How can I decode in PHP a json string that has a json string as a value inside?

Question

I need to decode a JSON string in PHP but the thing is that the string has a JSON encoded string as one of the value. Something like that :

{
    "key1": "value1",
    "key2": "{\"key3\":\"{\\\"key4\\\":\\\"value4\\\"}\"}"
}

I'm trying to use json_decode to parse it into an array but it always fail with the error that the syntax is incorrect. If I put it in any validator you can find online, it says the string is valid. Should I use another method to parse the string?

PS: I am fine with the fact that once decoded, the value of key2 won't be an array but instead a string that I would have to decode again.

Thanks!

How did you get this JSON value “into” your script? If you wrote this as string in code, then you need to pay attention to the backslashes … Start by making a debug output of the variable you are passing to json_encode, and see if you still got as many backslashes as you thought you did … — 04FS
– 04FS, Commented Feb 11, 2019 at 13:38
this is a hornet's nest. Post the code where this (valid but unwieldy) json is produced, you might have more help available here. — YvesLeBorg
– YvesLeBorg, Commented Feb 11, 2019 at 13:40
I got this from a server to server call from an external endpoint I don't control unfortunately. The other end gives me a big json which at some point contains some sort of receipt that it is json encoded and put into a string. The rest of the JSON is a usual JSON structure but one of the value is the json encoded receipt put as a string value. I have to decode everything before processing it properly. — Shex
– Shex, Commented Feb 11, 2019 at 13:42
What you have posted is not valid JSON. It would be valid JSON if all the PHP was removed and the content between ' and ' was left as it is. That way the \ characters would be treated as JSON escape characters and not PHP escape characters. Likely you introduced a new problem in constructing your reduced test case. — Quentin
– Quentin, Commented Feb 11, 2019 at 13:42

mitkosoft · Accepted Answer · 2019-02-11 13:48:40Z

2

Based on provided input - you must replace all those nasty backslashes and double quotes in order to get proper JSON:

<?php
$s = '{
    "key1": "value1",
    "key2": "{\"key3\":\"{\\\"key4\\\":\\\"value4\\\"}\"}"
}';

$s = str_replace('\\', '', $s);
$s = str_replace('"{', '{', $s);
$s = str_replace('}"', '}', $s);

print_r(json_decode($s, true));
?>

Output:

Array
(
[key1] => value1
[key2] => Array
    (
        [key3] => Array
            (
                [key4] => value4
            )
    )
)

answered Feb 11, 2019 at 13:48

mitkosoft

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Zeljka · Accepted Answer · 2019-02-11 13:48:04Z

0

Your json is bad, you have unnecessary " around {}

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$a = '{
    "key1": "value1",
    "key2": {"key3":{"key4":"value4"}}
}';

var_dump(json_decode($a));

answered Feb 11, 2019 at 13:48

Zeljka

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Comments

programmanip · Accepted Answer · 2019-02-11 14:42:44Z

0

You can prepare your string with this line of code:

    $json_string.= '{"id":"'.$variable.'","name":"'.preg_replace("~[\"]~","\\\\\"",$variable_name).'","other_name":"'.preg_replace("~[\"]~","",$other_name_variable).'"},';

You have to convert string to JSON array then. Don't forget about right quotes and string format. More about JSON: https://www.json.org/ -here you can find explanations and examples More for preg_replace function: http://php.net/manual/en/function.preg-replace.php

answered Feb 11, 2019 at 14:42

programmanip

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Comments

sami.ghaffari · Accepted Answer · 2019-02-11 14:51:13Z

-1

you can use json_decode() twice or more

$a = {
   "key1": "value1",
   "key2": "{\"key3\":\"{\\\"key4\\\":\\\"value4\\\"}\"}"
 }
 $b = json_decode($a);
 $c = json_decode($b->key2);

It's give you a object with key3. and again you can use json_decode().

It's the correct string for your structure if It's diffrence from it , it may be wrong .

{"key1":"value1","key2":"{\"key3\":\"{\\\"key4\\\":\\\"value4\\\"}\"}"}

NOTE : I think I should write a edit on this . when you use \ in string it put \ in your string if you want copy exactly this you should replace them . if you use json_last_error() you give syntax error in this string .

NOTE : If you do like the above answer from @mitkosoft , you change your structure because of two problem . First , you change your object to to array . Second , if you decode in first level you should get your "key2" value a string not an array . It's a json string that you can do it decode again .

edited Feb 11, 2019 at 14:51

answered Feb 11, 2019 at 13:44

sami.ghaffari

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5 Comments

04FS Over a year ago

$a = { ... } is not even JSON to begin with.

Shex Over a year ago

That's where my problem is. I tried json_decode but it returns a json error saying that the synthax is invalid. I can't do json_decode($a) in your example

sami.ghaffari Over a year ago

{"key1":"value1","key2":"{\"key3\":\"{\\\"key4\\\":\\\"value4\\\"}\"}"} it's right json that created by json_encode in php .

YvesLeBorg Over a year ago

that's going nowhere : $b is null and the next line will fail.

sami.ghaffari Over a year ago

Yes you don't copy this exactly because in php when you use \\ it put \ in string and if use json_last_eroor() you get wrong syntax . so you should replace them if you want copy this

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