Question

我有一个PHP脚本几乎可以完成我想要的所有操作。我从API中提取了一些信息并准备了一份CSV文档 - 文档创建得很好，但是有几列是空白的，我不知道如何从对象中调用它们（stdClass）我是进入我的vardump。我试图从下面的每个numkey对象中获取以下索引：

名称
VariantCustomIds-＆GT;编号
的updated_at

我真的在努力解决我的解码JSON的var_dump问题。以下是来自相关API的JSON响应：

{
"623587": {
  "caption": "",
  "created_at": "2015-06-30T17:59:24+00:00",
  "deleted": "0",
  "id": "623587",
  "links": {
    "VariantCustomIds": [
      {
        "id": "108601807"
      }
    ]
  },
  "name": "SKU",
  "updated_at": "2015-06-30T17:59:24+00:00"
},
"840664": {
  "caption": "",
  "created_at": "2015-09-21T15:04:18+00:00",
  "deleted": "0",
  "id": "840664",
  "links": {
    "VariantCustomIds": [
      {
        "id": "144012064"
      }
    ]
  },
  "name": "Headband Style",
  "updated_at": "2015-09-21T15:04:18+00:00"
},
"840684": {
  "caption": "",
  "created_at": "2015-09-21T15:04:18+00:00",
  "deleted": "0",
  "id": "840684",
  "links": {
    "VariantCustomIds": [
      {
        "id": "144012074"
      }
    ]
  },
  "name": "Ink Type",
  "updated_at": "2015-09-21T15:04:18+00:00"
},
"840694": {
  "caption": "",
  "created_at": "2015-09-21T15:04:18+00:00",
  "deleted": "0",
  "id": "840694",
  "links": {
    "VariantCustomIds": [
      {
        "id": "144012084"
      }
    ]
  },
  "name": "Fabric Type",
  "updated_at": "2015-09-21T15:04:18+00:00"
}
}

我通过PHP中的json_decode函数传递它，这是我得到的var_dump：

 object(stdClass)#1308 (1) { ["623587"]=> object(stdClass)#1310 (7) { ["caption"]=> string(0) "" ["created_at"]=> string(25) "2015-06-30T17:59:24+00:00" ["deleted"]=> string(1) "0" ["id"]=> string(6) "623587" ["links"]=> object(stdClass)#1309 (1) { ["VariantCustomIds"]=> array(1) { [0]=> object(stdClass)#1303 (1) { ["id"]=> string(9) "108601798" } } } ["name"]=> string(3) "SKU" ["updated_at"]=> string(25) "2015-06-30T17:59:24+00:00" } }

我已经尝试过为每个循环做一个，但是当我这样做时只有一个对象。我尝试过调用$ myVariable-＆gt; id，$ myVariable-＆gt; links和$ myVariable-＆gt; name。当我通过var_dump（）传递它们时，所有这些都返回NULL。

任何帮助都会非常感激。

Answer 1

只需按下括号/大括号，记住在JS中，{}是一个对象，而[]是一个数组：

json = {
|      | "623587": {
|      |           |    "caption": "",
|      |           |    "created_at": "2015-06-30T17:59:24+00:00",
|      |           |    "deleted": "0",
|      |           |    "id": "623587",
|      |           |    "links": {
|      |           |    |        |   "VariantCustomIds": [
|      |           |    |        |   |                   |    {
|      |           |    |        |   |                   |    |    "id": "108601807"
|      |           |    |        |   |                   |    |    |
$obj  ->  623587    ->  links   ->   variantcustomids    [0]  ->   id

所以你的最终PHP路径，使用适当的访问方法和大写

$obj[623587]->links->VariantCustomIds[0]->id

Answer 2

看看json_decode documentation。有第二个（optinal）参数告诉json_decode返回asoc数组。所以你的代码应该是这样的：

$decoded = json_decode($jsonString, true);

Answer 3

我知道，你已经解决了。但你也可以去StdClass对象方法：

<?php

$sample = file_get_contents('http://host/api/sample.json');
$data = json_decode($sample);

foreach ($data as $id => $object) {

    $name       = $object->name;        
    $id         = $object->links->VariantCustomIds[0]->id; #OR $id = $id       
    $updated_at = $object->updated_at;
}

PHP了解我的var_dump

3 个答案: