select doc->>"$.providers.abc.provider" from x1;
+----------------------------------+
| doc->>"$.providers.abc.provider" |
+----------------------------------+
| foo                              |
+----------------------------------+

一行一组（0.0008秒）
您的路径是providers..provider

您需要使用传递的表/子查询来解析出所需的部分。
是的，只是我不知道是第一家供应商。
这将解析出第一个provider元素。。
查询

SELECT 
 *
 , records_json_array.json_array->'$[0].token'
FROM (
SELECT 
 records.json->'$.providers.*' AS json_array
FROM (
SELECT 
 '{
  "providers": {
    "abc": {
      "provider": "foo",
      "token": "abc123"
    },
    "def": {
      "provider": "bar",
      "token": "def567"
    }
  }
}' AS json
FROM 
 DUAL
) AS records
) AS records_json_array

结果

| json_array                                                                       | records_json_array.json_array->'$[0].token' |
| -------------------------------------------------------------------------------- | ---------------------------------------------- |
| [{"token": "abc123", "provider": "foo"}, {"token": "def567", "provider": "bar"}] | "abc123"                                       |

演示

我最终解决了这个问题（使用mysql 5.7，这就是为什么我没有 ->> 语法）：

SET @j = '{
  "providers": {
    "abc": {
      "provider": "foo",
      "token": "abc123"
    },
    "def": {
      "provider": "bar",
      "token": "def567"
    }
  }
}';
SELECT JSON_UNQUOTE(
  JSON_EXTRACT(
    JSON_EXTRACT(@j, "$.providers.*"),
    "$[0].token"
  )
) AS token;