Retrieve certain row from array depending on a value - php

Question

Echo out the right row from an array compiled from a mysql database.
I have extracted information from a database (locations) containing three fields: id, name, city into an array called $array. I want to loop through another database (events) in which the id's from the first database (locations) are stored in a field. When looped I want to display the corresponding name and city from the locations database.
Is this possible without having to fetch information every loop?
This is my first try
$query = "Select id, name, city FROM locations WHERE typ = '1'";
$result = mysqli_query($conn, $query);
$row = array();
while ($row = mysqli_fetch_assoc($result)) {
$array[] = $row;
}
And then I thought I could specify the key myself like this:
$query = "Select id, name, city FROM locations WHERE typ = '1'";
$result = mysqli_query($conn, $query);
$row = array();
while ($row = mysqli_fetch_assoc($result)) {
$array[$row['id']] = $row;
}
But I couldn't figure out how to echo the right row.

Answer 1

You can join both tables in a single query, using something like this:
Select locations.id, locations.name, locations.city, events.name
FROM locations
JOIN events ON locations.id = events.id
WHERE locations.typ = '1'
The events.id on the JOIN statement is assuming that this is the column name of the id in the events table. I also made the assumption that these are the two fields that will match between the two tables. Adjust accordingly if your matching criteria is different.
The SELECT statement was modified to pull columns from both tables. Add whichever fields are relevant to your needs.