sqlfiddle

is there a limit to what sql fiddle can handle? sql fiddle doesn't compile anything and returns no error messages

半世苍凉 提交于 2019-12-10 13:09:09
问题 I recently created a query that compiles successfully and returns the desired result. When I used that piece of code as a subquery in another piece of code that a user on stackoverflow came up with for me, I encountered a few problems, which were ultimately solved. I attempted to use this query as a subquery in that piece of code given to me. However, sql fiddle doesn't return anything. No errors or compiled messages. When I tried putting in a syntax error on purpose-like a random + sign,

Can't use column names in select query on sqlfiddle (oracle)

懵懂的女人 提交于 2019-12-04 05:49:09
问题 I'm not if am I using sqlfidle incorrectly or if this is missing functionality? Steps to reproduce: Select oracle option (top left) create table and insert data: CREATE TABLE products ("P_Id" int, "ProductName" varchar2(10), "UnitPrice" numeric, "UnitsInStock" int, "UnitsOnOrder" int) // INSERT ALL INTO products ("P_Id", "ProductName", "UnitPrice", "UnitsInStock", "UnitsOnOrder") VALUES (1, 'Jarlsberg', 10.45, 16, 15) INTO products ("P_Id", "ProductName", "UnitPrice", "UnitsInStock",

Different results in sqlfiddle.com 5.5.30 and MariaDB 5.5.31

耗尽温柔 提交于 2019-12-02 10:06:47
问题 sqlfiddle: http://sqlfiddle.com/#!2/9a8b3/1 Taking the structure and data and query from the fiddle, importing into my MariaDB 5.5.31, I get different results: sqlfiddle PID NAME LEAGUEPOINTS TOTALLEAGUEPOINTS 2 Peter 16,13,9,4,2 44 1 Daniel 3425,543,234,43,29,22,21,21,19,17,13,12,12,12,11,9,9,9,8,7 4476 mariadb pid name leaguepoints totalleaguepoints 2 Peter 16,13,9,4,2 44 1 Daniel 3425,543,234,43,29,22,21,21,19,17,13,12,12,12,11,9,9,9,8,7,7,6,5,5,4,4,4,3,3,2,1 4520 Query: SELECT p.pid, p

Can't use column names in select query on sqlfiddle (oracle)

夙愿已清 提交于 2019-12-02 09:17:47
I'm not if am I using sqlfidle incorrectly or if this is missing functionality? Steps to reproduce: Select oracle option (top left) create table and insert data: CREATE TABLE products ("P_Id" int, "ProductName" varchar2(10), "UnitPrice" numeric, "UnitsInStock" int, "UnitsOnOrder" int) // INSERT ALL INTO products ("P_Id", "ProductName", "UnitPrice", "UnitsInStock", "UnitsOnOrder") VALUES (1, 'Jarlsberg', 10.45, 16, 15) INTO products ("P_Id", "ProductName", "UnitPrice", "UnitsInStock", "UnitsOnOrder") VALUES (2, 'Mascarpone', 32.56, 23, NULL) INTO products ("P_Id", "ProductName", "UnitPrice",

Is SQLFiddle broken? Errors for Oracle, SQL Server, …? [closed]

随声附和 提交于 2019-11-28 20:05:33
I'm trying to create a simple table in SQLFiddle for Oracle database and keep getting the cryptic error: Create script error. Here is a simple script it failed on: create table t1 (f1 number, f2 number, f3 number); Could anyone can help with that? Steve Chambers UPDATE: All seems to be working again. See the answer from Jake Feasel (the creator of SQL Fiddle). To summarise info from the comments as an answer and bring up to date as of writing: SQLFiddle was broken for about a year but as of 2017-07-07 it is working for Oracle 11g R2 . It also seems to be broken for MS SQL Server 2008 , MS SQL

LEFT OUTER JOIN query not returning expected rows

元气小坏坏 提交于 2019-11-28 01:09:33
My aim is to do exactly what a LEFT OUTER JOIN intends to do using the 4th venn diagram: SQL Diagrams : My query isn't returning any values at all, where in fact, it should be returning all within the Consultant_Memberships minus the one that is stored within Consultant_Memberships_Lists. Please see the SQL Fiddle for an easier understanding: SELECT * FROM consultant_memberships LEFT OUTER JOIN consultant_memberships_list ON consultant_memberships.`id` = consultant_memberships_list.membership_id WHERE consultant_memberships_list.consultant_id = $id AND consultant_memberships_list.membership_id

Execute triggers stored procedures on SqlFiddle. Mysql

浪尽此生 提交于 2019-11-27 13:39:06
Does SQL-fiddle facilitate execution of triggers/stored procedures? I have been unable to execute even the simplest form of stored procedure on sqlfiddle DELIMITER $$ DROP PROCEDURE IF EXISTS myProc $$ CREATE PROCEDURE myProc() BEGIN END$$ DELIMITER ; Sqlfiddle does not allow executing this(above) sql in build schema, but allows create table etc Note: The same syntax is working for me on my localhost using wamp with mysql 5.5.24 Can anyone guide please? Jake Feasel Instead of using the delimiter option (which is not a real SQL statement, but rather only a command for the mysql command prompt)

Is SQLFiddle broken? Errors for Oracle, SQL Server, …? [closed]

时光怂恿深爱的人放手 提交于 2019-11-27 12:56:14
问题 I'm trying to create a simple table in SQLFiddle for Oracle database and keep getting the cryptic error: Create script error. Here is a simple script it failed on: create table t1 (f1 number, f2 number, f3 number); Could anyone can help with that? 回答1: UPDATE: All seems to be working again. See the answer from Jake Feasel (the creator of SQL Fiddle). To summarise info from the comments as an answer and bring up to date as of writing: SQLFiddle was broken for about a year but as of 2017-07-07

SQL Fiddle Output Error

对着背影说爱祢 提交于 2019-11-27 09:33:10
Actually I am quite new to PL/SQL I created the following table using oracle pl/sql in SQL Fiddle create table Employee(name varchar2(100),id integer, salary integer,PRIMARY KEY(id)); insert into Employee(name,id,salary) values('sa',94,100); insert into Employee(name,id,salary) values('pr',88,150); insert into Employee(name,id,salary) values('ji',33,900); insert into Employee(name,id,salary) values('na',24,880); insert into Employee(name,id,salary) values('po',65,770); insert into Employee(name,id,salary) values('ri',69,910); insert into Employee(name,id,salary) values('uj',12,650); insert

LEFT OUTER JOIN query not returning expected rows

て烟熏妆下的殇ゞ 提交于 2019-11-26 21:50:42
问题 My aim is to do exactly what a LEFT OUTER JOIN intends to do using the 4th venn diagram: SQL Diagrams: My query isn't returning any values at all, where in fact, it should be returning all within the Consultant_Memberships minus the one that is stored within Consultant_Memberships_Lists. Please see the SQL Fiddle for an easier understanding: SELECT * FROM consultant_memberships LEFT OUTER JOIN consultant_memberships_list ON consultant_memberships.`id` = consultant_memberships_list.membership