PHP / MYSQL：在WAMP服务器中未创建表

Question

when I run this php file in my wamp server it connects to database, selects database but not create table . the output is this:

connected to database succussfully.
connected to database store_db succussfully.
table users was not created.

what is the problem???

php code:

<?php
$sql='CREATE TABLE users
(
PID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(PID),
NAME CHAR(20),
PASSWORD CHAR(15),
)';

mysql_connect("localhost","root","") or die('could not connect to database.');
echo 'connected to database succussfully.<br>';

mysql_select_db('store_db') or die('database store_db not found.');
echo 'connected to database store_db succussfully.<br>';

mysql_query($sql) or die('table users was not created');
echo 'table users was created in database store_db succussfully.<br>';

?>

Answer 1

<?php
$sql='CREATE TABLE users
(
PID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(PID),
NAME CHAR(20),
PASSWORD CHAR(15),
)';

remove the comma after PASSWORD CHAR(15),

in the event of an error, you can always retrieve errors using mysql_error() .

however. you should not be using the old mysql functions. much better and more secure options exist in the PDO and MySQLi extensions.

Answer 2

Your $sql:

$sql='CREATE TABLE users
(
PID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(PID),
NAME CHAR(20),
PASSWORD CHAR(15),
)';

should be:

$sql='CREATE TABLE users
(
PID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY(PID),
NAME CHAR(20),
PASSWORD CHAR(15)
)';

You should remove comma right after PASSWORD CHAR(15)

Answer 3

要获取正确的错误消息，请始终使用以下代码，以便将来找出确切的错误。

mysql_query($sql) or die(mysql_error());

Answer 4

Make sure the MySQL user you are attempting to create the table with has permissions to do so. (eg the user has been granted the 'CREATE' privileges for that database.)