CODEWITHSUNDEEP
phpjson
Outcast
Outcast

Reputation: 5117

How to produce a valid json output with array of json objects

I created a mySQL database with phpMyAdmin in my local server. In this database I stored the names and the location of my friends (along with an id as a primary key for the database). I wrote and run the following php script to retrieve these data from the database and project them on my local web server (XAMPP):

<?php

$dbServername = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'Friends';

$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);

header('Content-Type: application/json');

$sql = 'SELECT * FROM friends;';
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);

if ($resultCheck > 0) {
    while ($row = mysqli_fetch_assoc($result)) {
        echo json_encode($row, JSON_PRETTY_PRINT);

    }
}

However, in this way I take this output: 

{
    "id": "1",
    "name": "David Belton",
    "location": "New Haven"
}{
    "id": "2",
    "name": "Alex Danson",
    "location": "New York"
}

which is not a valid json output overall. I would like to have the following output:

[{
        "id": "1",
        "name": "David Belton",
        "location": "New Haven"
    }, {
        "id": "2",
        "name": "Alex Danson",
        "location": "New York"
  }]

(which is also a valid json output)

How can I do this?

Upvotes: 2

Views: 155

Answers (4)

Pankaj Makwana
Pankaj Makwana

Reputation: 3050

Create an array and push those records in while loop. At the end echo json_encode(); so the output will be in proper JSON format

$records = array();
if ($resultCheck > 0) {
    while ($row = mysqli_fetch_assoc($result)) {

        $records[] = $row;
    }
}
echo json_encode($records, JSON_PRETTY_PRINT);

Upvotes: 1

IcedAnt
IcedAnt

Reputation: 454

Or, instead of fetching results one-by-one, just use mysqli_fetch_all

if ($resultCheck > 0)
{
    $row = mysqli_fetch_all($result);
    echo json_encode($row, JSON_PRETTY_PRINT);
}

which is actually faster and uses less memory: http://php.net/manual/en/mysqli-result.fetch-all.php

Upvotes: 1

Mitya
Mitya

Reputation: 34576

This is happening because you're echoing out each object in isolation from the others, inside your loop.

while ($row = mysqli_fetch_assoc($result))
    echo json_encode($row, JSON_PRETTY_PRINT); //<-- echo out an isolated JSON object

Instead, just store everything in an array until you're done then echo out the JSON once, not iteratively.

$arr = [];
while ($row = mysqli_fetch_assoc($result)) $arr[] = $row;
json_encode($arr, JSON_PRETTY_PRINT);

Upvotes: 1

Eric Amshukov
Eric Amshukov

Reputation: 243

Store the result in the array, then echo the encoded/formatted array: 

<?php

$dbServername = 'localhost';
$dbUsername = 'root';
$dbPassword = '';
$dbName = 'Friends';

$conn = mysqli_connect($dbServername, $dbUsername, $dbPassword, $dbName);

header('Content-Type: application/json');

$sql = 'SELECT * FROM friends;';
$result = mysqli_query($conn, $sql);
$resultCheck = mysqli_num_rows($result);

$arr = array();

if ($resultCheck > 0) {
    while ($row = mysqli_fetch_assoc($result)) {
        array_push($arr, $row);
    }
}

echo json_encode($arr, JSON_PRETTY_PRINT);

Upvotes: 1

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