For example i have json strings like this ( from the first place ). And It's not formatted.
{"data":[{"id":"14","memo_kondisi":"Kekurangan
pekerjaan","total_row":"5","nilai_temuan":"1.000.000","data_sebab":[{"id":"15","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[{"id":"25","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"},{"id":"26","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"},{"id":"31","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"}]},{"id":"16","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[{"id":"34","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"},{"id":"35","id_rekomendasi":"10","id_sub_rekomendasi":"","id_s_sub_rekomendasi":"","nilai_rekomendasi":"0"}]}]},{"id":"15","memo_kondisi":"Kekurangan
pekerjaan","total_row":"2","nilai_temuan":"1.000.000","data_sebab":[{"id":"5","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]},{"id":"10","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]}]},{"id":"16","memo_kondisi":"","total_row":"2","nilai_temuan":"0","data_sebab":[{"id":"9","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]},{"id":"12","id_sebab":"","id_sub_sebab":"","memo_sebab":"coba","data_rekomendasi":[]}]}]}
I see some similar question that you have to use json_decode and i have to encode again and using json_encode($json,JSON_PRETTY_PRINT)
Is there a way for make json readable without decode the JSON first and encode it again in PHP ?
Note : I expect the result is still in JSON
Not really. Using someone else's parser lib won't make any difference, as they'll call json_decode() too.
You could create a little function that you could call:
function prettify($json)
{
$array = json_decode($json, true);
$json = json_encode($array, JSON_PRETTY_PRINT);
return $json;
}
Then echo prettify($jsonString); would be easier than constantly decoding and re-encoding. See here https://3v4l.org/CcJlf
Only a parser can understand the JSON, so you can either do what you proposed or write your own parser. If you have access to the origin of the JSON, make it pretty in the first place.
Related
I need to use a customer's API for loading JSON which only contains something like:
{"html" : "foo"}
The API is being used from other services so I'm pretty sure it's valid.
However, when trying to decode it using json_decode i'm always getting an empty string which means it's not valid. I found out i need to "fix" the JSON-String by replacing:
$json = str_replace("\\>", "\\\\>", $json); // \> = invalid json
It works mainly on each request but not on certain others but it's very tricky to debug and i can't imagine that replacing is the proper method.
How would i do it the easy way for converting the json string into a valid one?
thanks
Ok i could find out what's wrong:
The HTML contains backslashes in the closing tags, for example <br\>
You need to replace them like this:
$json = str_replace("\\>", "\\\\>", $json);
and json_decode will work
I am trying to decode json with PHP but dont know where am i wrong. Here is my code
$rr ='var modelsGlobal = [{"value":"FAFW3801LW","productdetailurl":"/Washers-Dryers/Washers/Front-Load/FAFW3801LW/"}{"value":"FAFW3801LW","productdetailurl":"/Washers-Dryers/Washers/Front-Load/FAFW3801LW/"}]';
$json = json_decode($rr, true);
foreach($json['modelsGlobal'] as $json){
$prod_id = $json["value"];
}
Please help
You are trying to decode (broken) JavaScript, not JSON.
JSON wouldn't include var modelsGlobal = and array members need a , between them.
Run your data through a linter.
After you fix the errors which are preventing the parsing, the JSON doesn't start with an object with a modelsGlobal, so loop over the array in $json directly.
Your JSON is incorrect. It is not JSON but JavaScript and it lacks a comma to separate the two objects of the array.
If you use PHP 5.3+ use json_last_error to check errors with json_encode/json_decode.
I have the following json output - I've tried multiple things however not having any luck - I just want to get the value for kind (software-package). Any ideas:
{"items":
[{"assets":
[{"kind":"software-package","url":"__URL__"}],
"metadata":{"bundle-identifier":"SimpleCalculator",
"bundle-version":"000","kind":"software",
"title":"com.work.demo","subtitle":"1.0"}
Thanks,
Try to decode it with json_encode like
$result_arr = json_decode($my_arr,true);
print_r($result_arr['items']['assets']['kind']);
That is not valid JSON.
Lets attempts to format that a little more sensibly.
{"items": [
{"assets": [
{"kind":"software-package","url":"__URL__"}
],
"metadata":{"bundle-identifier":"SimpleCalculator",
"bundle-version":"000","kind":"software",
"title":"com.work.demo","subtitle":"1.0"
}
You're missing a lot of closing brackets. :(
How are you acquiring this JSON. Is it hand formatted or generated by some software? They're may be bugs in the source, and not your code.
If you are using a version of PHP above 5.3 then there are built-in functions to decode and encode JSON strings already.
$some_json_string = {......};
$json_string_as_object = json_decode($some_json_string);
$json_string_as_array = json_decode($some_json_string, true);
You can also do the reverse:
$some_array = array(...);
$json_string = json_encode($some_array);
Json Decode : http://php.net/manual/en/function.json-decode.php
Json Encode: http://www.php.net/manual/en/function.json-encode.php
I have the following json
country_code({"latitude":"45.9390","longitude":"24.9811","zoom":6,"address":{"city":"-","country":"Romania","country_code":"RO","region":"-"}})
and i want just the country_code, how do i parse it?
I have this code
<?php
$json = "http://api.wipmania.com/jsonp?callback=jsonpCallback";
$jsonfile = file_get_contents($json);
var_dump(json_decode($jsonfile));
?>
and it returns NULL, why?
Thanks.
<?php
$jsonurl = "http://api.wipmania.com/json";
$json = file_get_contents($jsonurl);
var_dump(json_decode($json));
?>
You just need json not jsonp.
You can also try using json_decode($json, true) if you want to return the array.
you're requesting jsonp with http://api.wipmania.com/jsonp?callback=jsonpCallback, which returns a function containing JSON like:
jsonpCallback({"latitude":"44.9718","longitude":"-113.3405","zoom":3,"address":{"city":"-","country":"United States","country_code":"US","region":"-"}})
and not JSON itself. change your URL to http://api.wipmania.com/json to return pure JSON like:
{"latitude":"44.9718","longitude":"-113.3405","zoom":3,"address":{"city":"-","country":"United States","country_code":"US","region":"-"}}
notice the second chunk of code doesn't wrap the json in the jsonpCallback() function.
The website doesn't return pure JSON, but wrapped JSON. This is meant to be included as a script and will call a callback function. If you want to use it, you first need to remove the function call (the part until the first paranthesis and the paranthesis at the end).
If your server implements JSONP, it will assume the callback parameter to be a JSONP signal and the result will be similar to a JavaScript function, like
jsonpCallback("{yada: 'yada yada'}")
And then, json_decode won't be able to parse jsonpCallback("{yada: 'yada yada'}") as a valid JSON string
If country_code( along with closing parenthesis are include in your json, remove them.
This is not a valid json syntax: json
You are being returned JSONP, not JSON. JSONP is for cross-domain-requests in JavaScript. You don't need to use it when using PHP because you aren't affected by cross-domain-policies.
Since you are getting a string from the file_get_contents() function you can do a replacement of the country_code( text (this is the JSONP specific part of the response):
<?php
$json = "http://api.wipmania.com/jsonp?callback=jsonpCallback";
$jsonfile = substr(file_get_contents($json)), 13, -1);
var_dump(json_decode($jsonfile));
?>
Note
This works but JKirchartz's solution looks better, just request the correct data rather than messing around with the incorrect data.
Obviously in this situation, using the correct URL to access the API will return pure jSON.
"http://api.wipmania.com/json"
A lot of people are providing an alternative to the API in use, rather than answering the OP's question, so here is a solution for those looking for a way of handling jSONp in PHP.
First, the API allows you to specify a callback method, so you can either use Jasper's method of getting the jSON sub string, or you can give a callback method of json_decode, and modify the result to use with a call to eval. This is my alternative to Jasper's code example since I don't like to be a copy cat:
$json = "http://api.wipmania.com/jsonp?callback=json_decode";
$jsonfile eval(str_replace("(", "('", str_replace(")", "')", file_get_contents($json)))));
var_dump($jsonfile);
Admittedly this seems a little longer, more insecure, and not as clear to read as Jasper's code:
$json = "http://api.wipmania.com/jsonp?callback=jsonpCallback";
$jsonfile = substr(file_get_contents($json)), 13, -1);
var_dump(json_decode($jsonfile));
Then the jSON "address":{"city":"-","country":"Romania","country_code":"RO","region":"-"} tells us to access the country_code like so:
$jsonfile->{'address'}->{'country_code'};
I am running a Debian box with PHP v5.2.17. I am trying to get around the cross-domain issue with an XML file and am using this got to fetch any xml and return json:
<?php
header('content-type: application/json; charset=utf-8');
if( strlen($_GET["feed"]) >= 13 ) {
$xml = file_get_contents(urldecode($_GET["feed"]));
if($xml) {
$data = #simplexml_load_string($xml, "SimpleXMLElement", LIBXML_NOCDATA);
$json = json_encode($data);
echo isset($_GET["callback"]) ? "{$_GET[’callback’]}($json)" : $json;
}
}
?>
The problem is, its not returning valid json to jquery.. The start character is "(" and the end is ")" where jquery wants "[" as the start and "]" as the end. I've taken the output and used several online validation tools to check it..
Is there a way I can change these characters prior to sending back or pass json_encode options?
You could change json_encode($data) to json_encode(array($data)) if it expects an array (like you're saying):
$json = json_encode(array($data));
EDIT: Also, I believe the SimpleXml call will result in a bunch of SimpleXmlElements, perhaps json_encode then thinks it should be objects, instead of arrays? Perhaps casting to an array will yield the correct results.
You cannot json_encode() SimpleXMLElements (that's the type that is returned by simplexml_load_string(). You have to convert the data from the XML file into some native PHP type (most likely an array).
SORRY that's wrong. json_encode() can in fact encode SimpleXMLElements (at least on my PHP version 5.3.4). So if your client-side code expects an array you must wrap your $data in an array:
$json = json_encode(array($data));
We can use json_encode() function most probably on array. so you first take XML content into PHP array and then apply json_encode().I think this will solve your problem..
It seems that you are sending an empty callback parameter or something, but the callback parameter in jQuery must look exactly like this: callback=?