I am working on a project, and I have to use sql. The variable $file_name needs to be the table name, but when i try this:
$sqlTableCreate = "CREATE TABLE ". $file_name . "(
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
The table does not create. I checked by using this:
if ($sqlConnection->query($sqlTableCreate) === TRUE) {
echo 'Created Sucessfully';
} else {
echo 'Table does not create.';
}
I get 'Table does not create' when trying to use this. Help would be greatly appreciated. Thanks in advance!
Your filename contains a extension, but I suspect you just want to use the name without the extension as the name of the table. You can use the basename function to remove the extension.
$sqlTableCreate = "CREATE TABLE ". basename($file_name, ".csv") . "(
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
If there can be different extensions, and you want to remove them more generally, see
How to remove extension from string (only real extension!)
I don't see any issue with your posted query but couple things may be wrong
Make sure that there is no table exists with that same name. You can use IF NOT EXISTS marker to be sure like
CREATE TABLE IF NOT EXISTS". $file_name . "(
make sure that the variable $file_name is not empty. Else, you are passing a null identifier in CREATE TABLE statement; which will not succeed.
Per your comment: you have $file_name = 'currentScan.csv';
That's the problem here. You are trying to create a table named currentScan.csv which your DB engine thinking that currentscan is the DB name and .csv is the table name which obviously doesn't exits and so the error.
first check your database connection and change your query with given below :
$sqlTableCreate = "CREATE TABLE ". $file_name . " (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
Related
$tablename = $wpdb->prefix . "name";
$sql = "CREATE TABLE IF NOT EXISTS $tablename(
time_id INT(11) NOT NULL AUTO_INCREMENT,
job VARCHAR(120) NOT NULL,
employee VARCHAR(100) NOT NULL,
date_entry DATETIME NOT NULL,
duration INT(11) NOT NULL,
PRIMARY KEY(time_id)
);";
require_once(ABSPATH . 'wp-admin/includes/upgrade.php');
dbDelta($sql);
creates two tables: 'wp_name' and 'name'. If I define $tablename = 'wp_name' instead it only creates one table. Also '{$wpdb->prefix}name' doesn't change the problem.
What's the problem here? - thanks!
Get rd of the brackets
$sql = "CREATE TABLE IF NOT EXISTS $tablename (
...
);";
They are only used to replace internal wp variables
Update:
your case can't create two tables by itself, there is no way, you have somewhere the same code twice.
Without the prefix, the second code checks that there is already a table named nameand does nothing, but when you add the prefix, it tries to add the tbale name which azt this time doesn't exist, because it is call wp_name
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" . $table . "(id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY, firstname VARCHAR(30) NOT NULL, lastname VARCHAR(30), email VARCHAR(50), reg_date TIMESTAMP DEFAULT CURRENT_TIMESTAMP)";
I've encountered a certain problem. I want to create a table in my database through PHP but it does not work properly.. Even though I copy the text which serves to create a database into PHPMYADMIN, it works perfectly there. Can anyone find any mistake? I have already checked my connection parameters.
if(empty($errors)) {
$namess = mysqli_real_escape_string($conn, $_POST['topicname']);
mysqli_select_db($conn,"d197243_games");
$sql = "CREATE TABLE {$namess} (id INT UNSIGNED AUTO_INCREMENT, topicname VARCHAR(255) NOT NULL, topicdescription VARCHAR(255), topictext TEXT ,primary key (id));";
$result = mysqli_query($conn, $sql);
if(!$result) {
$errors[] = 'Unknown error while creating a post.';
}
}
In your query have you used
{$namess}
if yes then remove those flower brackets
and simply right $namess
as below
$sql = "CREATE TABLE $namess (id INT UNSIGNED AUTO_INCREMENT, topicname VARCHAR(255) NOT NULL, topicdescription VARCHAR(255), topictext TEXT ,primary key (id));";
and hope you have defined your table name in $namess varible
Alright, so I've solved it myself, and thanks for that ''show error'' command. It showed my problem. I have not had enough permissions to access the database.. I had to use different login credentials.. Anyways, thank you for your help & answers. Appriciate it.
I have variable whos value is a random number between 0 and 1000, I would like to use this as the name when creating a new table. I have tried to do this by concatenating my sql with the variable that stores the random number, this hasn't worked, is there a way of doing this? Thanks
include 'includes/db_connect_ssg.php';
if (isset($_POST['new_user_name'])&&isset($_POST['new_user_password'])) {
$username = $_POST['new_user_name'];
$password = $_POST['new_user_password'];
$randID = rand(0,1000);
$sql = "INSERT INTO `Users`(`id`, `username`, `password`, `admin`, `href`) VALUES ('$randID','$username','$password','0','ssgprofile.php?id=$randID')";
$query = mysqli_query($dbc, $sql);
$id = (string)$randID;
$q = "CREATE TABLE CONCAT('userTable_',$id) (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
$qquery = mysqli_query($dbc, $q);
if ($query&&$qquery) {
include 'admin_loadUsers.php';
}else{
echo "Could not connect sorry please try again later, for more info please contact BB Smithy at 0838100085";
}
}
You could just use:
$q = "CREATE TABLE `userTable_".$id."` (
id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
firstname VARCHAR(30) NOT NULL,
lastname VARCHAR(30) NOT NULL,
email VARCHAR(50),
reg_date TIMESTAMP
)";
But beware, creating a table with name containing a number is nearly always a sign of bad database design.
Instead of creating many tables with just one row, simply add columns firstname, lastname, email and reg_date to your table Users. Also your way of generating user ID by calling rand(0,1000) will result in collisions (rand will return a value which is already used as an ID in Users table). Use AUTO_INCREMENT for generating user IDs.
You do not have a valid table name
From the Mysql docs:
Identifiers may begin with a digit but unless quoted may not consist solely of digits
I am trying to create a table whenever the user submits his/her domain name. It is definitely going to be a .com or a .net or a .something
The problem is that my code does not create a table when it contains a .anything
it works fine for names and characters without a .something
$domain=$_POST['domain_name'];//it is dramatainment.com
$table = mysqli_query($connection, "CREATE TABLE $domain (
user_id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
user_name VARCHAR(100) NOT NULL,
user_domain VARCHAR(50) NOT NULL,
user_email VARCHAR(50),
user_password VARCHAR(50),
user_date date
) ");
if(!$table)
{
die('Could not create table: ' . mysqli_error($connection));
}
Does the creat table command have this limitation? Can this be solved?
You could try to replace the dot by an underscore.
Here you can find which character is allowed in a table name.
Additionally, it is possible to query using this syntax, which would conflict with your table name :
SELECT * FROM dbname.table_name;
UPDATE : it is possible using backticks to enclose table name :
$sql = 'CREATE TABLE `$tableName` (
user_id INT(6) UNSIGNED AUTO_INCREMENT PRIMARY KEY,
user_name VARCHAR(100) NOT NULL,
user_domain VARCHAR(50) NOT NULL,
user_email VARCHAR(50),
user_password VARCHAR(50),
user_date date
)';
Some special characters are invalid for identifiers such as table names.
See http://dev.mysql.com/doc/refman/5.7/en/identifiers.html