json array response for PHP web service

Question

In this service I am getting output but I want different output. Let me explain:

    $customer_id = $_POST['customer_id'];
    $response = array();
    $qry="SELECT category FROM nesbaty_customer where c_id='".$customer_id."' ";
    $qry_res=mysqli_query($con,$qry);
    $jsonData = array();
    while ($array = mysqli_fetch_assoc($qry_res)) 
    {
        $r= $array['category'];

        $jsonData[]=explode(",",$r);

    }
    echo json_encode(array('data' =>$jsonData));
    mysqli_close($con);

I am getting Output like this:

{
"data": {
    "category": [
        "Hotel",
        "Saloon"
    ]
}

}

But I want output like this!

{"data":[{"category":"Hotel"},{"category":"Saloon"}]}

As you wrote it down, it has duplicated key category, so you cant get it — Eakethet
– Eakethet, Commented May 10, 2018 at 7:42
Danger: You are vulnerable to SQL injection attacks that you need to defend yourself from. — Quentin
– Quentin, Commented May 10, 2018 at 8:30

Nick · Accepted Answer · 2018-05-10 13:08:10Z

2

You can't get exactly what you want because you would have an array with two identical keys. You can get something similar by changing:

$r= $array['category'];
$jsonData[]=explode(",",$r);

to

foreach (explode(',', $array['category']) as $cat) {
    $jsonData[]=array('category' => $cat);
}

$jsonData will look like this:

Array
(
    [data] => Array
        (
            [0] => Array
                (
                    [category] => Hotel
                )
            [1] => Array
                (
                    [category] => Saloon
                )
        )
)

and the output from json_encode will be:

{"data":[{"category":"Hotel"},{"category":"Saloon"}]}

edited May 10, 2018 at 13:08

answered May 10, 2018 at 8:01

Nick

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8 Comments

Jayvirsinh Vaghela Over a year ago

{ "data": { "category": [ "Hotel", "Saloon" ] } }

Jayvirsinh Vaghela Over a year ago

this is my output

Jayvirsinh Vaghela Over a year ago

BUT i want This {"data":[{"category":"Hotel"},{"category":"Saloon"}]}

Nick Over a year ago

But that is the output of this code... are you sure you copied it correctly?

Jayvirsinh Vaghela Over a year ago

again tell me above two line will replace by your code?

|

Community · Accepted Answer · 2021-10-07 11:39:35Z

1

The keys should be unique, you shouldn't repeat "category" as a key.

"An object whose names are all unique is interoperable in the sense that all software implementations receiving that object will agree on the name-value mappings. When the names within an object are not unique, the behavior of software that receives such an object is unpredictable."

See https://www.rfc-editor.org/rfc/rfc7159 for JSON definition.

As you can see, json_encode uses this rfc: "PHP implements a superset of JSON as specified in the original » RFC 7159. "

If you really want to write non-unique keys, you should write your custom json encode function, but the decode function will have unpredictable behaviour.

edited Oct 7, 2021 at 11:39

CommunityBot

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answered May 10, 2018 at 12:14

IgrewupwithSlackware

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2 Comments

Jayvirsinh Vaghela Over a year ago

brother firstly understand my query! multiple data imploded in single row and want to explode all that data like this {"data":[{"category":"Hotel"},{"category":"Saloon"}]}

IgrewupwithSlackware Over a year ago

We are not speaking about the database key, but about the JavaScript Object keys. You are trying to use the same key "category" two times inside this object.data

Mike Doe · Accepted Answer · 2018-05-10 12:21:52Z

0

Add each category as separate json object then:

while ($array = mysqli_fetch_assoc($qry_res)) {
    foreach ($array['category'] as $categoryName) {
        $jsonData[] = ['category' => $categoryName];
    }
}

This way you end up with an array of objects, thus you should be fine.

answered May 10, 2018 at 12:21

Mike Doe

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Comments

user9763435 · Accepted Answer · 2018-05-10 08:06:08Z

-1

The key of the array must be unique

answered May 10, 2018 at 8:06

user9763435

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Comments

Hosain Ahmed · Accepted Answer · 2018-05-10 08:26:08Z

-2

while ($array = mysqli_fetch_assoc($qry_res)) {

    $jsonData[]=$array;

}
echo json_encode(array('data' =>$jsonData));

answered May 10, 2018 at 8:26

Hosain Ahmed

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1 Comment

Jayvirsinh Vaghela Over a year ago

oh men ! i want to explode two category it will return in one!!

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