Right now, i've created database with having 10 columns in my database but i can't create the database table and i got "#1067 - Invalid default value for 'primary_key" error.
Here is my sql code:
CREATE TABLE `quiz`.`user_details` ( `primary_key` INT(6) UNSIGNED NOT NULL DEFAULT '1' AUTO_INCREMENT , `foreign_key` INT(6) UNSIGNED NOT NULL DEFAULT '1' AUTO_INCREMENT , `firstname` VARCHAR(20) NOT NULL , `lastname` VARCHAR(20) NOT NULL , `username` VARCHAR(20) NOT NULL , `password` VARCHAR(20) NOT NULL , `email` VARCHAR(20) NOT NULL , `phone_number` INT(10) NOT NULL , `address` CHAR(50) NOT NULL , `exam_taking` INT(3) NOT NULL AUTO_INCREMENT , PRIMARY KEY (`primary_key`(6)), UNIQUE `foreign_key` (`foreign_key`)) ENGINE = InnoDB;
Don't add DEFAULT value because you already defined primary key as AUTO INCREMENT.
Possible duplicate.
#1067 - Invalid default value for 'bonusid' how can i fix this error?
You cannot have a default value for auto incrementing fields. Simply change the definition to:
CREATE TABLE `quiz`.`user_details`
(`primary_key` INT(6) UNSIGNED NOT NULL AUTO_INCREMENT,
`foreign_key` INT(6) UNSIGNED NOT NULL AUTO_INCREMENT ,
`firstname` VARCHAR(20) NOT NULL ,
`lastname` VARCHAR(20) NOT NULL ,
`username` VARCHAR(20) NOT NULL ,
`password` VARCHAR(20) NOT NULL ,
`email` VARCHAR(20) NOT NULL ,
`phone_number` INT(10) NOT NULL ,
`address` CHAR(50) NOT NULL ,
`exam_taking` INT(3) NOT NULL AUTO_INCREMENT ,
PRIMARY KEY (`primary_key`),
UNIQUE `foreign_key` (`foreign_key`)) ENGINE = InnoDB;
Note also that a name foreign_key for an attribute which is not a foreign key is really misleading.
Related
CREATE TABLE `chuchutvlogin`. (
`id` INT NOT NULL AUTO_INCREMENT ,
`username` VARCHAR(20) NOT NULL ,
`gender` CHAR(1) NOT NULL ,
`email` VARCHAR(50) NOT NULL ,
`password` VARCHAR(20) NOT NULL ,
`number` BIGINT(10) NOT NULL ,
PRIMARY KEY (`id`)
) ENGINE = InnoDB;
what's issue?
i was creating a login databse and error #1064 came syntax error in localhost phpmyadmnin
Fix syntax error in your query, remove dot char after table name:
CREATE TABLE chuchutvlogin (
id INT NOT NULL AUTO_INCREMENT,
username VARCHAR(20) NOT NULL,
gender CHAR(1) NOT NULL,
email VARCHAR(50) NOT NULL,
password VARCHAR(20) NOT NULL,
number BIGINT(10) NOT NULL,
PRIMARY KEY (id)
) ENGINE = InnoDB;
dbfiddle
I am trying to create a table in a database in phpMyAdmin and I keep getting the "a symbol name was expected" error and I cannot figure out what is going on. Is my syntax wrong? I am new to this and I'm at a loss.
Column names, table names should be surrounded by backticks(``)
CREATE TABLE IF NOT EXISTS `sales` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50) NOT NULL,
`item` varchar(50) NOT NULL,
`date` varchar(50) NOT NULL,
`amount` int(11) NOT NULL,
PRIMARY KEY (`id`)
)
Or you can go without backticks as well:
CREATE TABLE IF NOT EXISTS sales (
id int(11) NOT NULL AUTO_INCREMENT,
name varchar(50) NOT NULL,
item varchar(50) NOT NULL,
date varchar(50) NOT NULL,
amount int(11) NOT NULL,
PRIMARY KEY (id)
)
Your code is wrong , So here is the correct Code :
CREATE TABLE IF NOT EXISTS `sales` (
`id` INT(11) NOT NULL AUTO_INCREMENT PRIMARY KEY,
`name` VARCHAR(50) CHARACTER SET utf8 NOT NULL,
`item` VARCHAR(50) CHARACTER SET utf8 NOT NULL,
`date` VARCHAR(50) NOT NULL,
`amount` int(11) NOT NULL
)
You used ' ' sign in your column properties. But mySQL allow you to use `` sign.
CREATE TABLE IF NOT EXISTS `sales` (
`id` int(11) NOT NULL AUTO_INCREMENT,
`name` varchar(50) NOT NULL,
`item` varchar(50) NOT NULL,
`date` varchar(50) NOT NULL,
`amount` int(11) NOT NULL,
PRIMARY KEY (`id`)
);
I have it set up to echo when my tables create but the users table just won't, I've looked at multiple solutions but none of them has helped here my code
<?
phpinclude_once("index.php");
$tbl_users = "CREATE TABLE IF NOT EXISTS users
(
id INT(11) NOT NULL AUTO_INCREMENT,
username VARCHAR(16) NOT NULL,
email VARCHAR(64) NOT NULL,
password VARCHAR(64) NOT NULL,
gender ENUM('m','f') NOT NULL,
website VARCHAR(64) NULL,
country VARCHAR(64) NULL,
userlevel ENUM('a','b','c','d') NOT NULL DEFAULT a,
avatar VARCHAR(64) NULL,
ip VARCHAR(64) NOT NULL,
signup DATETIME NOT NULL,
lastlogin DATETIME NOT NULL,
notescheck DATETIME NOT NULL,
activated ENUM('0','1') NOT NULL DEFAULT 0,
PRIMARY KEY (id)),
UNIQUE KEY username (username,email))";
$query = mysqli_query($db_spooky, $tbl_users);
if ($query === TRUE)
{
echo "<h3>user table created OK :) </h3>";
}
else
{
echo "<h3>user table NOT created :( </h3>";
}
?>
CREATE TABLE IF NOT EXISTS users (id INT(11) NOT NULL AUTO_INCREMENT
,username VARCHAR(16) NOT NULL
,email VARCHAR(64) NOT NULL,password VARCHAR(64) NOT NULL
,gender ENUM('m','f') NOT NULL
,website VARCHAR(64) NULL
,country VARCHAR(64) NULL
,userlevel ENUM('a','b','c','d') NOT NULL DEFAULT 'a'
,avatar VARCHAR(64) NULL
,ip VARCHAR(64) NOT NULL
,signup DATETIME NOT NULL
,lastlogin DATETIME NOT NULL
,notescheck DATETIME NOT NULL,activated ENUM('0','1') NOT NULL DEFAULT 0
,PRIMARY KEY (id)
,UNIQUE KEY username (username,email))
Learn to organize your code instead of writing it in one long line for one, and two, ENUM default for a needs to be in single quotes and you had an extra closing bracket. The code above is the corrected version.
This question already has answers here:
What is wrong with my SQL here? #1089 - Incorrect prefix key
(15 answers)
Closed 7 years ago.
I have a problem creating a table with phpmyadmin, which gives me the following error:
#1089 - Incorrect prefix key; the used key part isn't a string, the used length is longer than the key part, or the storage engine doesn't support unique prefix keys
This is the query that I do:
CREATE TABLE `b2b`.`users` ( `id` BIGINT NOT NULL AUTO_INCREMENT ,
`name` VARCHAR(30) NOT NULL ,
`surnames` VARCHAR(80) NOT NULL ,
`birthdate` DATE NOT NULL ,
`drivingdoc` VARCHAR(20) NOT NULL ,
`acdate` DATE NOT NULL ,
`countrydoc` VARCHAR(20) NOT NULL ,
`province` VARCHAR(20) NOT NULL ,
`locality` VARCHAR(35) NOT NULL ,
`address` VARCHAR(150) NOT NULL ,
`number` VARCHAR(20) NOT NULL ,
`flat` VARCHAR(20) NOT NULL ,
`door` VARCHAR(20) NOT NULL ,
`description` VARCHAR(2000) NOT NULL ,
PRIMARY KEY (`id`(7))) ENGINE = InnoDB;
Using MariaDB in ubuntu minimal.
The problem is:
PRIMARY KEY (`id`(7))
You cannot use part of a number as a key, you have to use the whole thing. Also, specifying lengths for numeric types is useless at best, and damaging at worst.
Change to:
PRIMARY KEY (`id`)
That Primary Key syntax is nothing I've ever seen before. Try this:
CREATE TABLE `b2b`.`users` (
`id` BIGINT NOT NULL AUTO_INCREMENT PRIMARY KEY,
....
) ENGINE = InnoDB;
Or if you want
CREATE TABLE `b2b`.`users` (
`id` BIGINT NOT NULL AUTO_INCREMENT,
....
PRIMARY KEY (id)
) ENGINE = InnoDB;
You have several errors in your sql, so try to find the difference with this sql
CREATE TABLE `b2b`.`users` ( `id` INT(7) NOT NULL AUTO_INCREMENT ,
`name` VARCHAR(30) NOT NULL ,
`surnames` VARCHAR(80) NOT NULL ,
`birthdate` DATE NOT NULL ,
`drivingdoc` VARCHAR(20) NOT NULL ,
`acdate` DATE NOT NULL ,
`countrydoc` VARCHAR(20) NOT NULL ,
`province` VARCHAR(20) NOT NULL ,
`locality` VARCHAR(35) NOT NULL ,
`address` VARCHAR(150) NOT NULL ,
`number` VARCHAR(20) NOT NULL ,
`flat` VARCHAR(20) NOT NULL ,
`door` VARCHAR(20) NOT NULL ,
`description` VARCHAR(255) NOT NULL ,
PRIMARY KEY (`id`)) ENGINE = InnoDB;
I want to create some table through PHP but it fails everytime, but the same code excutes perfectly when excuted through MYsql console or PHPMyAdmin
The SQLs are
$sql = <<<SQL_CODE
CREATE TABLE IF NOT EXISTS `bid` (
`aid` int(11) unsigned NOT NULL,
`uid` int(11) unsigned NOT NULL,
`name` varchar(20) NOT NULL,
`amount` smallint(6) unsigned NOT NULL,
`time` timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP
);
CREATE TABLE IF NOT EXISTS `item` (
`aid` int(11) unsigned NOT NULL AUTO_INCREMENT,
`name` varchar(40) NOT NULL,
`description` varchar(120) NOT NULL,
`img` int(11) unsigned NOT NULL,
`amount` smallint(6) unsigned NOT NULL,
`strtdate` date NOT NULL,
`enddate` date NOT NULL,
`uid` int(11) unsigned NOT NULL,
`uname` varchar(20) NOT NULL,
`uamount` int(11) unsigned NOT NULL,
PRIMARY KEY (`aid`)
);
CREATE TABLE IF NOT EXISTS `user` (
`uid` int(11) unsigned NOT NULL AUTO_INCREMENT,
`email` varchar(40) NOT NULL,
`password` varchar(40) NOT NULL,
`nameF` varchar(20) NOT NULL,
`nameL` varchar(20) NOT NULL,
`sex` varchar(1) NOT NULL,
`img` int(11) unsigned NOT NULL,
`country` varchar(10) NOT NULL,
`state` varchar(20) NOT NULL,
`address` varchar(120) NOT NULL,
`code` varchar(8) NOT NULL,
`isAdmin` varchar(1) NOT NULL,
PRIMARY KEY (`uid`),
UNIQUE KEY `email` (`email`)
);
SQL_CODE;
$sql2 = 'CREATE DATABASE `'.$db.'`;';
$sql3 = 'USE `'.$db.'`; ';
$sql4 = 'drop Database `'.$db.'``;';
if (!mysql_query($sql2)) echo mysql_error();
if (!mysql_query($sql3)) echo mysql_error();
//sleep(1);
if (!mysql_query($sql)) echo mysql_error();
echo $sql2.' '.$sql3.' '.$sql;
Error I'm getting
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 'CREATE TABLE IF NOT EXISTS item ( aid int(11) unsigned NOT NULL AUTO_INCREMENT,' at line 1CREATE DATABASE auction; USE auction; CREATE TABLE IF NOT EXISTS bid ( aid int(11) unsigned NOT NULL, uid int(11) unsigned NOT NULL, name varchar(20) NOT NULL, amount smallint(6) unsigned NOT NULL, time timestamp NOT NULL DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP); CREATE TABLE IF NOT EXISTS item ( aid int(11) unsigned NOT NULL AUTO_INCREMENT, name varchar(40) NOT NULL, description varchar(120) NOT NULL, img int(11) unsigned NOT NULL, amount smallint(6) unsigned NOT NULL, strtdate date NOT NULL, enddate date NOT NULL, uid int(11) unsigned NOT NULL, uname varchar(20) NOT NULL, uamount int(11) unsigned NOT NULL, PRIMARY KEY (aid));CREATE TABLE IF NOT EXISTS user ( uid int(11) unsigned NOT NULL AUTO_INCREMENT, email varchar(40) NOT NULL, password varchar(40) NOT NULL, nameF varchar(20) NOT NULL, nameL varchar(20) NOT NULL, sex varchar(1) NOT NULL, img int(11) unsigned NOT NULL, country varchar(10) NOT NULL, state varchar(20) NOT NULL, address varchar(120) NOT NULL, code varchar(8) NOT NULL, isAdmin varchar(1) NOT NULL, PRIMARY KEY (uid), UNIQUE KEY email (email));
From the mysql_query manual;
mysql_query() sends a unique query (multiple queries are not
supported)...
You need to split your triple CREATE TABLE statement into 3 separate statements and it will work.
I think that is a problem with multiple queries. I think that is solution.
http://php.net/manual/en/function.mysql-query.php
*mysql_query() sends a unique query (multiple queries are not supported) to the currently active database on the server that's associated with the specified link_identifier*
Just split them up and it will work.