I have created a table and now find that when I try and delete a row or change a field, I get the following message:
Unexpected update count received (Actual: 0, Expected: 1). All changes will be rolled back.
I saw another question about this that said to create a unique field and try again. I dropped the table, recreated it with a unique field and am still getting the same error.
The code to create the table is:
create table 'transcribercertifications' (
id int(11) not null unique auto_increment,
tid int(11) not null,
certId varchar(32) not null,
regNum int(11) not null,
expiration date
);
I have been deleting and updating 'by hand' in PhpStorm. Any help fixing this would be much appreciated.
You need to make your id column as a primary key. Try this one:
create table 'transcribercertifications' (
id int(11) not null primary key auto_increment,
tid int(11) not null,
certId varchar(32) not null,
regNum int(11) not null,
expiration date
);
Related
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 am getting one error while creating table using MySQL. I am explaining my query below.
$sql="CREATE TABLE db_cron( ".
"id INT NOT NULL AUTO_INCREMENT, ".
"user_alert INT(11) NOT NULL DEFAULT NONE, ".
"status INT(11) NOT NULL DEFAULT NONE, ".
"PRIMARY KEY ( id )); ";
I am getting the below error:
#1064 - 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 'NONE, status INT(11) NOT NULL DEFAULT NONE, PRIMARY KEY ( id ))'
at line 1
Here I need to create table with default value as none.
You have used too many quotes just to connect multiple string, it does not required.
Please remove those quotes and you also have to provide default value instead of NULL something like 0 OR any other value that you want.
However you can use below query to create table.
$sql = "CREATE TABLE `db_cron`( id INT NOT NULL AUTO_INCREMENT,
user_alert INT(11) NOT NULL DEFAULT 0,
status INT(11) NOT NULL DEFAULT 0,
PRIMARY KEY ( id ))";
above query will create table with default value 0 for user_alert and status
Hope this will helps you
Thanks & Regards.
1.You have applied unnecessary quotes.
2.You have NOT NULL along with DEFAULT NONE, which is not going to happen(You have to provide some value as default when using NOT NULL)
Do like below:-
$sql="CREATE TABLE db_cron(
id INT(11) NOT NULL AUTO_INCREMENT,
user_alert INT(11) NOT NULL DEFAULT 0,
status INT(11) NOT NULL DEFAULT 0,
PRIMARY KEY ( id )
)";
I'm trying to make some SQL commands in XAMPP. My query returns following error:
You have an error in your SQL syntax; check the manual that
corresponds to your MariaDB server version for the right syntax to use
near 'CREATE TABLE IF NOT EXISTS kayttajat ( id INT(10) NOT NULL
AUTO_INCREMENT, tunn' at line 2
I can't find the syntax error. And when I run the exact same command by copy-paste in command line, it works. So do I need some different syntax in PHP code?
Also if I remove the first command, the error message moves to [...] right syntax to use near 'CREATE TABLE IF NOT EXISTS rivit [...]. If I remove second command the error comes from third command and so on. I really don't understand where the error is.
$query='
CREATE DATABASE IF NOT EXISTS asdgfhj;
CREATE TABLE IF NOT EXISTS kayttajat
(
id INT(10) NOT NULL AUTO_INCREMENT,
tunnus1 varchar(32) NOT NULL,
tunnus2 varchar(32) NOT NULL,
nimi varchar(32),
nimi2 varchar(32),
oikeus INT(10) NOT NULL DEFAULT 1,
PRIMARY KEY (id)
);
CREATE TABLE IF NOT EXISTS rivit
(
id INT(10) NOT NULL AUTO_INCREMENT,
sivu INT(10) NOT NULL,
kayttaja INT(10) NOT NULL,
sana varchar(500),
kommentti varchar(1000),
aika TIMESTAMP DEFAULT CURRENT_TIMESTAMP,
muutos TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP,
PRIMARY KEY(id)
);
CREATE TABLE IF NOT EXISTS sivut
(
id INT(10) NOT NULL AUTO_INCREMENT,
nimi varchar(32) NOT NULL,
ohje varchar(2000) NOT NULL,
salaisuus INT(10) NOT NULL DEFAULT 1,
PRIMARY KEY(id)
);
';
$mysqli->query($query) or die($mysqli->error);
The query() method is designed to execute a single query not multiple ones.
What you are looking for is multi_query() to execute multiple queries separated by a semicolon.
$query = "SELECT CURRENT_USER();";
$query .= "SELECT Name FROM City ORDER BY ID LIMIT 20, 5";
/* execute multi query */
if ($mysqli->multi_query($query)) {
...
}
You should use multi_query if you want execute many queries with one command.
Also you don't select database which will used for insert statements.
You should add asdgfhj prefix for all tables or use USE asdgfhj after create table statement.
example with prefix:
<?php
$query='
CREATE DATABASE IF NOT EXISTS asdgfhj;
CREATE TABLE IF NOT EXISTS asdgfhj.kayttajat
(
id INT(10) NOT NULL AUTO_INCREMENT,
tunnus1 varchar(32) NOT NULL,
tunnus2 varchar(32) NOT NULL,
nimi varchar(32),
nimi2 varchar(32),
oikeus INT(10) NOT NULL DEFAULT 1,
PRIMARY KEY (id)
);
';
$mysqli->multi_query($query) or die($mysqli->error);
example with use statement:
<?php
$query='
CREATE DATABASE IF NOT EXISTS asdgfhj;
USE asdgfhj;
CREATE TABLE IF NOT EXISTS kayttajat
(
id INT(10) NOT NULL AUTO_INCREMENT,
tunnus1 varchar(32) NOT NULL,
tunnus2 varchar(32) NOT NULL,
nimi varchar(32),
nimi2 varchar(32),
oikeus INT(10) NOT NULL DEFAULT 1,
PRIMARY KEY (id)
);
';
$mysqli->multi_query($query) or die($mysqli->error);
I am trying to create a table in a php script and I am getting an SQL error. I can't find it or figure out how to fix it. I do have $table defined earlier in the script. Thanks in advance!
$newtable = "CREATE TABLE $table (id INT(11) NOT NULL AUTO_INCREMENT, time datetime NOT NULL,punchtype VARCHAR(255) NOT NULL,groupname VARCHAR(255) NOT NULL, dept VARCHAR(255) NOT NULL, notes VARCHAR(255))";
Error Message: Incorrect table definition; there can be only one auto column and it must be defined as a key
try this, it works
$newtable = "CREATE TABLE $table
(id INT(11) NOT NULL AUTO_INCREMENT,
PRIMARY KEY(id),
time datetime NOT NULL,
punchtype VARCHAR(255) NOT NULL,
groupname VARCHAR(255) NOT NULL,
dept VARCHAR(255) NOT NULL,
notes VARCHAR(255))";
I am trying to make a revisions page, but when I insert a new revision, it sets the id to 0 instead of 1 more than the last entry, so I can make 1 new revision, but no more. Here is my mySQL code:
("INSERT INTO `revisions` (version, author, content, `date`) VALUES ('".$version."', '".$_SESSION['username']."', '".$content."', '".$date."')");
Edit:
Wow that was embarrassing... I made the table id default 0...
Schema:
CREATE TABLE `revisions` (
`id` int(11) NOT NULL,
`version` varchar(255) NOT NULL,
`author` varchar(255) NOT NULL,
`content` varchar(255) NOT NULL,
`status` int(255) NOT NULL,
`date` varchar(255) NOT NULL,
PRIMARY KEY (`id`)
) ENGINE=InnoDB DEFAULT CHARSET=latin1
:) it saids clearly
id int(11) NOT NULL DEFAULT '0'
it will be inserted as default value 0 if you dont make any value , then it will be 0 as default.
if you wanna be 1 as default then use this
id int(11) NOT NULL DEFAULT '1'
or if you want it be incremented then use this
id int(11) NOT NULL AUTO_INCREMENT
//this will be automatically 1,2,3,4,...values incremented everytime you inseret new value
CREATE TABLE `revisions` (
`id` int(11) NOT NULL AUTO_INCREMENT,
......
you need to set id with auto increment
check out the mysql documentation: http://dev.mysql.com/doc/refman/5.0/en/example-auto-increment.html
You should be adding AUTO_INCREMENT to your id instead. That way, it'll just increment automatically (as the name suggests) each time you insert a row.
CREATE TABLE `revisions` (
`id` int(11) NOT NULL AUTO_INCREMENT,
...