PHP: Automated data insertion to two connected tables from one form - php

Question

So, I've been looking for a solution to my case, but I've kept finding only partial and not quite solving-the-matter kind of answers.
First, let me describe what I'm trying to achieve.
In my database I have two tables: PLACES and PLACES_CATEGORIES which are connected by a third table PLACES_A_CATEGORIES in an entity many to many. That is because a PLACE can be characterised by one or more CATEGORIES (but it can also have no CATEGORIES at all).
I want to add data send in one form to two tables: PLACES and PLACES_A_CATEGORIES. The user has all the categories listed with checkboxes and he may (but doesnt have to) check one or more of them.
I automated the display of those checkboxes so it reacts accordingly to changes in database (like adding or removing categories). This part works just fine, but let me show the code for you as it may be useful in solving the real issue:
$query = "SELECT name FROM places_categories";
$result = $connection->query($query);
$category_no = $result->num_rows;
echo "Categories of places:";
for ($j = 0; $j < $category_no; ++$j)
{
$category = $result->fetch_assoc()['name'];
echo '<br><input type="checkbox" id="'.$category.'" name="places_categories" value="'.$category.'"><label for="'.$category.'">'.$category.'</label><br>';
}
So, let's return to the problem. I want to:
always add data (only one row) to the table PLACES
add as many rows of data to the table PLACES_A_CATEGORIES as many checkboxes have been checked
So, let me now show you how I've tried to solve the matter and below I'll explain what and why I've done.
if ($everything_OK==true)//Hurra, everything is ok, lets add the place to the database
{
mysqli_query($connection, "SET NAMES utf8");//need it for special characters
//Adding multiple rows of data to database
$query = "SELECT name FROM places_categories";
$result = $connection->query($query);
$category_no = $result->num_rows;
for ($j = 0; $j < $category_no; ++$j)
{
$category[$j] = $result->fetch_assoc()['name'];
if ($_POST['places_categories'] == $category[$j])
{
//counts number of records in table PLACES
$query1 = "SELECT name FROM places";
$result1 = $connection->query($query1);
$places_no = $result1->num_rows;
$places_no += 1;
//looks for category_id in table places_categories where the name matches the current value from form
$query2 = "SELECT category_id FROM places_categories WHERE name='$category[$j]'";
$result2 = $connection->query($query2);
$what_category_id = mysqli_fetch_array($result2);
$connection->query("INSERT INTO places_a_categories VALUES ('$places_no', '$what_category_id')");
}
}
if ($connection->query("INSERT INTO places VALUES (NULL, 0, 0, 0, '$name', '$wysokosc', '$zajawka', '$zatloczenie', '$data_dodania', '$data_edycji', '$szer_geo', '$dlu_geo', '$tytul', '$opis', '$adres', '$tresc')"))
{
echo "Test!";
}
else
{
throw new Exception($connection->error);
}
}
Okay, explanations:
The part which inserts data to the table PLACES works just fine. It
adds data to the database according to what user has added in a form.
No help needed here.
Because of the before-mentioned automation of
table CATEGORIES I want to check how many of categories actually are
in the database. The first part of the code was supposed to do this.
with instruction FOR I assign every existing category to an array with a value equal to the name of the category in the database
then with instruction IF I want to add ass many rows of data to the table PLACES_A_CATEGORIES as many checkboxes have been checked
first value $places_no equals to id of the place which is being added to another table
second value $what_category_id looks for category_id in table PLACES_CATEGORIES where the name matches the current value got from the checkbox
And what are the results? Data is added to the table PLACES with no problem at all. But there is nothing added to the second table. Furthermore, I get no error message of any kind. It's probably some stupid error I just can't see... Any ideas? What have I done wrong?