mysqli cannot create a table [duplicate] - php

Question

This question already has an answer here:
What to do with mysqli problems? Errors like mysqli_fetch_array(): Argument #1 must be of type mysqli_result and such
(1 answer)
Closed 5 years ago.
I have a trouble with my mysql. It connects to database but it doesn't make a table. Here is an example of my code:
$db = new mysqli('localhost', 'user.name', 'user.pass', 'db');
if ($db === FALSE) {
echo "ERRROR";
}
$sql = "CREATE TABLE IF NOT EXISTS db (
ID INT NOT NULL AUTO_INCREMENT,
NAME VARCHAR(20) NULL
)";
$db->query($sql);
if (mysqli_query($db, $sql)) {
echo "TABLE CREATED SUCCESSFULLY";
} else {
echo "TABLE CREATED UNSUCCESSFULLY";
}
I appreciate every answer, thank you for help!

Answer 1

MySQL only allows auto-increment columns if they are defined as a key. Given that the column is labelled ID, which means you probably want it as a primary key, try this:
$sql = "CREATE TABLE IF NOT EXISTS db (
ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
NAME VARCHAR(20) NULL
)";

Answer 2

Because there is error in MySQL query
CREATE TABLE IF NOT EXISTS db (
ID INT NOT NULL AUTO_INCREMENT,
NAME VARCHAR(20) NULL;
Incorrect table definition; there can be only one auto column and it must be defined as a key
fix :
CREATE TABLE IF NOT EXISTS db (
ID INT NOT NULL AUTO_INCREMENT,
NAME VARCHAR(20) NULL,
key idx_ID(ID)
);
PRIMARY KEY is not compulsory, normal key can work too.

Answer 3

Can you please try this.
$db = new mysqli('localhost', 'user.name', 'user.pass', 'db');
if ($db === FALSE) {
echo "ERRROR";
}
$sql = "CREATE TABLE IF NOT EXISTS db (
ID INT NOT NULL AUTO_INCREMENT PRIMARY KEY,
NAME VARCHAR(20) NULL
)";
$db->query($sql);
if (mysqli_query($db, $sql)) {
echo "TABLE CREATED SUCCESSFULLY";
} else {
echo "TABLE CREATED UNSUCCESSFULLY";
}