Python return json to PHP and use json_decode

Question

According to this mine question Python find value in array by filter

This is what I have done

data = {}

for result in results:
    if 'stackoverflow.com' in result['serp_url']:
        data['url'] = result['serp_url']
        data['rank'] = result['serp_rank']
        data['query'] = result['query']
        print(data)
        exit

This is PHP code

$test = shell_exec("python3 py.py");
var_dump($test);

And this is output

/home/user/Desktop/pyphp/index.php:4:string '{'url': 'https://stackoverflow.com/', 'rank': 1, 'query': 'stackoverflow'}
{'url': 'https://www.quantcast.com/stackoverflow.com', 'rank': 36, 'query': 'stackoverflow'}
' (length=168)

When I use json_decode($test) I'm getting null as output.

What's the best way to use as json or array output from Python in PHP?

This is output `{'url': 'https://stackoverflow.com/', 'rank': 1, 'query': 'stackoverflow'} {'url': 'https://www.quantcast.com/stackoverflow.com', 'rank': 36, 'query': 'stackoverflow'}` as one string — Nicolas Marek, Sep 27 '18 at 10:45
That's not valid json, that's your problem. You should wrap it into an array. — Federico klez Culloca, Sep 27 '18 at 10:45
so this is not a correct JSON syntax... you got like 2 objects in one... `{'url': 'https://stackoverflow.com/', 'rank': 1, 'query': 'stackoverflow'} ` and `{'url': 'https://www.quantcast.com/stackoverflow.com', 'rank': 36, 'query': 'stackoverflow'}` — Flash Thunder, Sep 27 '18 at 10:46
From Python side? How to do that? I tried json.dumps(data) but still not working — Nicolas Marek, Sep 27 '18 at 10:46
seems that you got some other `print()` in the script, as the part you gave seems to be valid — Flash Thunder, Sep 27 '18 at 10:47
i think use `exec()` is better than `shell_exec` try this `exec('YOUR_COMMAND',$output)` and at last `print_r(output)` — nima, Sep 27 '18 at 10:47
@dogano it won't help :) `json_decode()` fails, doesn't really matter if it returns object or array, as it fails — Flash Thunder, Sep 27 '18 at 10:48
@NicolasMarek you can create json string as per this [link](https://stackoverflow.com/questions/23110383/how-to-dynamically-build-a-json-object-with-python) — Jigar Shah, Sep 27 '18 at 10:49
@FlashThunder null means that your JSON string is invalid. Have you validated the syntax? — Doğan Uçar, Sep 27 '18 at 10:50
I would concentrate on what python returns... it seems that after this function it return some other object — Flash Thunder, Sep 27 '18 at 10:50
@dogano but `(...,true)` param only makes it an array from stdClass — Flash Thunder, Sep 27 '18 at 10:51
Other reason its Invalid JSON. JSON specifies that you must use DoubleQuotes and not single quotes — RiggsFolly, Sep 27 '18 at 10:54
@RiggsFolly you must also ensure that your string is UTF8 encoded — Doğan Uçar, Sep 27 '18 at 10:56

score 1 · Answer 1 · answered Sep 27 '18 at 11:08

Thank you all for comments! According to that, I kind figure out solution.

Python script

data = {}
for result in results:
    if 'stackoverflow.com' in result['serp_url']:
        data['url'] = result['serp_url']
        data['rank'] = result['serp_rank']
        data['query'] = result['query']
        print(json.dumps(data))
        exit

PHP script

exec("python3 py.py", $output);

$test = [];

foreach($output as $key => $out) {
    $test[$key] = json_decode($out, true);
    print_r("Rank: " . $test[$key]['rank'] ." - ". $test[$key]['url']."<br>");
}

Output

Rank: 1 - https://stackoverflow.com/
Rank: 36 - https://www.quantcast.com/stackoverflow.com

Shailendra2014 · Answer 2 · 2018-09-27T11:11:33.363

0

You don't even have to even use Json package of Python. Here is little tweek. In Python and PHP code.

# Your py.py Looks Below: 
data = {}
data["aa"] = 100
# You can create any Valid Dictionary
print data

and Your Php File will have this code

<?php
    $test = shell_exec("python py.py");
    echo json_decode(json_encode($test));
    // $test itself is a json String you even dont need encode/decode
?>

And The Result in my Terminal got {'aa': 100} Which is Expected and you want.

Now Important Point is that shell_exec command will give you the string which is converted from Dictionary and fortunately, That string itself is a JSON, That why json_decode is returning NULL.

Above both Snippet works fine. Check them.

edited Sep 27 '18 at 11:11

answered Sep 27 '18 at 11:06

Shailendra2014

789
1
7
23

Thanks a lot! This will be useful as well! – Nicolas Marek Sep 27 '18 at 11:12
Great I hope it should solve your problem. Can you accept as the answer? – Shailendra2014 Sep 27 '18 at 11:13

score 0 · Answer 3 · answered Dec 25 '22 at 11:04

Python file:

import json
import base64

j_data = {
"name": "wick",
"age": 24,
}

jso_en = json.dumps(j_data , indent=4)
#encode with base64
res_arr_bytes = jso_en.encode("ascii")  
base64_bytes = base64.b64encode(res_arr_bytes)
base64_enc = base64_bytes.decode("ascii")
print(base64_enc)

PHP file:

$command = escapeshellcmd("python3 test.py");
$json_out = shell_exec($command);
$json_data = json_decode(base64_decode($json_out), true);
echo $json_data['name'];

output from php:

wick

Federico klez Culloca · Answer 4 · 2018-09-27T10:57:33.010

-1

The json you're generating is not valid (free objects outside of an array and strings between single quotes).

In your python code you should append your data objects to an array as you're creating them. Here's a rough attempt at solving your issue

import json

data = [] # use a list to keep your objects

for result in results:
    if 'stackoverflow.com' in result['serp_url']:
        record = {}
        record['url'] = result['serp_url']
        record['rank'] = result['serp_rank']
        record['query'] = result['query']
        data.append(record)
        # exit # <-- this doesn't seem correct, TBH
json.dumps(data)

edited Sep 27 '18 at 10:57

answered Sep 27 '18 at 10:51

Federico klez Culloca

26,308
17
56
95

In my opinion it's not a reason why it returns 2 objects at once... ofc objects won't aggregate, but still the problem is somewhere else. – Flash Thunder Sep 27 '18 at 10:54
@FlashThunder IMHO, the problem is that what OP posted is not what they're actually using to get that output (that's why I commented out the `exit`). I'll keep my answer here anyway just in case and I will understand and accept any downvote that comes my way :) – Federico klez Culloca Sep 27 '18 at 10:55
You may be right, that's why I only posted a comment. – Flash Thunder Sep 27 '18 at 10:57
Thanks for this, but didn't work, since I got null when executed script. – Nicolas Marek Sep 27 '18 at 11:06

Python return json to PHP and use json_decode

4 Answers4

Linked