How to convert php json format

php json format conversion method: 1. Convert php array and object into json format through the json_encode function; 2. Convert the json text into the corresponding PHP data structure through the json_decode function.

The operating environment of this article: Windows 7 system, PHP version 7.1, DELL G3 computer.

php json format conversion

php natively provides json_encode($str) and json_decode($str).

1.json_encode()

This function converts php array and object into json format.

eg：array
　　$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);　　echo json_encode($arr);
result：{"a":1,"b":2,"c":3,"d":4,"e":5}
eg：object
　　$obj->body = 'another post';
　　$obj->id = 21;
result：　　{　　　　"body":"another post",
　　　　"id":21,
　　}

2. Indexed arrays and associative arrays

PHP supports two types of arrays, one is an indexed array that only saves "value" (value) , the other is an associative array that stores name/value pairs.

Since javascript does not support associative arrays, json_encode() only converts the indexed array to array format, and converts the associative array to object format.

For example, now there is an index array

$arr = Array('one', 'two', 'three');

echo json_encode($ arr);

The result is:

　["one","two","three"]

If you change it For an associative array:

　$arr = Array('1'=>'one', '2'=>'two', '3'=>'three');

echo json_encode($arr);

The result changes:

　{"1":"one","2":"two ","3":"three"}

Note that the data format has changed from "[]" (array) to "{}" (object).

If you need to force "index array" into "object", you can write like this

　json_encode( (object)$arr);

or

json_encode ( $arr, JSON_FORCE_OBJECT );

3. Class conversion

The following is a PHP class:

class Foo {

const ERROR_CODE = '404';

public $public_ex = 'this is public';

private $private_ex = 'this is private!';

　　 protected $protected_ex = 'this should be protected';
#　　}

　　}

Now, perform json conversion on instances of this class:

$foo = new Foo;

$foo_json = json_encode($foo);

　echo $foo_json;

The output result is

　{"public_ex":"this is public"}

You can view it Arrive, In addition to public variables (public), other things (constants, private variables, methods, etc.) are lost.

[Recommended learning: PHP video tutorial]

4. json_decode()

This function is used to convert json text Convert to corresponding PHP data structure. Here is an example:

$json = '{"foo": 12345}';

$obj = json_decode($json);

print $ obj->{'foo'}; // 12345

Normally, json_decode() always returns a PHP object, not an array. For example:

　$json = '{"a":1,"b":2,"c":3,"d":4,"e":5}' ;

var_dump(json_decode($json));

The result is to generate a PHP object:

object(stdClass)#1 (5) {

　　["a"] => int(1)
　　["b"] => int(2)
　　["c"] => int(3)
　　["d"] => int(4)
　　["e"] => int(5)

　}

If you want To force the generation of a PHP associative array, json_decode() needs to add a parameter true：

　$json = '{"a":1,"b":2,"c":3 ,"d":4,"e":5}';

var_dump(json_decode($json),true);

The result is generated An associative array:

array(5) {

　["a"] => int(1)
　["b"] => int( 2)
　 ["c"] => int(3)
　 ["d"] => int(4)
　 ["e"] => int(5)

　}

5. Common errors of json_decode()

The following three ways of writing json are all wrong. Can you see where the mistake is? ?

　$bad_json = "{ 'bar': 'baz' }";

　$bad_json = '{ bar: "baz" }';

　$ bad_json = '{ "bar": "baz", }';

Executing json_decode() on these three strings will return null and report an error.

The first mistake is that the json delimiter (delimiter) only allows the use of double quotes, not single quotes. The second error is that the "name" (the part to the left of the colon) of the json name value pair must use double quotes in any case. The third error is that you cannot add a trailing comma after the last value.

In addition, json can only be used to represent objects and arrays. If json_decode() is used on a string or value, null will be returned.

var_dump(json_decode("Hello World")); //null

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