Error creating table: Incorrect table name - php

I cannot access the textbox content to create the table in Mysql and the error is
"Error creating table: Incorrect table name ''
<?php
$conn=mysqli_connect("localhost","root","","abc");
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$email = isset($_POST['email']) ? $_POST['email'] : '';
// sql to create table
$sql = "CREATE TABLE `$email` (
`id` INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
`firstname` VARCHAR(30) NOT NULL,
`lastname` VARCHAR(30) NOT NULL,
`email` VARCHAR(50),
`regdate` TIMESTAMP
)";
if ($conn->query($sql) === TRUE) {
echo "Table MyGuests created successfully";
} else {
echo "Error creating table: " . $conn->error;
}
$conn->close();
?>

If error is "Error creating table: Incorrect table name" and you query has:
CREATE TABLE `$email`
Most probably $email has an # sign in it, that shouldn't be part of a table name - use only [0-9,a-z,A-Z$_] (basic Latin letters, digits 0-9, dollar, underscore).
It is not likely that a table should be created each form sublimation, and that its name should be dynamic (you will not know it to query data from this table).
Not less important - never use user input directly in your SQL (to avoid SQL injection). Reference:
PHP MySQLi Prepared Statements Tutorial to Prevent SQL Injection

You can try as follows-
$email = $_POST['email'];
// sql to create table
$sql = "CREATE TABLE IF NOT EXISTS $email(
`id` INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
`firstname` VARCHAR(30) NOT NULL,
`lastname` VARCHAR(30) NOT NULL,
`email` VARCHAR(50),
`regdate` TIMESTAMP
)";

Related

Create TABLE using PHP, according to input [duplicate]

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)";

Create Table in mySQL with a name that includes a hyphon, percent symbol, full stop etc

I want to create tables in mySQL with names that include hyphons, full stops, the % symbol and others. The name will be in a variable called image_name. How do I go about this? Of course I could replace all with underscores, but that is suboptimal. Is the default collation wrong?
My Code:
$image_name = 'abc_def.ghi%';
$sql = "CREATE TABLE $image_name(
id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
film_name VARCHAR(70) NOT NULL,
...
)";
if(mysqli_query($link, $sql)){
echo "Table of image $image_name created successfully.";
} else{
echo "ERROR: Could not execute $sql. " . mysqli_error($link);
}
You should surround the table names with backticks - ``
The table name has to surrounded by backticks.
SQL DEMO
CREATE TABLE `abc_def.ghi%` (
id INT NOT NULL PRIMARY KEY AUTO_INCREMENT,
film_name VARCHAR(70) NOT NULL
)

Cannot create table with PHP

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

How to create multiple MySQL tables via PHP using a single query?

I am trying to create a "setup script" for my website. I would like to create the database, adding tables and some content at the same time. So far this is how I did it, but it seems kind off messy using multiple queries:
<?php
$servername = "localhost";
$username = "root";
$password = "password";
// Create connection
$conn = new mysqli($servername, $username, $password);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
// Create database
$sql = "CREATE DATABASE MYDB";
if ($conn->query($sql) === TRUE) {
echo "1. Database created successfully <br/>";
$conn->select_db("MYDB");
$sql_members = "CREATE TABLE MEMBERS (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
USERNAME VARCHAR(30) NOT NULL,
EMAIL VARCHAR(40) NOT NULL,
DISCOUNT VARCHAR(5),
PASSW CHAR(128),
ROLE VARCHAR(9)
)";
if ($conn->query($sql_members) === TRUE) {
echo "2. Table MEMBERS created successfully <br/>";
} else {
echo "Error creating table: " . $conn->error;
}
$sql_content = "CREATE TABLE CONTENT (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
TITLE VARCHAR(30) NOT NULL,
TEXT VARCHAR(30) NOT NULL
)";
if ($conn->query($sql_content) === TRUE) {
echo "3. Table CONTENT created successfully <br/>";
} else {
echo "Error creating table: " . $conn->error;
}
} else {
echo "Error creating database: " . $conn->error;
}
$conn->close();
?>
Is there a better way?
Thanks!
== UPDATE ==
I have tried to export the database and use the resulted .sql file as my setup query, but something is wrong, I get:
Error creating tables: 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 'INSERT INTO CONTACTS (ID, NAME, PHONE,
EMAIL, ADDRESS, CITY, `COUN' at line 12
CREATE TABLE IF NOT EXISTS `CONTACTS` (
`ID` int(11) NOT NULL AUTO_INCREMENT,
`NAME` varchar(25) COLLATE utf8_romanian_ci NOT NULL,
`PHONE` varchar(16) COLLATE utf8_romanian_ci NOT NULL,
`EMAIL` varchar(35) COLLATE utf8_romanian_ci NOT NULL,
`ADDRESS` text COLLATE utf8_romanian_ci NOT NULL,
`CITY` varchar(16) COLLATE utf8_romanian_ci NOT NULL,
`COUNTRY` varchar(16) COLLATE utf8_romanian_ci NOT NULL,
PRIMARY KEY (`ID`)
) ENGINE=InnoDB DEFAULT CHARSET=utf8 COLLATE=utf8_romanian_ci AUTO_INCREMENT=2 ;
INSERT INTO `CONTACTS` (`ID`, `NAME`, `PHONE`, `EMAIL`, `ADDRESS`, `CITY`, `COUNTRY`) VALUES
(1, 'Peter Brown', '0742062307', 'office#shop.com', 'Avenue 13.', 'Santaclaus', 'Austria');
== SOLUTUION ==
I needed "multi_query()" for executing my multiple queries.
You can try this too :p
$errors = [];
$table1 = "CREATE TABLE MEMBERS (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
USERNAME VARCHAR(30) NOT NULL,
EMAIL VARCHAR(40) NOT NULL,
DISCOUNT VARCHAR(5),
PASSW CHAR(128),
ROLE VARCHAR(9)
)";
$table2 = "CREATE TABLE CONTENT (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
TITLE VARCHAR(30) NOT NULL,
TEXT VARCHAR(30) NOT NULL
)";
$tables = [$table1, $table2];
foreach($tables as $k => $sql){
$query = #$conn->query($sql);
if(!$query)
$errors[] = "Table $k : Creation failed ($conn->error)";
else
$errors[] = "Table $k : Creation done";
}
foreach($errors as $msg) {
echo "$msg <br>";
}
You could export the whole database including all tables using the command line or using PhPMyAdmin. Then query the content of the file in php to create the database.
you can create a file and put all your sql queries in it..
CREATE TABLE MEMBERS (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
USERNAME VARCHAR(30) NOT NULL,
EMAIL VARCHAR(40) NOT NULL,
DISCOUNT VARCHAR(5),
PASSW CHAR(128),
ROLE VARCHAR(9)
);
CREATE TABLE CONTENT (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
TITLE VARCHAR(30) NOT NULL,
TEXT VARCHAR(30) NOT NULL
);
then in your php code:
$query = file_get_contents ('queries.sql');
if ($conn->query($query) === TRUE) {
echo "all tables created successfully <br/>";
} else {
echo "Error creating tables: " . $conn->error;
}

users storing own data . Mysql table name taken from session variable [duplicate]

This question already has answers here:
PHP parse/syntax errors; and how to solve them
(20 answers)
Closed 6 years ago.
Can someone help me with that code, any directions welcome
I want every user who passed session to create table(every user creates own table -others cant see it).
<?php
session_start();
if($_SESSION['user']==''){
header("Location:login.php");
}else{
$dbh=new PDO('mysql:dbname=something;host=127.0.0.1', 'something', 'something');
$sql=$dbh->prepare("SELECT * FROM users WHERE id=?");
$sql->execute(array($_SESSION['user']));
while($r=$sql->fetch()){
$sql = "CREATE TABLE .'$r['username'].'" (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
$conn->exec($sql);
echo " created successfully";
}
catch(PDOException $e)
{
echo $sql . "<br>" . $e->getMessage();
}
$conn = null;
}
}
?>
You can try to fix your php code and get something like this:
<?php
session_start();
if($_SESSION['user']==''){
header("Location:login.php");
}else{
$dbh=new PDO('mysql:dbname=something;host=127.0.0.1', 'something', 'something');
$sql=$dbh->prepare("SELECT * FROM users WHERE id=?");
$sql->execute(array($_SESSION['user']));
while($r=$sql->fetch()){
$sql = "CREATE TABLE `".$r['username']."` (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
try {
$dbh->exec($sql);
echo " created successfully";
} catch(PDOException $e) {
echo $sql . "<br>" . $e->getMessage();
}
}
$conn = null;
}
?>
You have some error in your query string. You have one double quote you should remove, and the table name should be wrapped with ` instead of '. So your code should look like:
$sql = "CREATE TABLE `{$r['username']}` (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
And even if its not really related to the topic you should ask yourself if create a table per user is the best solution.

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