创建MySQL表不起作用

Question

I don't understand where my error in this code to create a simple Table in a MySQL database is:

<?php

// Create connection
$con=mysqli_connect("localhost", "administrator", "199992", "test");

// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}

// Select DB
mysqli_select_db($con, "test");

// Create table
$sql = "CREATE TABLE IF NOT EXISTS Notizen(Benutzer TEXT,Datum TEXT,Notiz TEXT)";  

This should actually work, shouldn't it?
But when I try to insert something later:

if(!(empty($_POST[vorname]) and empty($_POST[nachname]) and empty($_POST[notiz]))) {
$sql = "INSERT INTO Notizen (Benutzer, Datum, Notiz)
    VALUES
    ('$_POST[vorname] $_POST[nachname]', 'datum', '$_POST[notiz]')"; 

if (!mysqli_query($con,$sql))
{
echo "Error: " . mysqli_error($con);
 }
}

I get an error:

"Error: Table 'test.notizen' doesn't exist".

Answer 1

1. As I see in your code, you are not executing the create query. execute it by mysqli_query($con,$sql) .
2. Give the space between Notizen (Benutzer table name and braces in create table query.
3. Check that your query is executing or not by using the error function to create table.

Answer 2

确保使用mysqli_query（）运行查询。在将表定义存储在数据库中之前无法插入

Answer 3

After this line

$sql = "CREATE TABLE IF NOT EXISTS Notizen(Benutzer TEXT,Datum TEXT,Notiz TEXT)";  

You must run a

if (!mysqli_query($con,$sql))
{
    echo "Error creating a database: " . mysqli_error($con);
}