PHP output json web service

Question

I am very new to the PHP, I am coming from the background of .net. I am trying to make web service which will take parameter from URL and process business logic then output it in jSON.

This is my code

<?php
    $data = '';
    $Site = $_GET['COUNTRY_SITE'];
    $Language = $_GET['LANGUAGE'];

    $data = "{\"Site\":" . "\"" . $Site . "\"" . ",\"Language\":" . "\"" . $Language . "\"" . "}";
    //header('Content-type: application/json');
    //echo $data;
    $x =  json_decode($data,true);
    var_dump($x);

?>

I am getting below output

array(2) {
  ["Site"]=>
  string(5) "India"
  ["Language"]=>
  string(2) "GB"
}

Why I am not getting output like this

{
"Site":"India",
"Language":"GB"
}

Can somebody help me and do explain if possible

Lajos Arpad · Accepted Answer · 2017-03-19 07:02:39Z

1

You are manually creating a JSON and then decode it. The output can be expected, in fact. I think you should use PHP tools to convert your object to JSON:

<?php
    $data = '';
    $Site = $_GET['COUNTRY_SITE'];
    $Language = $_GET['LANGUAGE'];

    //Associative array
    $data = array("Site" => $Site, "Language" => $Language);    

    $x =  json_encode($data, JSON_PRETTY_PRINT);
    var_dump($x);

?>

answered Mar 19, 2017 at 7:02

Lajos Arpad

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Comments

Sahil Gulati · Accepted Answer · 2017-03-19 06:42:33Z

1

Replace var_dump($x); to echo $x = json_encode(json_decode($data,true),JSON_PRETTY_PRINT);

Notes:

json_decode($data,true) This will output an array.
json_encode(json_decode($data,true),JSON_PRETTY_PRINT); This will change array into json format in pretty view

Output:

{
    "Site": "s",
    "Language": "ss"
}

answered Mar 19, 2017 at 6:42

Sahil Gulati

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Comments

Christopher · Accepted Answer · 2017-03-19 06:59:54Z

What you're echoing out is a php array based on the JSON string you built, $data.

However you should reverse your flow and build a php array and then use json_encode() to build the string for you. This will ensure the json is always formatted correctly.

<?php
$data = [
  'Site' => $_GET['COUNTRY_SITE'],
  'Language' => $_GET['LANGUAGE']
];

header('Content-type: application/json');
echo json_encode($data, JSON_PRETTY_PRINT);

The JSON_PRETTY_PRINT option will format it with spaces so that it's more pleasant to read but you can remove it to condense the output.

4EACH · Accepted Answer · 2017-03-19 07:01:10Z

1

<?php
    $data = '';
    $Site = $_GET['COUNTRY_SITE'];
    $Language = $_GET['LANGUAGE'];

    $data = "{\"Site\":" . "\"" . $Site . "\"" . ",\"Language\":" . "\"" . $Language . "\"" . "}";
    //header('Content-type: application/json');
    //echo $data;
    $x =  json_encode(json_decode($data,true),JSON_PRETTY_PRINT);
    var_dump($x);

?>

If you want dynamic JSON result:

<?php
echo json_encode($_GET,JSON_PRETTY_PRINT);
?>

edited Mar 19, 2017 at 7:01

answered Mar 19, 2017 at 6:54

4EACH

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Comments

Muktesh Kumar · Accepted Answer · 2017-03-19 07:08:01Z

1

Try this code

<?php
    $data = '';
    $Site = $_GET['COUNTRY_SITE'];
    $Language = $_GET['LANGUAGE'];
    $data =array('Site'=>$Site,'Language'=>$Language);

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

    $x =  json_encode($data);
    var_dump($x);

?>

answered Mar 19, 2017 at 7:08

Muktesh Kumar

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