Array json from php is not output in the corect way

Question

I'm trying to make a json output from my db in php with all of users from sql table but i don't know how to do that..i'm new with this. I want the code to output like this.

[
    {
        "name": "Maybell",
        "avatar": "zeldman/128.jpg",
        "data": {
            "company": "Alibaba",
            "title": "Engineer"
        }
    }
]

but the my code does the following output and is not ok.

{
    "name": "Maybell",
    "avatar": "zeldman\/128.jpg",
    "company": "alibaba"
    "title": "Engineer"

}

Here is the Php code:

<?php
header("Content-type: text/json");
$db = mysqli_connect("localhost","root","","testest");
//MSG
$query = "SELECT * FROM mls_users LIMIT 20";
$result = mysqli_query($db, $query);
//Add all records to an array
$rows = array();
while($row = $result->fetch_array()){
    $name = $row['name'];
    $avatar = $row['avatar'];
    $company= $row['company'];
    $title= $row['title'];
}
//Return result to jTable
$qryResult = array();
$qryResult['name'] = $name;
$qryResult['avatar'] = $avatar;
$qryResult['company'] = $company;
$qryResult['title'] = $title;

 echo json_encode($qryResult,JSON_PRETTY_PRINT);
mysqli_close($db);

?>

Ethan · Accepted Answer · 2017-11-03 16:08:52Z

4

Just change:

//Add all records to an array
$rows = array();
while($row = $result->fetch_array()){
    $name = $row['name'];
    $avatar = $row['avatar'];
    $company= $row['company'];
    $title= $row['title'];
}
//Return result to jTable
$qryResult = array();
$qryResult['name'] = $name;
$qryResult['avatar'] = $avatar;
$qryResult['company'] = $company;
$qryResult['title'] = $title;

To:

//Add all records to an array
$qryResult = [];
while ($row = $result->fetch_array()) {
    $qryResult[] = [
        'name'   => $row['name'],
        'avatar' => $row['avatar'],
        'data'   => [
            'company' => $row['company'],
            'title'   => $row['title'],
        ],
    ];
}

You can skip setting the intermediate variables and just add onto the $qryResult array directly.

edited Nov 3, 2017 at 16:08

answered Nov 3, 2017 at 15:59

Ethan

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5 Comments

Ethan Over a year ago

No problem @Alex :). If my answer helped you, don't forget to accept it.

Nick Over a year ago

Not sure why you're getting downvoted, this is correct.

Sarah Groß Over a year ago

Won't this still only output the last row? Because the $qryResult array is being initialized after the loop and only the last state of the variables being added to it.

kscherrer Over a year ago

take a mix of this answer and mine. i did correct the fetching of all rows but oversaw the data array. edit: i edited my answer, so it has the data array too

Ethan Over a year ago

@Connum Yes, you're right, I've updated my answer. Thanks for your feedback :)

kscherrer · Accepted Answer · 2017-11-03 16:09:09Z

2

try this:

<?php
header("Content-type: text/json");
$db = mysqli_connect("localhost","root","","testest");
//MSG
$query = "SELECT * FROM mls_users LIMIT 20";
$result = mysqli_query($db, $query);
//Add all records to an array
$users = array();
while($row = $result->mysqli_fetch_assoc()){
    $users[] = [
        'name' => $row['name'],
        'avatar' => $row['avatar'],
        'data' => [
            'company' => $row['company'],
            'title' => $row['title']
        ]
    ]
}

echo json_encode($users,JSON_PRETTY_PRINT);
mysqli_close($db);

?>

I changed fetch_row() to mysqli_fetch_assoc() and then actually made an array with all fetched rows. You kind of wanted to do that i can see that in comments but you didnt. Then just json encode it.

The outlining brackets [ and ] should come automatically when the array has multiple elements in it.

edited Nov 3, 2017 at 16:09

answered Nov 3, 2017 at 16:03

kscherrer

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1 Comment

Alex Over a year ago

Thank you guys! u're the best :)

nzajt · Accepted Answer · 2017-11-03 16:02:15Z

0

You are almost there, you just need to wrap it in an array and make data an array.

$qryResult = array();
$qryResult['name'] = $name;
$qryResult['avatar'] = $avatar;
// Now make an array to hold data
$data = array();
$data['company'] = $company;
$data['title'] = $title;
// Add data to qryResult
$qryResult['data'] = $data
// Wrap qryResult in array so output will be wrapped in array
$outPutResults = array($qryResult);
echo json_encode($outPutResults,JSON_PRETTY_PRINT);

answered Nov 3, 2017 at 16:02

nzajt

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Comments

pr1nc3 · Accepted Answer · 2017-11-03 16:11:13Z

0

$qryResult = array();
$qryResult['name'] = $name;
$qryResult['avatar'] = $avatar;
$qryResult['data']['company'] = $company;
$qryResult['data']['title'] = $title;

Try that way. You are asking for a multidimensional-array so you need to set it up as one.

answered Nov 3, 2017 at 16:11

pr1nc3

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