Having a bit of a struggle here with adding JOINs to a query. I am connecting to two separate databases (on the same server). For this reason, I am writing this mysqli simply and will convert to a prepared statement once it's working.
// REMOVED: DB VARIABLES
$conn = new mysqli($servername, $username, $password, $db_connective_data);
if ($conn->connect_error) { die("Connection failed: " . $conn->connect_error); }
$conn2 = new mysqli($servername, $username, $password, $db_resources);
if ($conn2->connect_error) { die("Connection failed: " . $conn2->connect_error); }
$sql = "SELECT * FROM downloads LEFT JOIN resource_data ON downloads.resource_id_REF=resource_data.resource_id WHERE downloads.user_basics_id_REF='$user_id'";
$result = $conn->query($sql);
$number_of_download_rows_returned = mysqli_num_rows($result) -1;
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
$resource_id_REF[] = $row['resource_id_REF'];
$download_date[] = date('Y-m-d', strtotime($row['download_date']));
$resource_title[] = $row['resource_title'];
$resource_title_link[] = str_replace(" ", "-", $row['resource_title']);
}
}
$conn->close();
A query without a JOIN works fine (albeit without returning the resource_title):
$sql = "SELECT * FROM downloads WHERE downloads.user_basics_id_REF='$user_id' ORDER BY downloads.download_date DESC";
What am I missing here? The first code sample will return no results. The second one will return three.
Any assistance is greatly appreciated.
Here is a list of the different database names that I reference. As I stated, some data is in the "connective_data" db and some is in the "resources" db.
$db_connective_data = "connective_data";
$db_lists = "lists";
$db_messaging = "messaging";
$db_resources = "resources";
$db_users = "users";
I can't seem to get two of them connected. Am I missing something strikingly obvious here?
There is no need to create 2 connections if the databases are located on the same mysql server. You can simply reference tables from another database as databasename.tablename.
As a result, you can join 2 tables from 2 different databases as:
$sql = "SELECT * FROM yourdatabase1.downloads LEFT JOIN yourdatabase2.resource_data ON yourdatabase1.downloads.resource_id_REF=yourdatabase2.resource_data.resource_id WHERE yourdatabase1.downloads.user_basics_id_REF='$user_id'";
Obviously, you need to substitute your real database names for yourdatabase1 and yourdatabase2 in the above query.
Update: Are you sure you need so many databases? These seem to be tables to me, not databases.
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