making table in mysql using php

Question

and again maybe im been dumb or not seeing something I should but I am having a problem making a table in MySQL using a php script here is the code

$maketable = "CREATE TABLE IF NOT EXISTS $scompany ( 
  id INT(11) NOT NULL AUTO_INCREMENT, 
  deskname VARCHAR(255) NOT NULL, 
  lfloor VARCHAR(255) NOT NULL,, 
  internet ENUM('y','n') NOT NULL DEFAULT 'n', 
  pc ENUM('y','n') NOT NULL DEFAULT 'n',  
  monitor ENUM('y','n') NOT NULL DEFAULT 'n', 
  extra TEXT NULL, 
  PRIMARY KEY (id))";

$tablequery = mysqli_query($db_x, $maketable);

but for some reason it wont work can you help please

there is syntax error in query, change your code to $tablequery = mysqli_query($db_x, $maketable) or die(mysqli_error($db_x)); to see it — Iłya Bursov
– Iłya Bursov, Commented Nov 18, 2013 at 22:02
Yeah, let's just die. Because that's what we all do when things don't go as planned, right? We just die. — Sherlock
– Sherlock, Commented Nov 18, 2013 at 22:03
@Sherlock - it depends if your boss is the forgiving type or not! — EmmyS
– EmmyS, Commented Nov 18, 2013 at 22:04

Kacer · Accepted Answer · 2013-11-18 22:06:00Z

1

You have double comma in part

lfloor VARCHAR(255) NOT NULL,,

What's the point of making separate table for each company? This is very bad idea and will give you many troubles.

answered Nov 18, 2013 at 22:06

Kacer

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1 Comment

Chris par Over a year ago

the reason is I was thinking it would be easier for each company to edit their table they need to. but every says its a bad idea so I might rethink it

rubo77 · Accepted Answer · 2013-11-18 22:05:39Z

1

There is a double komma in your Query at the end of this part:

 lfloor VARCHAR(255) NOT NULL,,

answered Nov 18, 2013 at 22:05

rubo77

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Comments

Sherlock · Accepted Answer · 2013-11-18 22:02:06Z

0

Be glad it doesn't work.

You don't want to do this.

Creating a table for every company, whatever the exact definition of company is in your application, is a Very Bad Thing to do. Try googling the term 'normalization' and see where that gets you.

answered Nov 18, 2013 at 22:02

Sherlock

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2 Comments

Iłya Bursov Over a year ago

normalization is of course great thing, but for performance reasons de-normalization can be considered as good option, also this is not an answer to problem

Sherlock Over a year ago

This clearly is no example of denormalizing because of performance issues. Also: this is the answer to the problem, but the problem is of a magnitude different from what the asker thinks it is.

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