PHP/MySQL Relationships / Echo Field - php

Question

I'm trying to build my 1st relationship database and it looks like this so far:
Relationships Table
Partners Table
Location Table
I want to echo all the row info including the related location names on my page, how would I get the following to echo?
ID: 2 Name: Salisbury Removals Locations: Salisbury
ID: 4 Name: Inbetween Removals Locations: Salisbury, Southampton
ID: 5 Name: Southampton Removals Locations: Southampton
=====SOLVED!=====
$sql = "SELECT partner_id, partner_name, email_address, active FROM partners WHERE active ='yes' ORDER BY partner_id ASC";
$connect->query($sql);
if ($partners = $connect->query($sql)) {
foreach ($partners as $partner) {
echo '<li><ul>';
echo '<li>' . $partner['partner_id'] . '</li>';
echo '<li>' . $partner['partner_name'] . '</li>';
echo '<li>' . $partner['email_address'] . '</li>';
echo '<li>' . $partner['active'] . '</li>';
// START GET LOCATIONS FROM RELATED TABLE
echo '<ul>';
$sql2 = "SELECT p.partner_name AS Name, p.partner_id AS ID, l.location_name AS Locations from partners_locations r, partners p, locations l WHERE p.partner_id = r.partner_id AND l.location_id = r.location_id AND r.partner_id =" . $partner['partner_id'] . "";
$connect->query($sql2);
if ($locations = $connect->query($sql2)) {
foreach ($locations as $location) {
echo '<li>' . $location['Locations'] . '</li>';
}
} else {
echo "Error: No Locations<br>";
}
echo '</ul>';
// END GET LOCATIONS FROM RELATED TABLE
echo '</ul></li>';
}
} else {
echo "Error: No Active Partners<br>";
}

Answer 1

The best thing you could do is make a new table to contain the partner_id and location_id.
tbl_relationships_new
Pros for this approach :-
1).When you need to remove a location from a partner, you wouldn't need to edit the column locations. You could simply delete an entry from this new table.
2). When you need to add more data in the locations field, you could simply just insert into the new table, which is rather easy than having to update partners.locations.
Now, you could use easy left joins to get the required data.
SQL query for my table solution.
SELECT t.*,p.*,l.* FROM tbl_relationships_new t, partners p, locations l LEFT JOIN
partners
ON
t.partner_id = p.partner_id
LEFT JOIN
locations
ON
l.location_id = t.location_id
WHERE
t.partner_id = 2
UPDATE
Here are the queries based on your table structure.
1). ID: 2 Name: Salisbury Removals Locations: Salisbury
SELECT p.partner_name AS Name, p.partner_id AS ID, l.location_name AS Locations from relationships r, partners p, locations l WHERE p.partner_id = r.partner_id AND l.location_id = r.location_id AND r.partner_id = 2
2). When there are 2 locations.
SELECT p.partner_name AS Name, p.partner_id AS ID, l.location_name AS Locations from relationships r, partners p, locations l WHERE p.partner_id = r.partner_id AND l.location_id = r.location_id AND r.partner_id = 5
UPDATE
Solution without explicitly mentioning an ID.
SELECT p.partner_name AS Name, p.partner_id AS ID, l.location_name AS Locations from relationships r, partners p, locations l WHERE p.partner_id = r.partner_id AND l.location_id = r.location_id

Answer 2

<?php
$dbconn = mysqli_connect(DBSERVER, DBUSER, DBPWD, DBNAME);
$sql = "SELECT partner_id, partner_name, locations FROM Partners";
$result = mysqli_query($dbconn, $sql);
while ($row = mysqli_fetch_assoc($result)) {
echo "ID: ".$row['partner_id']." Name: ".$row['partner_name']." Locations: ";
$locs = explode(",",$row['locations']);
$first = true;
foreach ($locs as $loc) {
$sql2 = "SELECT location_name FROM Locations WHERE location_id = '".$loc."'";
$result2 = mysqli_query($dbconn, $sql2);
$row2 = mysqli_fetch_assoc($result2);
$separator = ($first?"":", ");
$first = false;
echo $separator.$row2['location_name'];
}
echo "<br/>";
}
?>

Answer 3

First your table Table: Partners entry should be like this.
partener_id | partner_name | email_address | active | location_id
2 Test Partner ....... yes 1
5 Good Removals ....... yes 1
5 Good Removals ....... yes 2
4 Special Removals ....... yes 2
So, you could make join with Table: Locations.
After that execute below sql query.
$sql = "select * from partners Left join locations on locations.location_id=partners.location_id";
For better understanding of query execution and data display. follow below link.
http://www.w3schools.com/php/php_mysql_select.asp

Answer 4

Your "locations" field in the Partners table is wrong, you must create another table to handle relations between Partners and Locations, called Locations_Partners par example. This table must have 2 fields, location_id and partner_id. This way, you can relate many partners with many locations. Then, you code should look like this:
<?php
$servername = "www.server.com";
$username = "username";
$password = "password";
$dbname = "your database name";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT partner_id, partner_name, location_name FROM Partners
INNER JOIN Locations_Partners ON Partners.partner_id = Locations_Partners.Partner_id
INNER JOIN Locations ON Locations_Partners.Location_id = Locations.location_id";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["partner_name"]. " " . $row["location_name"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>