Creating an associative array in a while loop from a query

Question

I get this example:

<?php 
$shop = array( array( Title => "rose", 
                  Price => 1.25,
                  Number => 15 
                ),
           array( Title => "daisy", 
                  Price => 0.75,
                  Number => 25,
                ),
           array( Title => "orchid", 
                  Price => 1.15,
                  Number => 7 
                )
         );
?>

But how can I make one of these with the following structure:

$result2 = mysqli_query($con, "SELECT r.id AS id, CONCAT(g.fname,' ',g.lname) AS name,
                                    r.arrival AS arrival, r.departure AS departure 
                                    FROM reservations r JOIN guests g ON r.guest = g.id
                                    WHERE r.unit = ".$unit_id."");

while ($row2 = mysqli_fetch_assoc($result2))
    {
        $reservations = array(  array (id => $row['id'],
                                      name => $row['name'],
                                      arrival => $row['arrival'],
                                      departure => $row['departure'],
                                    ));

    }

mseifert · Accepted Answer · 2014-01-29 21:18:12Z

3

Are you looking for this? This will give $reservations with a structure like the one you show.

$result2 = mysqli_query($con, "SELECT r.id AS id, CONCAT(g.fname,' ',g.lname) AS name,
                                    r.arrival AS arrival, r.departure AS departure 
                                    FROM reservations r JOIN guests g ON r.guest = g.id
                                    WHERE r.unit = ".$unit_id."");

$reservations = array();  // makes sure the array exists in case result set is empty
while ($row = mysqli_fetch_assoc($result2))
    {
        $reservations[] =   array (id => $row['id'],
                                      name => $row['name'],
                                      arrival => $row['arrival'],
                                      departure => $row['departure'],
                                    );

    }

Access as follows:

foreach ($reservations as $row){
    echo $row["name"]." arrival ". $row["arrival"]; 
    echo "<br />"; 
}

You can also access:

for ($row = 0; $row<count($reservations); $row++){
    echo $reservations[$row]["name"]." arrival ". $reservations[$row]["arrival"]; 
    echo "<br />"; 
}

edited Jan 29, 2014 at 21:18

answered Jan 29, 2014 at 21:05

mseifert

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4 Comments

Burning Hippo Over a year ago

The query will return x number of rows. So I want to account for that. To me, that just looks like the array will be re created each iteration and will only contain the last row from the query when finished.

mseifert Over a year ago

It will create $reservations with x number of elements if that is how many rows are returned from the query.

Burning Hippo Over a year ago

okay. How does one access a row? For instance, in my first example one could access the data like so:

for ($row = 0; $row < 3; $row++) {     echo $shop[$row]["Title"]." costs ".$shop[$row]["Price"]." and you get ".$shop[$row]["Number"];     echo "<br />"; }

Burning Hippo Over a year ago

Okay, I'll give it a try later and see if all works well. Looks as if it will. Thanks a lot.

Djidel · Accepted Answer · 2014-01-29 21:20:56Z

1

What about this:

$result2 = mysqli_query($con, "SELECT r.id AS id, CONCAT(g.fname,' ',g.lname) AS name,
                               r.arrival AS arrival, r.departure AS departure 
                               FROM reservations r JOIN guests g ON r.guest = g.id
                               WHERE r.unit = ".$unit_id."");
while($row = mysqli_fetch_assoc($result2) $reservations[] =   $row;

this instruction:

$reservations[] = $row;

means that $reservations is an array and we will append this array with $row on the next key ($reservations[ next key ] = $row)

edited Jan 29, 2014 at 21:20

answered Jan 29, 2014 at 21:11

Djidel

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3 Comments

Burning Hippo Over a year ago

How would you access the data?

Djidel Over a year ago

for instance: $reservations[0]['name']

Burning Hippo Over a year ago

Thanks. answer goes to first-responder but I might end up using yours out of simplicity anyway.

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