CODEWITHSUNDEEP
phpmysqljoin
Chris Patten
Chris Patten

Reputation: 129

Properly display results of mySQL one-to-many query

I have two tables:

      TRIPS
-----------------
tripID | clientID

and 

              LEGS
--------------------------------
legID | depart | arrive | tripID

TRIPS has a one-to-many relationship with LEGS in that there are several legID's per tripID. I need to display them in the following format:

Trip tripID1:
    Leg legID1: depart1 - arrive1
    Leg legID2: depart2 - arrive2

Trip tripID2:
    Leg legID3: depart3 - arrive3
    Leg legID4: depart4 - arrive4

etc...

I've been able to iterate through the legIDs via a WHILE() loop, but I'm having trouble embedding a LEGS loop inside the TRIPS loop. My query is:

<?php
$legsQuery = "SELECT  trips.tripID, legs.depart, legs.arrive FROM legs, trips WHERE `trips`.tripID = `legs`.tripID";
$legsQueryResult = mysql_query($legsQuery) or die("QUERY LEG ERROR: " . mysql_error());
while($row = mysql_fetch_assoc($legsQueryResult)) {
    print_r($row);
}
?>

Upvotes: 2

Views: 1669

Answers (1)

binaryLV
binaryLV

Reputation: 9122

  1. add order by clause to sort by trip ID
  2. create $lastTripID variable to check when you get "legs" from "new trip"
  3. [recommended] use join to select data from multiple tables

Code:

<?php
    $legsQuery = "
        select  
            trips.tripID, 
            legs.depart, 
            legs.arrive 
        from 
            legs
            inner join trips on trips.tripID = legs.tripID
        order by
            trips.tripID
    ";
    $legsQueryResult = mysql_query($legsQuery) or die("QUERY LEG ERROR: " . mysql_error());
    $lastTripID = null;
    while ($row = mysql_fetch_assoc($legsQueryResult)) {
        if ( $row['tripID'] !== $lastTripID ) {
            echo $row['tripID'], "\n";
            $lastTripID = $row['tripID'];
        }
        print_r($row);
    }

Upvotes: 1

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