运算符不存在:json = json

时间:2015-09-29 11:49:06

标签: sql json postgresql postgresql-9.4 jsonb

当我尝试从表格中选择一些记录时

    SELECT * FROM movie_test WHERE tags = ('["dramatic","women", "political"]'::json)

sql代码抛出错误

LINE 1: SELECT * FROM movie_test WHERE tags = ('["dramatic","women",...
                                        ^
HINT:  No operator matches the given name and argument type(s). You might      need to add explicit type casts.

********** 错误 **********

ERROR: operator does not exist: json = json
SQL 状态: 42883
指导建议:No operator matches the given name and argument type(s). You might need to add explicit type casts.
字符:37

我是否遗漏了某些内容或者我可以在哪里了解这个错误。

1 个答案:

答案 0 :(得分:20)

您无法比较json值。您可以改为比较文本值:

SELECT * 
FROM movie_test 
WHERE tags::text = '["dramatic","women","political"]'

但请注意,json类型的值以文本格式存储。因此,比较的结果取决于您是否始终如一地应用相同的格式:

SELECT 
    '["dramatic" ,"women", "political"]'::json::text =  
    '["dramatic","women","political"]'::json::text      -- yields false!

在Postgres 9.4+中,您可以使用类型jsonb来解决此问题,该类型以分解的二进制格式存储。可以比较此类型的值:

SELECT 
    '["dramatic" ,"women", "political"]'::jsonb =  
    '["dramatic","women","political"]'::jsonb           -- yields true

所以这个查询更可靠:

SELECT * 
FROM movie_test 
WHERE tags::jsonb = '["dramatic","women","political"]'::jsonb

详细了解JSON Types