how to use JSON.stringify and json_decode() properly

Question

Im trying to pass a mulitidimensional Javascript array to another page on my site by:

• using JSON.stringify on the array

• assigning the resultant value to an input field

• posting that field to the second page

• using json_decode on the posted value

• then var_dump to test

• (echo'ing the posted variable directly just to see if it came through at all)

Javascript on page one:

var JSONstr = JSON.stringify(fullInfoArray);
document.getElementById('JSONfullInfoArray').value= JSONstr;

php on page two:

$data = json_decode($_POST["JSONfullInfoArray"]);
var_dump($data);

echo($_POST["JSONfullInfoArray"]);

The echo works fine but the var_dump returns NULL

What have I done wrong?

---

This got me fixed up:

$postedData = $_POST["JSONfullInfoArray"];
$tempData = str_replace("\\", "",$postedData);
$cleanData = json_decode($tempData);
var_dump($cleanData);

Im not sure why but the post was coming through with a bunch of "\\" characters separating each variable in the string

Figured it out using json_last_error() as sugested by Bart which returned JSON_ERROR_SYNTAX

Answer 1

When you save some data using JSON.stringify() and then need to read that in php. The following code worked for me.

json_decode( html_entity_decode( stripslashes ($jsonString ) ) );

Thanks to @Thisguyhastwothumbs

Answer 2

When you use JSON stringify then use html_entity_decode first before json_decode.

$tempData = html_entity_decode($tempData);
$cleanData = json_decode($tempData);

Answer 3

You'll need to check the contents of $_POST["JSONfullInfoArray"] . If something doesn't parse json_decode will just return null . This isn't very helpful so when null is returned you should check json_last_error() to get more info on what went wrong.

Answer 4

None of the other answers worked in my case, most likely because the JSON array contained special characters. What fixed it for me:

Javascript (added encodeURIComponent)

var JSONstr = encodeURIComponent(JSON.stringify(fullInfoArray));
document.getElementById('JSONfullInfoArray').value = JSONstr;

PHP (unchanged from the question)

$data = json_decode($_POST["JSONfullInfoArray"]);
var_dump($data);

echo($_POST["JSONfullInfoArray"]);

Both echo and var_dump have been verified to work fine on a sample of more than 2000 user-entered datasets that included a URL field and a long text field, and that were returning NULL on var_dump for a subset that included URLs with the characters ?&# .

Answer 5

stripslashes(htmlspecialchars(JSON_DATA))

Answer 6

jsonText = $_REQUEST['myJSON'];

$decodedText = html_entity_decode($jsonText);

$myArray = json_decode($decodedText, true);`

Answer 7

我不知道这是如何工作的，但它确实有效。

$post_data = json_decode(json_encode($_POST['request_key']));