For example I have php array as:
$arr = ["a"=>[],"b"=>[]];
for which I need json object as:
{"a":[],"b":{}}
I have huge array for which json_encode function is applied. If I put option as json_encode($data,JSON_FORCE_OBJECT) then it will make every [] object to {} object which is highly undesirable.
Can I apply encode option specific to array keys? Because one should be able to make {"a":[],"b":{}} from php array which is valid json.
you can use anonymous classes for this, like(DEMO):
$arr = ["a"=>[],"b"=>new class{}];
This way you can collectively turn some keys into objects and others into arrays.
You can also use the typecast (object), like(DEMO):
$arr = ["a"=>[],"b"=>(object)[]];
I would personally prefer the typecasting method.
Related
To create an empty JSON object I do usually use:
json_encode((object) null);
casting null to an object works, but is there any other preferable way and/or any problem with this solution?
Recommended method
json_decode ("{}") will return a stdClass per default, using the below should therefor be considered safe/portable and correct.
json_encode (new stdClass);
Your solution could work..
The documentation specifies that (object) null will result in an empty object, some might therefor say that your code is valid and that it's the method to use.
PHP: Objects - Manual
If a value of any other type is converted to an object, a new instance of the stdClass built-in class is created. If the value was NULL, the new instance will be empty.
.. but, try to keep it safe!
Though you never know when/if the above will change, so if you'd like to be 100% certain that you will always will end up with a {} in your encoded data you could use a hack such as:
$empty = json_decode ("{}");
$result = json_encode($empty); // "{}"
Even though it's tedious and ugly I do assume/hope that json_encode/json_decode is compatible with one another and always will evaluate the following to true:
$a = <something>;
$a === json_decode (json_encode ($a));
If you use objects as dynamic dictionaries (and I guess you do), then I think you want to use an ArrayObject.
It maps into JSON dictionary even when it's empty. It is great if you need to distinguish between lists (arrays) and dictionaries (associative arrays):
$complex = array('list' => array(), 'dict' => new ArrayObject());
print json_encode($complex); // -> {"list":[],"dict":{}}
You can also manipulate it seamlessly (as you would do with an associative array), and it will keep rendering properly into a dictionary:
$complex['dict']['a'] = 123;
print json_encode($complex); // -> {"list":[],"dict":{"a":123}}
unset($complex['dict']['a']);
print json_encode($complex); // -> {"list":[],"dict":{}}
If you need this to be 100% compatible both ways, you can also wrap json_decode so that it returns ArrayObjects instead of stdClass objects (you'll need to walk the result tree and recursively replace all the objects, which is a fairly easy task).
Gotchas. Only one I've found so far: is_array(new ArrayObject()) evaluates to false. You need to find and replace is_array occurrences with is_iterable.
Well, json_encode() simply returns a string from a PHP array/object/etc. You can achieve the same effect much more efficiently by doing:
$json = '{}';
There's really no point in using a function to accomplish this.
UPDATE
As per your comment updates, you could try:
$test = json_encode(array('some_properties'=>new stdClass));
Though I'm not sure that's any better than what you've been doing.
json_encode($array, JSON_FORCE_OBJECT) will do it too. see https://www.php.net/manual/en/function.json-encode.php
When doing a json_encode a multidimensional array in PHP, I'm noticing a different output simply by naming one of the arrays, as opposed to not naming them. For Example:
$arrytest = array(array('a'=>1, 'b'=>2),array('c'=>3),array('d'=>4));
json_encode($arrytest)
gives a single array of multiple json objects
[{"a":1,"b":2},{"c":3},{"d":4}];
whereas simply assigning a name to the middle array
$arrytest = array(array('a'=>1, 'b'=>2),"secondarray"=>array('c'=>3),array('d'=>4));
json_encode($arrytest)
creates a single json object with multiple json objects inside
{"0":{"a":1,"b":2},"secondarray":{"c":3},"1":{"d":4}};
why would the 1st option not return the same reasults as the 2nd execpt with "1" in place of "secondarray"
In JSON, arrays [] only every have numeric keys, whereas objects {} have string properties. The inclusion of a array key in your second example forces the entire outer structure to be an object by necessity. The inner objects of both examples are made as objects because of the inclusion of string keys a,b,c,d.
If you were to use the JSON_FORCE_OBJECT option on the first example, you should get back a similar structure to the second, with the outer structure an object rather than an array. Without specifying that you want it as an object, the absence of string keys in the outer array causes PHP to assume it is to be encoded as the equivalent array structure in JSON.
$arrytest = array(array('a'=>1, 'b'=>2),array('c'=>3),array('d'=>4));
// Force the outer structure into an object rather than array
echo json_encode($arrytest , JSON_FORCE_OBJECT);
// {"0":{"a":1,"b":2},"1":{"c":3},"2":{"d":4}}
Arrays with continuous numerical keys are encoded as JSON arrays. That's just how it is. Why? Because it makes sense.
Since the keys can be expressed implicitly through the array encoding, there is no reason to explicitly encoded them as object keys.
See all the examples in the json_encode documentation.
At the first option you only have numeric indexes (0, 1 and 2). Although they are not explicitly declared, php automatically creates them.
At the second option, you declare one of the indexes as an string and this makes PHP internally transform all indexes to string.
When you json encode the first array, it's not necessary to show the integers in the generated json string because any decoder should be able to "guess" that they are 0, 1 and 2.
But in the second array, this is necessary, as the decoder must know the key value in your array.
It's pretty simple. No indexes declared in array? Them they are 0, 1, 2, 3 and so on.
output of this as in json form is year1{a,b},year2{c}, year3{d}
**a has value 1 ,b=2,c=3,d=4 stored in array of year1's a,b years2's c and years3's d respectivily
$array1 = array('a'=>1, 'b'=>2);
$array2 = array('c'=>3);
$array3 = array('d'=>4)
$form = array("year1" =>$array1,
"year2" =>$array2,
"year3" =>$array3,
);
$data = json_encode($form);
To create an empty JSON object I do usually use:
json_encode((object) null);
casting null to an object works, but is there any other preferable way and/or any problem with this solution?
Recommended method
json_decode ("{}") will return a stdClass per default, using the below should therefor be considered safe/portable and correct.
json_encode (new stdClass);
Your solution could work..
The documentation specifies that (object) null will result in an empty object, some might therefor say that your code is valid and that it's the method to use.
PHP: Objects - Manual
If a value of any other type is converted to an object, a new instance of the stdClass built-in class is created. If the value was NULL, the new instance will be empty.
.. but, try to keep it safe!
Though you never know when/if the above will change, so if you'd like to be 100% certain that you will always will end up with a {} in your encoded data you could use a hack such as:
$empty = json_decode ("{}");
$result = json_encode($empty); // "{}"
Even though it's tedious and ugly I do assume/hope that json_encode/json_decode is compatible with one another and always will evaluate the following to true:
$a = <something>;
$a === json_decode (json_encode ($a));
If you use objects as dynamic dictionaries (and I guess you do), then I think you want to use an ArrayObject.
It maps into JSON dictionary even when it's empty. It is great if you need to distinguish between lists (arrays) and dictionaries (associative arrays):
$complex = array('list' => array(), 'dict' => new ArrayObject());
print json_encode($complex); // -> {"list":[],"dict":{}}
You can also manipulate it seamlessly (as you would do with an associative array), and it will keep rendering properly into a dictionary:
$complex['dict']['a'] = 123;
print json_encode($complex); // -> {"list":[],"dict":{"a":123}}
unset($complex['dict']['a']);
print json_encode($complex); // -> {"list":[],"dict":{}}
If you need this to be 100% compatible both ways, you can also wrap json_decode so that it returns ArrayObjects instead of stdClass objects (you'll need to walk the result tree and recursively replace all the objects, which is a fairly easy task).
Gotchas. Only one I've found so far: is_array(new ArrayObject()) evaluates to false. You need to find and replace is_array occurrences with is_iterable.
Well, json_encode() simply returns a string from a PHP array/object/etc. You can achieve the same effect much more efficiently by doing:
$json = '{}';
There's really no point in using a function to accomplish this.
UPDATE
As per your comment updates, you could try:
$test = json_encode(array('some_properties'=>new stdClass));
Though I'm not sure that's any better than what you've been doing.
json_encode($array, JSON_FORCE_OBJECT) will do it too. see https://www.php.net/manual/en/function.json-encode.php
I don't understand... I'm doing something and when I do print_r($var); it tells me that I have an array, so naturally I think I have an array yet when I do
if(is_array($xml->searchResult->item))
it returns false
I use this array with foreach(); in documentation it says that foreach() won't work with anything else but array, so assuming that this is an array that I'm working...
plus, if I try to access it via
echo $xml->searchResult->item[3];
i will get 4th element of my array
print_r will also print objects as though they are arrays.
well, is_array() returns true if your variable is an array, otherwise, it returns false. In your case, $xml->searchResult->item seems not to be an array. What is the output for
var_dump($xml->searchResult->item)
? Another hint: You can determine the type of a variable via gettype().
is_array() returns true only for real php arrays. It is possible to create a "fake" array by using the ArrayAccess class. That is, you can use normal array semantics (such as item[3]) but it is not a real array. I suspect your $item is an object. So use
if($x instanceof ArrayAccess || is_array($x))
Instead.
plus, if I try to access it via echo $xml->searchResult->item[3]; i will get 4th element of my array
That's right, the first element is always 0 unless you specifically change it.
The manual does mention that foreach works on objects as well, and it will iterate over properties.
In your case, the situation is slightly different because I guess you're using SimpleXML, which is yet another special case. SimpleXMLElement has its own iterator, which I assume is hardcoded as it doesn't seem to implement any of SPL's Iterator interfaces.
Long story short, some objects can be used as an array, but they are not one.
Just be careful to check what has been returned by your array. You might have an object with class or stdClass which is empty and you cannot get element from it.
i have an simple array:
array
0 => string 'Kum' (length=3)
1 => string 'Kumpel' (length=6)
when I encode the array using json_encode(), i get following:
["Kum","Kumpel"]
My question is, what is the reason to get ["Kum","Kumpel"] instead of { "0" : "Kum", "1" : "Kumpel" }?
"{}" brackets specify an object and "[]" are used for arrays according to JSON specification. Arrays don't have enumeration, if you look at it from memory allocation perspective. It's just data followed by more data, objects from other hand have properties with names and the data is assigned to the properties, therefore to encode such object you must also pass the correct property names. But for array you don't need to specify the indexes, because they always will be 0..n, where n is the length of the array - 1, the only thing that matters is the order of data.
$array = array("a","b","c");
json_encode($array); // ["a","b","c"]
json_encode($array, JSON_FORCE_OBJECT); // {"0":"a", "1":"b","2":"c"}
The reason why JSON_FORCE_OBJECT foces it to use "0,1,2" is because to assign data to obeject you must assign it to a property, since no property names are given by developer (only the data) the encoder uses array indexes as property names, because those are the only names which would make sense.
Note: according to PHP manual the options parameters are only available from PHP 5.3.
For older PHP versions refer to chelmertz's answer for a way to make json_encode to use indexes.
As Gumbo said, on the JS-side it won't matter. To force PHP into it, try this:
$a = new stdClass();
$a->{0} = "Kum";
$a->{1} = "Kumpel";
echo json_encode($a);
Not that usable, I'd stick with the array notation.
Just cast as an object and it will work fine...the JSON_FORCE_OBJECT parameter does exactly the same thing.
json_encode((object)$array);
Don't forget to convert it back into a php array so you can access its values in php:
$array = (object)$array;
$array = (array)$array;
json_encode($array);
Since you’re having a PHP array with just numeric keys, there is no need to use a JavaScript object. But if you need one, try Maiku Mori’s suggestion.
I personally think this is a bug that needs to be fixed in PHP. JSON_FORCE_OBJECT is absolutely not an answer. If you try to do any sort of generic programming you get tripped up constantly. For example, the following is valid PHP:
array("0" => array(0,1,2,3), "1" => array(4,5,6,7));
And should be converted to
{"0": [0,1,2,3], "1": [4,5,6,7]}
Yet PHP expects me to either accept
[[0,1,2,3],[4,5,6,7]]
or
{"0":{"0":1,"1":1,"2":2,"3":3},"1":{"0":4,"1":5,"2":6,"3":7}}
Neither of which are right at all. How can I possibly decode an object like that? What possible reason is there to ever change something that is clearly using strings as indexes? It's like PHP was trying to be clever to help out idiotic people who can't differentiate strings from ints, but in the process messed up anyone legitimately using strings as indexes, regardless of what the value COULD be turned into.