运算符不存在:json = json

Question

当我尝试从表中选择一些记录时

    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.

********** ?? **********

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

我错过了什么或者我可以在哪里了解这个错误.

Answer 1

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,类型的值以json给定它们的格式存储为文本.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,您可以使用类型来解决此问题,类型jsonb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

阅读有关JSON类型的更多信息.