How to merge 2 arrays (PHP, Form)

Question

I have multiple adresses in a database, for each row 1 departure 1 arrival.

I would like to add these in a "form/select"

I can display all of these with :

echo $arraydep[0].$arrayarr[0];

but

$adresses = array($arraydep[0], $arrayarr[0]);

or

$adresses = array_merge($arraydep[0], $arrayarr[0]);

or

$adresses[] = array($arraydep[0], $arrayarr[0]);

does not display all in my form, only the departure and the arrival for only 1 row

It would really nice if you could help me :)

Here is my complete code :

<?php
include "./cfg/db.php";

$sql = 'SELECT * FROM commandes WHERE phone="555-1234"';
$list = mysqli_query($base, $sql) or die("Erreur SQL !" . $sql . "<br />" . mysqli_error());
mysqli_query($base, $sql) or die("Erreur SQL !" . $sql . "<br />" . mysqli_error());

while ($data = mysqli_fetch_array($list))
    {
    $arraydep = array(
        $data[depadresse]
    );
    $arrayarr = array(
        $data[arradresse]
    );
    echo $arraydep[0] . $arrayarr[0]; // echo all adresses from database for this phone number (for testing)
    $adresses = array(
        $arraydep[0],
        $arrayarr[0]
    );
    }

?>
    <form action="test2.php" method="post">
    <select name="test2">
        <?php

foreach($adresses as $adresse)
    {
?>
        <option value="<?php
    echo $adresses . "<br />"; ?>">
        <?php
    echo $adresse . "<br />"; // only echo 1 departure and 1 arrival for 1 row in the drowdown menu
    }

?>
        </option>
    </select>
    <input type="submit" value="Valider" />
</form>

Regards.

reference your array elements using quotation marks, e.g. $data['depadresse'] — lovelace
– lovelace, Commented Jan 10, 2019 at 17:35

Abdur Rahman · Accepted Answer · 2019-01-10 17:34:01Z

3

You are only inserting the first row in the array. Try this.

$i = 0;
    while ($data = mysqli_fetch_array($list))
        {
        $arraydep = array(
            $data[depadresse]
        );
        $arrayarr = array(
            $data[arradresse]
        );
        echo $arraydep[$i] . $arrayarr[$i]; // echo all adresses from database for this phone number (for testing)
        $adresses = array(
            $arraydep[$i],
            $arrayarr[$i]
        );
        $i++
        }

answered Jan 10, 2019 at 17:34

Abdur Rahman

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Comments

Abdullah · Accepted Answer · 2019-01-10 21:29:29Z

0

define your array at outside loop otherwise that does not work on next statement.

look usages at below of my code

$arraydep = array();
$arrayarr = array();
$adresses = array();
while ($data = mysqli_fetch_array($list))
{
    $arraydep[] =  $data['depadresse'];
    $arrayarr[] =  $data['arradresse'];
    $adresses[] = array( $arraydep[0], $arrayarr[0]);
}

var_dump($adresses);

echo $adresses[0][0] . ' ' . $adresses[0][1];

echo $adresses[1];

foreach ($oneOfDep as $arraydep){
    echo $oneOfDep . ' ';
}

answered Jan 10, 2019 at 21:29

Abdullah

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Comments

Guillermo Phillips · Accepted Answer · 2019-01-11 10:23:20Z

You need to append each set of arrive/depart to the array each time around the loop.

$adresses = array();
while ($data = mysqli_fetch_array($list,MYSQLI_ASSOC))
{
    $adresses[] = array(
        'dep' => $data['depadresse'],
        'arr' => $data['arradresse']
    );
}

If you use MYSQLI_ASSOC to retrieve results, you'll have the results keyed by the field names. I don't know what your fields are called, but they need to go inside quotes when accessing $data.

It's also good practice to initialise arrays before using them.

You should also avoid SELECT * usage and instead always list the fields you are interested in using. This tells other developers your intentions.

You should also consider using mysqli_fetch_all($list, MYSQLI_ASSOC) and then you won't need to use the above loop.

To output results just use the below. Although I'm not sure exactly how you want to output the result:

foreach($adresses as $adresse) 
{
    echo "<option value='{$adresse['dep']}'>{$adresse['dep']}</option>";
    echo "<option value='{$adresse['arr']}'>{$adresse['arr']}</option>";

}

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