I want to create a table with variables passed into my php file. However, the SQL does not work when I pass in '12345' and works when I pass in 'a12345' instead.
This is my error that is given.
Error creating the table
query was
CREATE TABLE 123456 ( humidity VARCHAR(50) NOT NULL, temperature VARCHAR(50)
NOT NULL, gasquality VARCHAR(50) NOT NULL, timestamp DATETIME NOT NULL
DEFAULT CURRENT_TIMESTAMP)
mysqlerror: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
'123456 ( humidity VARCHAR(50) NOT NULL, temperature VARCHAR(50) NOT NULL,
gasq' at line 1
Creating database failed!
and my function that creates the table
function CreateTableNode(&$formvars)
{
$serialno = $formvars['serialno'];
$qry = "CREATE TABLE ".$serialno." (".
" humidity VARCHAR(50) NOT NULL, ".
" temperature VARCHAR(50) NOT NULL, ".
" gasquality VARCHAR(50) NOT NULL, ".
" timestamp DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP)";
if(!mysqli_query($this->connection,$qry))
{
$this->HandleDBError("Error creating the table \nquery was\n $qry");
return false;
}
return true;
}
I want to be able to create tables with numeric names like '12345' or '154124' for other purposes. Thanks alot!
My suggestion:
Provide a prefix to the table you created.
Moreover, I couldn't
see the primary key in your table. However, it is not necessary to
have it but if your table design doesn't have a primary key, you need
to rethink your design. It plays a vital role to join tables.
Your code can be rewritten as:
function CreateTableNode (&$formvars) {
$host = 'localhost';
$database = 'test';
$dbuser = 'root';
$dbpass = '';
try {
$pdo = new PDO('mysql:host=localhost; dbname=test', $dbuser, $dbpass);
} catch (PDOException $e) {
print "ERROR! : " . $e->getMessage() . "<br/>";
die();
}
$serialno = $formvars['serialno'];
$qry = "CREATE TABLE ".$serialno." ("."
`id` INT NOT NULL AUTO_INCREMENT ,
`humidity` VARCHAR(50) NOT NULL ,
`temperature` VARCHAR(50) NOT NULL ,
`gasquality` VARCHAR(50) NOT NULL ,
`timestamp` DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP ,
PRIMARY KEY (`id`)
)";
$stmt = $pdo->prepare($qry);
$stmt->execute();
$pdo = null;
return true;
}
You just need to wrap some elements in the query with quotes as the duplicated thread mentioned by underscore_d says:
$qry = "CREATE TABLE '$serialno' (
'humidity' VARCHAR(50) NOT NULL,
'temperature' VARCHAR(50) NOT NULL,
'gasquality' VARCHAR(50) NOT NULL,
'timestamp' DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP)";
This will fix your syntax errror in the query.
Marking to close the question as duplicated
The name of the entity was expected. (near "123456" at position 13)
Try adding a prefix to the table name as such
"t_12345"
CREATE TABLE t_12345
MySql does not allow numeric values as table name.
MySQL doesn't allow the creation of tables with names made solely of digits unless the name is quotes. See here
Identifiers may begin with a digit but unless quoted may not consist solely of digits.
Try quoting the name with backticks (`) or prefix the table name.
The error says "Creating database failed!".
So I assume you haven't selected the database in the connection query. You should do that or select it with "use mydatabase;" first. Of course, you may need to create the database first.
With PDO it would look like:
$conn = new PDO("mysql:host=$servername;dbname=myDB", $username, $password);
Please see dbname=myDB which preselects the right db for you.
Reference: https://www.w3schools.com/php/php_mysql_connect.asp
Using mysql functions, you can use:
mysql_select_db($dbname)
Reference: http://php.net/manual/en/function.mysql-select-db.php
Related
This question already has answers here:
When to use single quotes, double quotes, and backticks in MySQL
(13 answers)
Closed 3 years ago.
I am about to create a table, but I want to declare it based on the user's input. thankyou for any response, all answers are appreciated, more power!
I am receiving this error (Error creating table: 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 ''2020-2021' ( id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, firstname VARCHAR' at line 1)
here's the sample code I am doing.
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "mias";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$table = $_POST['usersinput'];
// sql to create table
$sql = "CREATE TABLE $table (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP ON UPDATE CURRENT_TIMESTAMP
)";
if ($conn->query($sql) === TRUE) {
echo "Table MyGuests created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close();
?>
Try this code by replacing your code. It will work. i have tried. Problem in your last line of your code.
CREATE TABLE $table(
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30),
email VARCHAR(50),
reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP)
As Nigel has said in the comment, it's definitely a bad idea to allow user input to create a table.
How I would think about doing this would be to use relationships between the Table Guests and the Table or Booking you want them to be added to.
You would just need to create two tables, one for the Booking and one for the Guests then in the Guests table, have a Booking_ID field which would contain the ID of the bookings the user should be added to.
This way, when you want to look for Guests for a specific table, you would be able to do SELECT * FROM MyGuests WHERE booking_id=[the booking id] and this would return the guests for that table.
Like other users stated there are several reasons (most importantly security) not to do that, but if you really want it you have to use concatenation for your string:
Option
$sql = "CREATE TABLE {$table}(id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30), email VARCHAR(50), reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP)";
Option
$sql = "CREATE TABLE" . $table . "(id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30), email VARCHAR(50), reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP)";
This question already has an answer here:
Syntax error due to using a reserved word as a table or column name in MySQL
(1 answer)
Closed 5 years ago.
Do you have any idea whats wrong with my code? I can't really figure it out.
Connection failed: SQLSTATE[42000]: Syntax error or access violation: 1064 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 'VARCHAR(50) NOT NULL, url TEXT(65535) NOT NULL, ip VARCHAR(150) NOT NULL)' at line 3
<?php
$host = "localhost";
$dbname = "nope";
$username = "nope";
$password = "nope";
if(isset($_GET["s"])){
try{
$pdo = new PDO("mysql:host=".$host.";dbname=".$dbname,$username,$password);
$pdo->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$table = "links";
$sql ="CREATE TABLE IF NOT EXISTS $table(
ID INT(11) AUTO_INCREMENT PRIMARY KEY,
key VARCHAR(50) NOT NULL,
url TEXT(65535) NOT NULL,
ip VARCHAR(150) NOT NULL)";
$pdo->exec($sql);
} catch (PDOException $e) {
echo 'Connection failed: ' . $e->getMessage();
}
/*$statement = $pdo->prepare("select url from $table where key = :key");
$statement->execute(array(':key' => $_GET["s"]));
$result = $statement->fetch();
echo $result;*/
}
else{
echo "error";
}
?>
Your column name key is a reserved keyword in mysql
The solution is to quote the column name in backticks
`key` VARCHAR(50) NOT NULL,
or just change the name of the column to whatever you like which is not a reserved word.
my_key VARCHAR(50) NOT NULL,
So your statement should be
$sql ="CREATE TABLE IF NOT EXISTS $table(
ID INT(11) AUTO_INCREMENT PRIMARY KEY,
`key` VARCHAR(50) NOT NULL,
url TEXT(65535) NOT NULL,
ip VARCHAR(150) NOT NULL)";
or
$sql ="CREATE TABLE IF NOT EXISTS $table(
ID INT(11) AUTO_INCREMENT PRIMARY KEY,
table_key VARCHAR(50) NOT NULL,
url TEXT(65535) NOT NULL,
ip VARCHAR(150) NOT NULL)";
Try the following query instead:
CREATE TABLE IF NOT EXISTS $table(
`ID` INT(11) AUTO_INCREMENT PRIMARY KEY,
`key` VARCHAR(50) NOT NULL,
`url` TEXT NOT NULL,
`ip` VARCHAR(150) NOT NULL)
Key is a Reserved Word in MariaDB. And thus needs to be enclosed in ``. And there is no need to define length for type TEXT
I am trying to learn using mysql in php. I started off trying to create a table in mysql, and using the mysqli extension.
My code:
<?php
$truemsg = "Table created successfully";
$falsemsg = "Error creating table: ";
$servername = "localhost";
$username = "myuser";
$password = "mypass";
$db = "mytable";
// Create database
$sql = "USE ".$db.";".
'CREATE TABLE IF NOT EXISTS Authentication (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
userid VARCHAR(30) NOT NULL,
password VARCHAR(30) NOT NULL,
role VARCHAR(20) NOT NULL,
email VARCHAR(50)
);';
print "Sql command is ".$sql;
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
print "<p></p>";
if ($conn->query($sql) === TRUE) {
echo $truemsg;
} else {
echo $falsemsg . $conn->error;
}
$conn->close();
?>
The error is:
Sql command is USE mytable;CREATE TABLE IF NOT EXISTS Authentication ( id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, userid VARCHAR(30) NOT NULL, password VARCHAR(30) NOT NULL, role VARCHAR(20) NOT NULL, email VARCHAR(50) );
Error creating table: 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 Authentication ( id INT(6) UNSIGNED AUTO_INCREMENT PR' at line 1
I tried pasting the same command on the mysql command line, and it works fine. What's the problem using this in php?
You are supposed to run queries one by one
$sql = "query one";
$conn->query($sql);
$sql = 'query two';
$conn->query($sql);
instead of coupling them all in one statement.
DO NOT use mysqi_multi_query() either, this asynchronous function is not intended for the everyday use.
Also, in this particular case USE query is superfluous. Database should go into constructor:
$conn = new mysqli($servername, $username, $password, $db);
^^^ here
Also, tell mysqli to throw errors by itself, automatically, instead of checking result of every database command manually:
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
This way you will get neat and clean code:
<?php
$servername = "localhost";
$username = "myuser";
$password = "mypass";
$db = "mytable";
// Create data table
$sql = 'CREATE TABLE IF NOT EXISTS Authentication (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
userid VARCHAR(30) NOT NULL,
password VARCHAR(30) NOT NULL,
role VARCHAR(20) NOT NULL,
email VARCHAR(50)
)';
// Create connection
mysqli_report(MYSQLI_REPORT_ERROR | MYSQLI_REPORT_STRICT);
$conn = new mysqli($servername, $username, $password, $db);
// Run a query
$conn->query($sql);
echo "Table created successfully";
This code will either report that table has been created successfully, or emit an error, with a detailed explanation on what went wrong.
This seems to be like a mysql multiple query problem
$conn->select_db($db);
you can use this function before the query to use the database and remove the use database statement from your query string , then you query string becomes
$sql = 'CREATE TABLE IF NOT EXISTS Authentication (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
userid VARCHAR(30) NOT NULL,
password VARCHAR(30) NOT NULL,
role VARCHAR(20) NOT NULL,
email VARCHAR(50)
)';
that may work for you ..
I am trying to automatically create the database and table so the website i am creating can be used on a fresh version of XXAMP. At the moment I used PHP myadmin to create the table. however when it loads on a fresh version of XXAMP the database will not be saved on the fresh computer. Therefore im trying to create PHP to automatically create the database and table so content can be added. This is my attempt at the moment but it doesn't seem to be working. Can anyone push me in the right direction?
<?php
$servername = "localhost";
$username = "root";
$password = "";
$dbname = "contentdatabase";
try {
$conn = new PDO("mysql:host=$servername;dbname=$dbname", $username, $password);
// set the PDO error mode to exception
$conn->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
// sql to create table
$sql = "CREATE TABLE items
id INT(10) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
itemName text NULL,
itemDescription text NULL,
itemPrice float NULL,
itemStock smallint(6) NULL,
itemImage VARCHAR(100) NULL,
)";
// use exec() because no results are returned
$conn->exec($sql);
echo "Table items created successfully";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
?>
What you want to do is known as database migration, and there are some frameworks available for it, please take a look at this which is a framework focused in DB migrations.
Replace the corresponding part of your code with the below. I have added comments in the relevant places. Remove the comments before running the code
// sql to create table
$sql = "CREATE TABLE items ( //yours lacks this opening parentheses
id INT(10) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
itemName text NULL,
itemDescription text NULL,
itemPrice float NULL,
itemStock smallint(6) NULL,
itemImage VARCHAR(100) NULL //I have removed the last comma (,)
)";
Thank you for your help!
I have got this semi working now thankfully! I can now create a new database and table. However I am now getting this error but i guess its some sort of if statement i need to add.
SQLSTATE[42S01]: Base table or view already exists: 1050 Table 'item' already existsCREATE DATABASE contentdatabase1
SQLSTATE[HY000]: General error: 1007 Can't create database 'contentdatabase1'; database exists
I saw a lot of same question but I couldn't solve my case.
If I run this code:
<?php
include_once($_SERVER['DOCUMENT_ROOT'].'/config.php');
$servername = HOST;
$username = USERNAME;
$password = PASSWORD;
$dbname = DB;
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// sql to create table
$sql = "CREATE TABLE IF NOT EXISTS Articls (
id INT(10) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
name VARCHAR(254) COLLATE utf8_persian_ci NOT NULL
) DEFAULT COLLATE utf8_persian_ci";
if ($conn->query($sql) === TRUE) {
echo "Table Articls created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
/////////////////////////////////////////////////////////////////////////
$sql = "CREATE TABLE IF NOT EXISTS Tags (
id INT(10) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
id_articls INT(10) UNSIGNED NOT NULL,
name VARCHAR(256) COLLATE utf8_persian_ci NOT NULL,
FOREIGN KEY(Tags.id_articls) REFERENCES Articls(Articls.id)
) DEFAULT COLLATE utf8_persian_ci";
if ($conn->query($sql) === TRUE) {
echo "Table Tags created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close();
?>
I get this error: ( If I remove FOREIGN KEY it works)
Table Articls created successfully Error creating table: Can't create
table 'admin_wepar.Tags' (errno: 150)
Edit
If a change into Articls.id and Tags.id_articls I got this error:
Table Articls created successfullyError creating table: 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 'FOREIGN KEY
(Tags.id_articls) REFERENCES Articls(Articls.id) ) DEFAULT COLLA' at
line 5
You need to declare both Articls.id and Tags.id_articls signed or unsigned
Tags.id_articls is a signed integer while Articl.id is an unsigned integer. MySQL requires referencing field to be exactly the same type. Make Tags.id_articls unsigned to have it work.
Additionally, the table names in the column lists are not allowed in MySQL. It is always clear which table is meant: first the referencing table and then the referenced table. So change
FOREIGN KEY(Tags.id_articls) REFERENCES Articls(Articls.id)
into
FOREIGN KEY(id_articls) REFERENCES Articls(id)
and it will work.