unable to create table mysql php

Posted on 2011-02-24
Last Modified: 2012-05-11
Hi experts, I have this simple php script below. It is not creating the table in to mysql.
Can you please check it out and possibly find where the error in the code is?

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "">
<html xmlns="" lang="gr" xml:lang="gr">

<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<link rel="StyleSheet" type="text/css" href="code/main.css" />
<div style="position:absolute; left:0px;  top:0px; right:0px;width:100%; height:100%; padding:0px; margin:0px; text-align:left;">

include "code/";

mysql_select_db($database) or die( "Unable to select database");

mysql_query("CREATE TABLE pediko_tbl (   // it does not create the table here
id int(6) NOT NULL auto_increment,
link_th varchar(100) NOT NULL,
link varchar(100) NOT NULL,
UNIQUE KEY (link_th))")or die( "Unable to create table");
mysql_query("SET NAMES 'utf8'");

if ($handle = opendir('path')) {
    while (false !== ($file = readdir($handle))) {
        if ($file != "." && $file != "..") {

            mysql_query("INSERT INTO pediko_tbl VALUES('','$file' , '$file')");
            echo "$file";


echo "all $i inserted";

Question by:ikon32
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LVL 17

Assisted Solution

psimation earned 125 total points
ID: 34976804
General tip for debugging SQL issues in PHP -

1. Echo your SQL query in your code after variable substitution.
2. Use PHPMyAdmin or MySQL Workbench to the run that exact query that is being echoe'd

In your code above, try to assign the mysql_query to a variable, ie, $result = mysql_query(..........) mysql_query() will not run "on it's own" - you have to assign to a variable.
LVL 17

Assisted Solution

by:Shinesh Premrajan
Shinesh Premrajan earned 125 total points
ID: 34976862

mysql_query("CREATE TABLE IF NOT EXISTS pediko_tbl (

Update the SQL query to this.

The better option is to first create the table in the database and then run the script for insert query.
LVL 11

Accepted Solution

Bruce Smith earned 125 total points
ID: 34976891
You must declare the id as the primary key:
CREATE TABLE thegamepoker.pediko_tbl (id int(6) NOT NULL auto_increment,link_th varchar(100) NOT NULL,link varchar(100) NOT NULL,Primary KEY (id), UNIQUE KEY (link_th));

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LVL 11

Expert Comment

by:Bruce Smith
ID: 34976913
Actually, my apologies, it doesn't HAVE to be the Primary key, it can be declared as UNIQUE instead also. Either way.

1. Primary

CREATE TABLE thegamepoker.pediko_tbl (id int(6) NOT NULL auto_increment,link_th varchar(100) NOT NULL,link varchar(100) NOT NULL, PRIMARY KEY (id), UNIQUE KEY (link_th));

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2. Unique

CREATE TABLE thegamepoker.pediko_tbl (id int(6) NOT NULL auto_increment,link_th varchar(100) NOT NULL,link varchar(100) NOT NULL, UNIQUE KEY (id), UNIQUE KEY (link_th));

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LVL 11

Expert Comment

by:Bruce Smith
ID: 34976924
PS. ignore the "thegamepoker." in front of the table name. I was testing it in my environment.  :)
LVL 34

Assisted Solution

by:Beverley Portlock
Beverley Portlock earned 125 total points
ID: 34978421
Change your DIE clause from

die( "Unable to create table");


die( "Unable to create table " . mysql_error() );

then at least you'll know why it failed.

Author Closing Comment

ID: 34978508
Big thanks to all, your comments were excellent. I had to accept all as correct answers cause the solution I guess  was a combination... found that out after I tried all your solutions. thanks so much!

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