$sql = "CREATE TABLE comments
(
ID INT NOT NULL AUTO_INCREMENT,
PosterName VARCHAR(32),
Title VARCHAR(32),
Content VARCHAR(500)
)";
$con->query($sql);
No errors, connection to database is successful. Does anyone know why it doesnt work?
You should have seen that error with that statement:
Incorrect table definition; there can be only one auto column and it must be defined as a key:
auto_increment column must have an UNIQUEindex on them, or more generally being the PRIMARY KEY:
$sql = "CREATE TABLE comments
(
ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
PosterName VARCHAR(32),
Title VARCHAR(32),
Content VARCHAR(500)
)";
Related
I want to check if the table in database exists, if not create a table with code
CREATE TABLE Topics (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
email VARCHAR(70) NOT NULL,
message TEXT
)
and if not create a table. I know I have to use it but I don't know what to use before...
If you're using MySQL, use this query:
CREATE TABLE IF NOT EXISTS Topics (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
email VARCHAR(70) NOT NULL,
message TEXT
)
It will create table only if it does not exist.
You can show tables before creating new table by run this commands together
USE database_name; -- write your database name
SHOW TABLES;
OR create table if not exit
CREATE TABLE IF NOT EXISTS Topics (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
email VARCHAR(70) NOT NULL,
message TEXT
)
I am creating a MySQL database and this doesn't seem to work, i was able to create a review table but now i'm trying to drop that table and create a reviews table but it doesn't seem to work. Please can someone take a look at this and help me check to see what's wrong here?
$reviewsTable = "CREATE TABLE reviews (
ID int NOT NULL AUTO_INCREMENT,
Name varchar(100) NOT NULL,
Website varchar(100) NOT NULL,
Review varchar(100) NOT NULL,
TimeOfYear varchar(50),
DayOfYear varchar(50),
PRIMARY KEY (website)
)";
$drop = "DROP TABLE review";
mysqli_query($connect,$drop);
mysqli_query($connect,$reviewsTable);
Just use if exists to drop the table if there is one then create your table.
Id has to be primary key because of the auto increment. all auto increments have to be primary key. You can index website though. but i set id as primary key below this should help.
$reviewsTable = "
DROP TABLE IF EXISTS review;
CREATE TABLE reviews (
ID int NOT NULL AUTO_INCREMENT,
Name varchar(100) NOT NULL,
Website varchar(100) NOT NULL,
Review varchar(100) NOT NULL,
TimeOfYear varchar(50),
DayOfYear varchar(50),
PRIMARY KEY (ID)
)";
I have read and searched all the site, but nothing I had found worked for me...so please help a newbie understand what he is doing wrong.
So I am trying to create an add to favorite function.
I need to create 3 tables. The first two worked like magic, but the 3rd one ...well I got in the trouble with the FOREIGN KEY
I get no error message, but it won't create the 3rd table.
Here are my codes:
<?php
$connect = mysql_connect("127.0.0.1","root","");
$db = mysql_select_db("mydb");
mysql_query("CREATE TABLE IF NOT EXISTS users
(
userid bigint,
firstname varchar(25),
lastname varchar(15),
email varchar(250),
gender varchar(10),
username varchar(15),
password varchar(15),
age int,
activ boolean,
date TIMESTAMP NULL default CURRENT_TIMESTAMP
)ENGINE=INNODB
");
mysql_query("CREATE TABLE IF NOT EXISTS products (
productid int(11) NOT NULL AUTO_INCREMENT,
productname varchar(100) NOT NULL,
productdescription varchar(250) NOT NULL,
price decimal(6,2) NOT NULL,
PRIMARY KEY (`productid`)
) ENGINE=INNODB DEFAULT CHARSET=latin1 AUTO_INCREMENT=7 ");
mysql_query("CREATE TABLE IF NOT EXISTS favorites
(
userid bigint NOT NULL,
productid int(11) NOT NULL,
PRIMARY KEY (userid, productid),
FOREIGN KEY (userid) REFERENCES user (userid) ON DELETE CASCADE ON UPDATE CASCADE,
FOREIGN KEY (productid) REFERENCES product (productid) ON DELETE CASCADE ON UPDATE CASCADE
)ENGINE=INNODB
");
echo "The DataBase was successfully created!";
mysql_close();
?>
The foreign keys refer to non-existing tables.
The tables are names users and products while the third table refers to them as user and product (singular).
"userid" of "user" table must be primary key.
Hi I'm having issues creating my post database. I'm trying to make a forenge key to link to my users database. Can someone please help?
Here's the code for my tables :
CREATE TABLE USERS(
UserID int NOT NULL AUTO_INCREMENT,
UserName varchar(255),
UserPassword varchar(255) NOT NULL,
UserEmailAddress varchar(255) NOT NULL,
Admin int DEFAULT 0,
PRIMARY KEY (userID,UserName)
)ENGINE=InnoDB;
CREATE TABLE POSTS(
postID int NOT NULL AUTO_INCREMENT,
postTitle varchar(255) NOT NULL,
postContent varchar(255) NOT NULL,
category varchar(255) NOT NULL,
postDate Date NOT NULL,
postAuthor varchar(255),
tag varchar(255),
PRIMARY KEY(postID),
FOREIGN KEY(postAuthor) REFERENCES USERS(UserName)
)ENGINE=InnoDB;
Here's the last InnoDB error message:
Error in foreign key constraint of table db/POSTS:
FOREIGN KEY(postAuthor) REFERENCES USERS(UserName)
)ENGINE=InnoDB:
Cannot find an index in the referenced table where the
referenced columns appear as the first columns, or column types
in the table and the referenced table do not match for constraint.
Note that the internal storage type of ENUM and SET changed in
tables created with >= InnoDB-4.1.12, and such columns in old tables
cannot be referenced by such columns in new tables.
See http://dev.mysql.com/doc/refman/5.5/en/innodb-foreign-key-constraints.html
for correct foreign key definition.
There really should not be a good reason to have a compound primary key on the first table. So, I think you intend:
CREATE TABLE USERS (
UserID int NOT NULL AUTO_INCREMENT,
UserName varchar(255),
UserPassword varchar(255) NOT NULL,
UserEmailAddress varchar(255) NOT NULL,
Admin int DEFAULT 0,
PRIMARY KEY (userID),
UNIQUE (UserName)
);
CREATE TABLE POSTS (
postID int NOT NULL AUTO_INCREMENT,
postTitle varchar(255) NOT NULL,
postContent varchar(255) NOT NULL,
category varchar(255) NOT NULL,
postDate Date NOT NULL,
postAuthor int,
tag varchar(255),
PRIMARY KEY(postID),
FOREIGN KEY(postAuthor) REFERENCES USERS(UserId)
);
Some notes:
An auto-incremented id is unique on every row. It makes a good primary key.
A primary key can consist of multiple columns (called a composite primary key). However, an auto-incremented id doesn't make much sense as one of the columns. Just use such an id itself.
If you use a composite primary key, then the foreign key references need to include all columns.
I chose UserId for the foreign key reference. You could also use UserName (because it is unique).
UserName is a bad choice for foreign keys, because -- conceivably -- a user could change his or her name.
The error is caused by incorrect foreign key definition. In the concrete case you are missing a complete column in your foreign key definition.
In the USERS table you have defined primary key as composite key of UserID and UserName columns.
CREATE TABLE USERS (
UserID int NOT NULL AUTO_INCREMENT,
UserName varchar(255),
UserPassword varchar(255) NOT NULL,
UserEmailAddress varchar(255) NOT NULL,
Admin int DEFAULT 0,
PRIMARY KEY (UserID,UserName)
) ENGINE=InnoDB;
note that it is good practice to respect case of the identifiers (column names)
In the POSTS table you declared your foreign key to reference only one column in the USERS table, the UserName column. This is incorrect as you need to reference entire primary key of the USERS table which is (UserID, UserName). So to fix the error you need to add one additional column to the POSTS table and change your foreign key definition like this:
CREATE TABLE POSTS(
postID int NOT NULL AUTO_INCREMENT,
postTitle varchar(255) NOT NULL,
postContent varchar(255) NOT NULL,
category varchar(255) NOT NULL,
postDate Date NOT NULL,
authorId int,
postAuthor varchar(255),
tag varchar(255),
PRIMARY KEY(postID),
FOREIGN KEY(authorId, postAuthor) REFERENCES USERS(UserID, UserName)
) ENGINE=InnoDB;
Please look at following fiddle: http://sqlfiddle.com/#!9/92ff1/1
NOTE: If you can you should re-architect this to not use the composite primary key in the USERS table as it does not make sense from what I can see in the displayed code. You can change the tables like this:
CREATE TABLE USERS (
UserID int NOT NULL AUTO_INCREMENT,
UserName varchar(255),
UserPassword varchar(255) NOT NULL,
UserEmailAddress varchar(255) NOT NULL,
Admin int DEFAULT 0,
PRIMARY KEY (UserID)
) ENGINE=InnoDB;
CREATE TABLE POSTS (
postID int NOT NULL AUTO_INCREMENT,
postTitle varchar(255) NOT NULL,
postContent varchar(255) NOT NULL,
category varchar(255) NOT NULL,
postDate Date NOT NULL,
postAuthorID int,
tag varchar(255),
PRIMARY KEY(postID),
FOREIGN KEY(postAuthorID) REFERENCES USERS(UserID)
) ENGINE=InnoDB;;
The exact error I keep seeing is:
Key column 'alarmID' doesn't exist in table
alarmID is my primary key field.
Here is the code I have:
$sql = "CREATE TABLE IF NOT EXISTS alarms (
alaramID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY (alarmID),
Title CHAR(30),
Description TEXT,
DT DATETIME
)";
Note: I am coding in PHP.
$sql = "CREATE TABLE IF NOT EXISTS alarms (
alaramID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY (alaramID),
Title CHAR(30),
Description TEXT,
DT DATETIME
)";
alaramID
The primary key in your table is alaramID and note the error its alarmID.So correct the spelling in the query like this
$sql = "CREATE TABLE IF NOT EXISTS alarms (
alaramID INT NOT NULL AUTO_INCREMENT,
PRIMARY KEY (alaramID),
Title CHAR(30),
Description TEXT,
DT DATETIME
)";