PHP + json: do not include empty fields

Question

The code below brings me an array of images. I have 15 fields: "img1", "img2", "img3" and so on.

Sometimes not all the fields have data. I don´t want to include them into the array. Is there a way to keep these empty fields outside the array?

<?php

include_once 'db_connect.php';

$i_id = $_POST["imovel_id"];

$sql = "SELECT * FROM iMoveis WHERE imovel_id='$i_id'";
$result = mysqli_query($mysqli, $sql);

$response = array();
$images = array();

while($row = mysqli_fetch_assoc($result)){

$images[] = array('images' => $row['img1']);
$images[] = array('images' => $row['img2']);
$images[] = array('images' => $row['img3']);
$images[] = array('images' => $row['img4']);
$images[] = array('images' => $row['img5']);
$images[] = array('images' => $row['img6']);
$images[] = array('images' => $row['img7']);
$images[] = array('images' => $row['img8']);
$images[] = array('images' => $row['img9']);
$images[] = array('images' => $row['img10']);
$images[] = array('images' => $row['img11']);
$images[] = array('images' => $row['img12']);
$images[] = array('images' => $row['img13']);
$images[] = array('images' => $row['img14']);
$images[] = array('images' => $row['img15']);

} 

  $response['posts'] = $images;

  echo json_encode($response, JSON_UNESCAPED_SLASHES);


?>

Community · Accepted Answer · 2020-06-20 09:12:55Z

3

Big Picture

You need to normalize your data instead of having fields with names like img15. Put the images in a different table with a reference to imovel_id. Normalized data would make this much easier.

Hacky Fix for Your Problem

With your current table structure, you have no choice but to do lots and lots of if statements (or a loop over the data). Something like this:

<?php

include_once 'db_connect.php';

$i_id = $_POST["imovel_id"];

$sql = "SELECT * FROM iMoveis WHERE imovel_id='$i_id'";
$result = mysqli_query($mysqli, $sql);

$response = array();
$images = array();

while($row = mysqli_fetch_assoc($result)){
    for ($index = 1; $index <= 15; $index++) { // loop over the columns
        if (!empty($row['img' . $index])) { // check for null or empty columns
            $images[] = array('images' => $row['img' . $index]); // add only if there is data
        }
    }
}
$response['posts'] = $images;
echo json_encode($response, JSON_UNESCAPED_SLASHES);
?>

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Nov 4, 2015 at 20:11

elixenide

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3 Comments

madsongr Over a year ago

I am building an Android prototype app to search a real estate office db, to get the images of apartments, houses, etc. Do you think these databases use tables with references?

elixenide Over a year ago

@madsongr I have no idea how that database might be structured. If that table is one you created then you should refactor it. If not, the loop in my answer will work perfectly well.

madsongr Over a year ago

Yes both answers worked. Thank you so much! As a prototype I built a simple db with images in the same table. Gotta work around with others situations now. Thanks again.

durbnpoisn · Accepted Answer · 2015-11-04 20:08:48Z

1

Just modify your array populator with an IF statement:

if($row['img1'])  $images[] = array('images' => $row['img1'];

So it will only add that one if there is something in it.

answered Nov 4, 2015 at 20:08

durbnpoisn

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PHP + json: do not include empty fields

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