARCH PRIVATE LIMITED"
  }
  7 => {#570
    +"companyID": "U72900KA2003PTC033042"
    +"companyName": "OPENGEAR NETWORKS PRIVATE LIMITED"
  }
  8 => {#569
    +"companyID": "U72200DL2005PTC170412"
    +"companyName": "OPEN SOLUTIONS SOFTWARE SERVICES PRIVATE LIMITED"
  }
  9 => {#560
    +"companyID": "U72112MP1985PTC002837"
    +"companyName": "OPEN END SPINNERS PVT LTD"
  }
]

但数组索引也即将到来。我希望它如下所示

[
    {"companyID":"U72200TG2005PTC045191","companyName":"OPEN TEXT TECHNOLOGIES INDIA PRIVATE LIMITED"},
    {"companyID":"U51909HR2002PTC034956","companyName":"OPEN WAYS MARKETING PRIVATE LIMITED"},
    {"companyID":"U00892HR2005PTC035863","companyName":"OPEN HOUSE DIGITAL MEDIA PRIVATE LIMITED"}
    ]

---

我如何做到这一点？我已经尝试过了，但我的最终JSON响应包含了我不想要的数组索引。

就我个人而言，我会以这种方式绕过迭代：

$data = (array)json_decode($result);
return reset($data);

这会将内部值保留为对象，但仍将内部值作为对象返回

如果必须迭代，则应调用

array\u values

作为返回：

// Everything that you have above, just replace the return with
return array_values($new_array);

这将消除键，并使其成为

json\u encode

将其视为数组，而不是专用对象

---

也就是说，如果您正在进行

var\u转储

或

var\u导出

，它将始终包含索引号。这只是格式的一部分。

就我个人而言，我会这样绕过迭代：

$data = (array)json_decode($result);
return reset($data);

这会将内部值保留为对象，但仍将内部值作为对象返回

如果必须迭代，则应调用

array\u values

作为返回：

// Everything that you have above, just replace the return with
return array_values($new_array);

这将消除键，并使其成为

json\u encode

将其视为数组，而不是专用对象

---

也就是说，如果您正在进行

var\u转储

或

var\u导出

，它将始终包含索引号。这只是格式化的一部分。

我相信这不是一个问题，这是php数组的行为，如果数组没有任何索引，那么默认情况下它将采用数字索引，请查看@Ajeesh，我认为

$data=json\u decode（$result，true）

和使用

$data[0]

将提供预期的输出。我相信这不是问题，这是php数组的行为，如果数组没有任何索引，那么默认情况下它将采用数字索引，请查看@Ajeesh，我想

$data=json\u decode（$result，true）

并使用

$data[0]

将获得预期的输出。