查询MySQL报错，不知道什么原因 [ 2.0 版本 ]

我用ContentAttr::findOne(["content_id"=>$model->id]);来获得一行记录。

结果报出以下错误：

PDOStatement::execute(): Unknown type 245 sent by the server. Please send a report to the developers
The SQL being executed was: SELECT * FROM `content_attr` WHERE `content_id`='1'

数据库是MySQL5.7.11，数据库的建表语句是：

CREATE TABLE `content_attr` (
  `content_id` int(10) unsigned NOT NULL,
  `content` longtext NOT NULL,
  `attr` json NOT NULL,
  UNIQUE KEY `content_id` (`content_id`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8

在mysql控制台直接执行这个SQL语句没有报错

mysql> SELECT * FROM `content_attr` WHERE `content_id`='1'\G;
*************************** 1. row ***************************
content_id: 1
   content: <p>下次</p>
      attr: {"content_attr": "{\"bt\":\"\\u53d1\",\"btx\":\"\\u8bdd\"}", "attr_data_model": [{"id": "1", "name": "bt", "type": "1", "label": "标题", "rules": "required", "value": "发", "created_at": "1457175722", "updated_at": "1457175722"}, {"id": "2", "name": "btx", "type": "1", "label": "标题1", "rules": "required", "value": "话", "created_at": "1457175722", "updated_at": "1457175722"}]}

我专门写了个测试

如果仅仅select content_id那么就没问题

如果select attr 就直接报错了

说明是PHP在处理MySQL JSON列时出的问题

怎么解决呢？

==============================================================

已经解决了，猜测原因应该是PHP不能直接处理JSON列，用CAST(attr as CHAR)转了一下就可以了

另外 据说是PHP升级到5.6.11以后就好了，我还没有，下载了5.6.19，还没部署

==============================================================