json_decode generates string instead of object

Question

I need to get a JSON Object from Database, convert it to php object and foreach the key=>value pairs.

The JSON Object is stored with JSON.stringify(obj) in MySQL text field. Afterwards i get the data with PDO, but i have same unterstanding problem how to decode the json object to an object in php. The Result is everytime a string.

$export = $data['document']->tax_rates;
$export = json_encode($export);
$export = json_decode($export);
var_dump($export);
echo($export)

// only test 
var_dump(json_decode($data['document']->tax_rates));
var_dump($data['document']->tax_rates)

$jsontest = json_decode('{"a":1,"b":2}');
var_dump($jsontest);

The result is:

string(42) "{"7":39729.69,"19":107.73}" // var_dump
{"7":39729.69,"19":107.73}  // echo

NULL // result of decode without encode before
string(42) "{"7":39729.69,"19":107.73}"  // var_dump raw data['document']

//test result as it should be
object(stdClass)#21 (2) {
  ["a"]=>
  int(1)
  ["b"]=>
  int(2)
}

Debugging the Json output the error: JSON_ERROR_SYNTAX. But is is a valid JSON Object.

How can i genereate an php object here and where is my issue in this case? If i understand correct, the output is a string because since it is stored in the text field, it is already a string. If it is so, is it possible to handle it this way or is it impossible? Thanks.

Well, you only need to encode (and decode) once. Data is already encoded, yet the first thing you do when you retrieve it is to encode it again. — Álvaro González, Jun 20 '20 at 13:02
If i only encode the result of var_dump ist still string: string(44) ""{"7":39729.69,"19":107.73}"" — Marco, Jun 20 '20 at 13:06
If i remove the line $export = json_encode($export); so the result is NULL — Marco, Jun 20 '20 at 13:09
I removed my comment because, honestly, I don't know what you're trying to illustrate with your test case so I can't give much advice. `{"7":39729.69,"19":107.73}` can't possibly have 42 bytes, not even in UTF-16, so it can't be the actual raw data. — Álvaro González, Jun 20 '20 at 13:15
The mysql field is utf8_general_ci and the pdo also fetch in UTF-8. The value in Database looks like {"7":39729.69,"19":107.73} — Marco, Jun 20 '20 at 13:21
There you are: that isn't valid JSON. Trying pasting `[{"7":3.5,"19":19,"20":20}]` that in [jsonlint.com/](https://jsonlint.com/) — Álvaro González, Jun 20 '20 at 13:22
Okay, so my problem is storing the json object with JSON.stringify(obj) in mysql. I already searched a solution why the quation marks stored as unicode but found no solution yet. Thank you, i have to check it again! — Marco, Jun 20 '20 at 13:25
Are you trying to implement a home-brew solution for SQL injection? — Álvaro González, Jun 20 '20 at 13:26
I only try to save the the json to database .... this works now, thanks! $html = $data['document']->tax_rates; $json = html_entity_decode($html); $json = json_decode($json); var_dump($json); — Marco, Jun 20 '20 at 13:31
Just use prepared statements as explained at [How can I prevent SQL injection in PHP?](https://stackoverflow.com/questions/60174/how-can-i-prevent-sql-injection-in-php). SQL doesn't have anything to do with HTML! — Álvaro González, Jun 20 '20 at 13:34
Well, there've never been any builtin feature in PHP to encode input as HTML automagically so you must be doing it yourself, somewhere :) — Álvaro González, Jun 20 '20 at 13:37

score 2 · Accepted Answer · answered Jun 20 '20 at 13:25

What you have is not JSON but HTML-encoded JSON. You can recover the original data with html_entity_decode():

$html = '[{&#34;7&#34;:3.5,&#34;19&#34;:19,&#34;20&#34;:20}]';
$json = html_entity_decode($html);
echo $json;

But it'd be better to identify and fix the bug that's causing invalid data form being stored in the first place.

json_decode generates string instead of object

1 Answers1