how to make some data in a variable not visible in the json array

Question

I created several views in the JSON array but my JSON displays empty data on PHP

This is my code in PHP:

<?php 

include 'koneksi.php';

$data = $koneksi -> query("SELECT * FROM pricelist");

$results = array();
while($line = mysqli_fetch_object($data))

{
    $results[]=$line;
}
echo json_encode($results);

 ?>

This is the result:

{
        "kategori": "Data 1",
        "id": "4",
        "nama": "Service One",
        "harga": "250000",
        "nama1": "Service Two",
        "harga1": "30000",
        "nama2": "Service Three",
        "harga2": "400000"
    },
    {
        "kategori": "Kapasitor Indoor",
        "id": "5",
        "nama": "",
        "harga": "200000",
        "nama1": "",
        "harga1": "",
        "nama2": "",
        "harga2": ""
    },
    {
        "kategori": "Kapasitor OutDoor",
        "id": "6",
        "nama": "",
        "harga": "300000",
        "nama1": "",
        "harga1": "",
        "nama2": "",
        "harga2": ""

How to hide the value in the nama1 and harga1, because the value is empty.

Thanks.

Dont use `*` in your query, select only the columns you want. — Lawrence Cherone, Nov 09 '18 at 08:55
while the data nama1 and harga1 of one in the previous row have data. so I want to hide only empty data — Chiko Frainstact, Nov 09 '18 at 09:13
[https://stackoverflow.com/questions/5285448/mysql-select-only-not-null-values](https://stackoverflow.com/questions/5285448/mysql-select-only-not-null-values) — Sfili_81, Nov 09 '18 at 09:20

score 1 · Answer 1 · answered Nov 09 '18 at 09:03

1

As Lawrence mentioned, extend your query with the columns you actually want to display:

$data = $koneksi -> query("SELECT kategori,id,nama,harga,nama2,harga2 FROM pricelist");

answered Nov 09 '18 at 09:03

Edwin Krause

1,766
1
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if I use this command. in the Kategori : Data 1 name1 and price1 will also not be visible? – Chiko Frainstact Nov 09 '18 at 09:17
Your question didn't specify that you want to select only columns with value initially. So this answer is invalid. You should checkout the link posted by @sfili_81 – Edwin Krause Nov 09 '18 at 09:42

mickmackusa · Answer 2 · 2018-11-09T11:46:20.477

0

So long as you don't have any 0 values that you need to preserve, you can use array_filter()'s default/greedy behavior.

while ($row = $data->fetch_assoc()) {  // maintain OO syntax
    $results[] = array_filter($row);   // this removes: empty/falsey values including zero
}
echo json_encode($results);

If you might need to preserver 0 values, while removing empty and null values, then tell array_filter() to use strlen(). (Demo)

while ($row = $data->fetch_assoc()) {           // maintain OO syntax
    $results[] = array_filter($row, 'strlen');  // only removes: false, null, and empty strings
}
echo json_encode($results);

edited Nov 09 '18 at 11:46

answered Nov 09 '18 at 10:33

mickmackusa

43,625
12
83
136

Even better, you can use array map `$filteredResults = array_map('array_filter', $results)); echo json_encode($filteredResults);` – Damian Dziaduch Nov 09 '18 at 11:26
That styling is not without risk. https://3v4l.org/0Tn1m But so long as you check it for a false result value first, that is fine. It's potato vs potatoe really. Regardless of the technique, you have to iterate every value to check it. I guess I could have written out half-a-dozen more different ways, but I feel that would be overkill. – mickmackusa Nov 09 '18 at 11:39

how to make some data in a variable not visible in the json array

2 Answers2