السلام عليكم ورحمة الله وبركاته
لإنشاء جدول قمت بكتابة كود
<?php
//include("db_connect.php");
//$link=mysql_dbconnect();
$link=mysql_connect('localhost','root');
if(!$link)
{
die('not connected :'.mysql_error());
}
else
echo 'connected sucussfuly';
//$link=mysql_dbconnect();
$query="CREATE DATABASE IF NOT EXISTS my_guest";
$resut=mysql_query($query);
$result='';
$query='';
$query="CREATE TABLE guest;
(
Name varchar(15) NULL ,
Password varchar(15) NULL,
primary key(Name)
)";
$result=mysql_query($query);
if(!$result)
{
echo "DB ERROR\n";
echo 'MYSQL Error:'.mysql_error();
exit;
}else echo"Table guest is successfully created";
?>وظهر خطأ
connected sucussfulyDB ERROR MYSQL Error:You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '; ( Name varchar(15) NULL , Password varchar(15) NULL, primary key(Name) )' at line 1
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وجزاكم الله خيرا
