PHP json_decode error: Array value found, but an object is required

Question

我正在尝试解码以下 JSON 文件

{
     "email": "mail@gmail.com",
     "password": "12345",
     "languageProficiency": {
            "language": "English",
             "proficiency": 4
   },
     "tags": [
         {
            "name": "singing"
         },
         {
           "name": "dance"
         }
   ]
}

当我这样做时

 $data = json_decode($jsonContent, true);
 echo $data;
 die();

我有以下错误：

找到数组值，但需要一个对象

问题

1) 如何查看来自 JSON 的数据 2) 如何访问 tags array

中每个对象的属性

我正在根据此模式验证 Json 内容

  {
    "$schema": "http://json-schema.org/draft-04/schema#",
    "type": "object",
    "properties": {
      "email": {
        "type": "string"
      },
      "password": {
        "type": "string"
      },
      "languageProficiency": {
        "type": "object",
        "properties": {
          "language": {
            "type": "string"
          },
          "proficiency": {
            "type": "integer"
          }
        },
        "required": [
          "language",
          "proficiency"
        ]
      },
      "tags": {
        "type": "array",
        "items": {
          "type": "object",
          "properties": {
            "name": {
              "type": "string"
            }
          },
          "required": [
            "name"
          ]
        }
      }
    },
    "required": [
      "email",
      "password",
      "languageProficiency",
      "tags"
    ]
  }

更新

我尝试关注查看json内容

 print_r($data)

但我仍然得到同样的错误

Answer 1

$data = json_decode($jsonContent, true);

上面这行代码会return数组，不能直接用echo打印数组。

要打印数组的特定值（例如电子邮件），请这样做：

echo $data["email"];

提示：

用print_r()知道数组结构是这样的，

echo "<pre>";
print_r($data);

Answer 2

您可以尝试 print_r($data) 来探索价值

PHP json_decode error: Array value found, but an object is required

PHP json_decode error: Array value found, but an object is required

php

arrays

json

jsonschema