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易语言如何处理json数据

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["id":"64","username":"西瓜都烂了","userll":"0","usernl":"0","usermj":"0","userzl":"0","userhp":"0","usermp":"0","uservocation":"2","usersxdian":"5","userjndian":"1","usergold":"100","userzhijing":"1","userexp":"0","usernextexp":"1000","vip":"0","userwuxing":"5","usergengu":"5","userjingjie":"0","usergongfa":"0","xx_users_id":"56"]

如上面这段,我需要在易语中分别取出不同段所对应的值。求大神帮忙。最后直接上代码。
要用直接json取值,不要分割文本啥的,那个效率不高。

参考技术A …I guess my answer would be determined by different conditions

Hive---如何处理json数据

问题:表中数据像如下这种情况(某个字段是JSON格式),想要单独取出JSON中的某个字段:id,ids,total_number 怎么办?

hive (first_test)> select * from user;OKuser.id user.infos      user.device_id1       {"id": 1701439105,"ids": [2154137571,3889177061],"total_number": 493}   2012       {"id": 1701439106,"ids": [2154137571,3889177061],"total_number": 494}   301Time taken: 0.035 seconds, Fetched: 2 row(s)

文章目录

  • get_json_object
  • json_tuple (官方推荐:效率高)

1、建表

create table user(id int,infos string,device_id int)row format delimitedfields terminated by "\t"lines terminated by "\n";

2、导入数据

user.txt

1       {"id": 1701439105,"ids": [2154137571,3889177061],"total_number": 493}   2012       {"id": 1701439106,"ids": [2154137571,3889177061],"total_number": 494}   301
load data local inpath "/opt/soft/hive/testjob/user.txt" into table user;

get_json_object

使用 get_json_object 函数,对json字符串进行解析
使用方式:
select get_json_object(json所在的列名,’$.json中字段名1’,…)

例如:

hive (first_test)> select get_json_object(infos,"$.id") from user;OK_c017014391051701439106Time taken: 0.071 seconds, Fetched: 2 row(s)

json_tuple (官方推荐:效率高)

使用json_tuple函数,对json字符串进行解析
使用方式:
select json_tuple(json所在的列名,“json中字段名1”,…) from 表名;

例如:

hive (first_test)> select json_tuple(infos,"total_number") from user_tmp;OKc0493494Time taken: 0.033 seconds, Fetched: 2 row(s)

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