CODEWITHSUNDEEP
phpjson
Dave Moz
Dave Moz

Reputation: 127

PHP: file_get_contents('php://input') returning string for JSON message

I am trying to read in a JSON message in my PHP app and this is my php code:

$json = file_get_contents('php://input');
$obj = json_decode($json, TRUE);
echo $obj->{'S3URL'};

When I do this I am getting the following error:

Trying to get property of non-object in setImage.php on line 25 (line 25 is the echo $obj->{'S3URL'}; line)

This is the request body of the request to the page:

Request Url: http://localhost:8888/setImage.php
Request Method: POST
Status Code: 200
Params: {
   "S3URL": "http://url.com"
}

This is the request headers:

Accept: application/json
Content-Type: application/json
Connection: keep-alive
Origin: chrome-extension: //rest-console-id
User-Agent: Mozilla/5.0 (Macintosh; Intel Mac OS X 10_9_1) AppleWebKit/537.36 (KHTML,

However, if I instead echo out the $json variable I get the following:

S3URL=http%3A%2F%2Furl.com

So it looks like file_get_contents('php://input'); is reading it in as a string, and not as JSON, which will make parsing it more difficult.

Any idea why it isn't being returned as JSON, or how I can get it to be returned as JSON?

Upvotes: 11

Views: 91817

Answers (7)

Ansh Mehta
Ansh Mehta

Reputation: 28

If you are using javascript to send JSON, to a php file

send_recieve.js

var myObject = JSON.stringify({"name":"John","age":30,"city":"New York"});
var xhr = new XMLHttpRequest();
xhr.open("POST","http://localhost/dashboard/iwplab/j-comp/receive_send.php",false);
xhr.setRequestHeader("Content-type","application/json");
xhr.onreadystatechange = function(){
    if(xhr.readyState==4){console.log("xhr response:"+xhr.response)}
    alert(xhr.responseText);
 };
 xhr.send(myObject);

recieve_send.php

<?php
    $var = json_decode(file_get_contents("php://input"),true);
    echo "Data recieved by PHP file.\n";
    if ($var["name"]=="John"){
        echo "a";
    }
    else{
        echo "b";
    }
    //Manipulate/validate/store/retrieve to database here
    //echo statements work as response
    echo "\nSent";

?>

echo statements work as a response.

Upvotes: 0

Shanaka Madusanka
Shanaka Madusanka

Reputation: 99

Try this https://stackoverflow.com/a/56801419/7378998

It works for JSON or jquery post

$.post(url, $('form').serialize());

Upvotes: 0

AKASH VERMA
AKASH VERMA

Reputation: 99

there are two type for executing this type of request

First : you can use it as an stdClassObject for this

$data = json_decode(file_get_contents('php://input'));

it will return a object and you can retrieve data from this like

$name = $data->name;

Second : you can use it as an array for this

$data = json_decode(file_get_contents('php://input'), true);

it will return a object and you can retrieve data from this like

$name = $data['name'];

Upvotes: 7

Ratanveer Singh
Ratanveer Singh

Reputation: 1

Use this one result will 

$chd = json_decode(file_get_contents('php://input'), TRUE);
$childs = implode("",$chd);

Upvotes: 0

Prithabai R
Prithabai R

Reputation: 58

The problem may be form php://input (is a read-only stream that allows you to read raw data from the request body). change some setting from php.ini , try to make "allow_url_fopen" on.

Upvotes: 0

ramin
ramin

Reputation: 17

$obj = json_decode($json);

Just remove the true

Upvotes: -1

elixenide
elixenide

Reputation: 44841

Your use of json_decode is creating an associative array, not an object. You can treat it like an array, instead of an object. If you want an object, use this, instead:

$obj = json_decode($json);

See the documentation on the second parameter to json_decode():

assoc When TRUE, returned objects will be converted into associative arrays.

Also, as Johannes H. pointed out in the comments, the output of echo $json; indicates that you are not actually receiving JSON, in the first place, so you will need to address that, as well. You asked why it isn't JSON; without seeing how you are requesting this script, it's impossible to say for sure.

Upvotes: 8

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