Question

我从API获得了以下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"}
  ]
]

我想将其转换为适当的对象数组

[
{"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"}
]

但我无法实现这个

这是我试过的代码

$data = json_decode($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_array返回了

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"
  }
]

但阵列索引也即将到来。我希望它像下面的

[
    {"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"}
    ]

我如何实现这一目标？我已经尝试了但是我的最终JSON响应包含了我不想要的数组索引。

Answer 1

就个人而言，我会以这种方式绕过迭代：

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

这将保留您的内部值作为对象，并仍然将内部值作为对象返回。

如果必须进行迭代，则应将array_values称为返回：

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

这将消除键，它将使json_encode将它视为数组，而不是专门的对象。

也就是说，如果您正在执行var_dump或var_export，它将始终包含索引号。这只是格式化的一部分。

无法在laravel中将索引数组转换为没有索引的对象数组

1 个答案: