如何使用 SQL Developer 提供的“技巧”将每个表从特定用户转换为 JSON 格式答案

【问题标题】：How to convert every Table from a specific User to JSON Format using the "trick" provided by SQL Developer如何使用 SQL Developer 提供的“技巧”将每个表从特定用户转换为 JSON 格式
【发布时间】：2020-01-23 20:30:32
【问题描述】：

我想将特定用户的所有表转换为 JSON（或 XML）格式。我读过 SQL Developer 提到的一个“技巧”。

也就是说，我已经开始创建一个带有两个参数的Procedure：

p_format：格式（在我的例子中是“json”）
p_user：用户名

作为 IDE，我使用 Oracle SQL Developer，我的数据库是 Oracle XE 数据库。

首先，该过程循环遍历给定用户的所有表，并且在循环中，它应该执行以下操作：

SELECT /*p_format*/ * FROM p_user || '.' || table

很遗憾，我不能使用上面提到的这个 SELECT 语句。我需要使用命令EXECUTE IMMEDIATE <Statement>。

我遇到的下一个问题是：我想输出EXECUTE IMMEDIATE 命令的结果。因此我使用了命令EXECUTE IMMEDIATE <Statement> INTO <Variable>。编译程序并执行后，我偶然发现了以下错误：

"inconsistent datatypes: expected %s got %s"

这是我的程序代码：

CREATE OR REPLACE PROCEDURE EXPORT_TABLE_TO_FORMAT_FROM(p_format VARCHAR2, p_user VARCHAR2) IS
/***************************************************************************
        Author: 
        Class:  
        School: 
        Date:   

        Function - EXPORT_TABLE_TO_JSON_FROM(p_user):
        Displays the data of every table from a given User as JSON
        Parameter: p_user ... User
***************************************************************************/ 
v_tableData VARCHAR2(32767);
v_sqlStatement VARCHAR2(200);
BEGIN
  FOR tablerec IN (SELECT *
                   FROM   ALL_TABLES
                   WHERE OWNER = p_user)
  LOOP
    v_sqlStatement := 'SELECT /*' || p_format || '*/ * FROM ' || p_user || '.' || tablerec.TABLE_NAME;
    EXECUTE IMMEDIATE v_sqlStatement INTO v_tableData;

    DBMS_OUTPUT.PUT_LINE (v_sqlStatement);
  END LOOP;
END;

您可以看到，我遍历了给定用户的所有表，并使用p_format 和p_user 以及tablerec.TABLE_NAME 创建了一个sql 语句。

想要的结果应该是这样的：

{"results":[{"columns":[{"name":"COUNTRY_ID","type":"CHAR"},
{"name":"COUNTRY_NAME","type":"VARCHAR2"},{"name":"REGION_ID","type":"NUMBER"}],"items":
[
{"country_id":"AR","country_name":"Argentina","region_id":2},
{"country_id":"AU","country_name":"Australia","region_id":3},
{"country_id":"BE","country_name":"Belgium","region_id":1},
{"country_id":"BR","country_name":"Brazil","region_id":2},
{"country_id":"CA","country_name":"Canada","region_id":2},
{"country_id":"CH","country_name":"Switzerland","region_id":1},
{"country_id":"CN","country_name":"China","region_id":3},
{"country_id":"DE","country_name":"Germany","region_id":1},
{"country_id":"DK","country_name":"Denmark","region_id":1},
{"country_id":"EG","country_name":"Egypt","region_id":4},
{"country_id":"FR","country_name":"France","region_id":1},
{"country_id":"IL","country_name":"Israel","region_id":4},
{"country_id":"IN","country_name":"India","region_id":3},
{"country_id":"IT","country_name":"Italy","region_id":1},
{"country_id":"JP","country_name":"Japan","region_id":3},
{"country_id":"KW","country_name":"Kuwait","region_id":4},
{"country_id":"ML","country_name":"Malaysia","region_id":3},
{"country_id":"MX","country_name":"Mexico","region_id":2},
{"country_id":"NG","country_name":"Nigeria","region_id":4},
{"country_id":"NL","country_name":"Netherlands","region_id":1},
{"country_id":"SG","country_name":"Singapore","region_id":3},
{"country_id":"UK","country_name":"United Kingdom","region_id":1},
{"country_id":"US","country_name":"United States of America","region_id":2},
{"country_id":"ZM","country_name":"Zambia","region_id":4},
{"country_id":"ZW","country_name":"Zimbabwe","region_id":4}]}]}

【问题讨论】：

标签： sql json oracle plsql oracle-sqldeveloper

【解决方案1】：

JSON 提示特定于 SQL Developer 和 SQLcl，而不是直接用于数据库。所以你需要在这些工具中运行整个事情。

最简单的方法是让你的脚本编写一个你可以运行的脚本，例如

spool /tmp/get_all_json.sql
select 'select /*json*/ * from '||table_name||';' 
from user_tables;
spool off
@/tmp/get_all_json.sql

【讨论】：