Designing "Friend" relationships across two mysql tables

Designing "Friend" relationships across two mysql tables

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Designing "Friend" relationships across two mysql tables
Tag : mysql , By : KL.
Date : November 28 2020, 08:01 AM

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MySql: "SHOW CREATE TABLE" doesn't work for tables named "x-..."?

Tag : mysql , By : Enrique Anaya
Date : March 29 2020, 07:55 AM
it should still fix some issue SHOW CREATE TABLE works even if there are no rows.
Could you please reproduce this with a command line client or phpMyAdmin?
SHOW CREATE TABLE `x-addresses`

MYSQL : Two tables "Signup" and "Login" ? Is it good to make them a single table or keep them two di

Tag : php , By : Adam May
Date : March 29 2020, 07:55 AM
help you fix your problem It is a good idea if there is a one-to-one mapping of login and signup information. You get better performance with one id lookup. Maybe a "Users" table?
On a sidenote, I hope by "password" is not a clear text password field and is instead a hash of the password. You should never have clear text passwords stored anywhere.

"Friend" relationships across two mysql tables

Tag : mysql , By : Adam Hill
Date : March 29 2020, 07:55 AM
it should still fix some issue I'm not quite sure how to construct an sql join inorder to find who a specific users "friends" are. , This uses a STRAIGHT join, but should work:
select u.u_Name
from friends f, user u
where (f.u_ID1 = '1' and u.u_ID = f.u_ID2) 
  or (f.u_ID2 = '1' and u.u_ID = f.u_ID1)

Two tables, many foreign key relationships with "On Update Cascade"

Tag : sql-server , By : Bo.
Date : March 29 2020, 07:55 AM
I wish this help you The problem stems from a limitation in T-SQL itself. From this Error message 1795 "a table cannot appear more than one time in a list of all the cascading referential actions that are started by either a DELETE or an UPDATE statement.
The solution is to use triggers on the lookup table to simulate an update cascade for the remaining relationships.

How can I make mysql "num_rows" work in "SELECT TABLE_NAME FROM information_schema.TABLES"?

Tag : php , By : scotta01
Date : March 29 2020, 07:55 AM
I hope this helps . Your problem is actually in your call to sprintf, it is trying to process %n as a conversion specification. You need to change the % in that string to %% i.e.
$query_Recordset4 = sprintf("SELECT TABLE_NAME FROM information_schema.TABLES WHERE TABLE_SCHEMA='message' and TABLE_NAME like '1a%%nd5'");
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