JSON_EXTRACT不适用于嵌套json数据 [英] JSON_EXTRACT not working for nested json data

问题描述

我想从我的表中选择json数据，以便显示我的表数据是这样的:

I want to select the data from my table which is json data so to show my table data is like this :

user_id:    1

metaname:  mymetaname

meta_value:    a:1:{i:0;a:10:{s:7:"street1";s:36:"shiv plaza";s:4:"city";s:5:"surat";s:5:"state";s:7:"gujarat";s:7:"zipcode";s:6:"395010";s:14:"dollet_country";s:2:"IN";s:10:"tostreet1l";s:5:"surat";s:7:"tocityl";s:5:"surat";s:8:"tostatel";s:5:"surat";s:10:"tozipcodel";s:6:"395000";s:17:"todollet_countryl";s:2:"IN";}}

我正在尝试运行此查询:

And i am trying to run this query :

SELECT user_id,JSON_EXTRACT(meta_value, '$."city"') FROM  `usermetatable`

但是显示错误:

[功能json_extract的参数1中的JSON文本无效:无效 值."位于位置0.]

[Invalid JSON text in argument 1 to function json_extract: "Invalid value." at position 0.]

我在表中的json数据无法更改为其他数据，并且可以肯定是正确的JSON，有人可以更正上面的查询吗?

My json data in table can not be changed to other and it's correct JSON for sure, Could anyone correct above query ?

推荐答案

这不是JSON数据.它看起来像一个序列化的PHP对象.参见 http://php.net/serialize

That's not JSON data. It looks like a serialized PHP object. See http://php.net/serialize

没有MySQL函数可以从该序列化对象中提取字段.您应该将整个对象提取到PHP应用程序中，然后调用 unserialize()在它上面，然后访问对象成员.

There's no MySQL function for extracting a field from that serialized object. You should fetch the whole object into a PHP app, and call unserialize() on it, then access the object members.

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