Sharing how to use json_decode and json_encode in PHP

json_decode and json_encode are both json strings. This article mainly shares with you how to use json_decode and json_encode in PHP. I hope it can help you.

1. json_encode()

This function is mainly used to convert arrays and objects into json format. First look at an example of array conversion:

$arr = array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);　　echo json_encode($arr);　

The result is

{"a":1,"b":2,"c":3,"d":4,"e":5}　

Then look at an example of object conversion:

$obj->body      = 'another post';　　$obj->id       = 21;　　$obj->approved    = true;　　$obj->favorite_count = 1;　　$obj->status     = NULL;　　
echo json_encode($obj);　

The result is

{　　　　
    "body":"another post",　　
　　"id":21,　　
　　"approved":true,　　
　　"favorite_count":1,　　
　　"status":null　　
　　}

Due to json Only UTF-8 encoded characters are accepted, so the parameters of json_encode() must be UTF-8 encoded, otherwise you will get empty characters or null. When Chinese uses GB2312 encoding, or foreign languages ​​use ISO-8859-1 encoding, special attention should be paid to this point.

Indexed array and associative array

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

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, there is now an index array

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

The result is:

["one","two","three"]

If It is changed to 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

json_encode( (object)$arr );　

or

json_encode ( $arr, JSON_FORCE_OBJECT );　

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'; 
 　　
　　　　public function getErrorCode() {　　
　　　　　　return self::ERROR_CODE;　　
　　　　}　　
　　}

Now, perform json conversion on the instance 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 see that, except for the public variables (public), other things (constants, private variables, methods, etc.) are lost.

2. json_decode()

This function is used to convert json text into the 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 An associative array is generated:

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

Common errors of json_decode()

The following three ways of writing json are all wrong. Can you see where the error 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 mistake is that the "name" (the part to the left of the colon) of the json name-value pair must be double quoted 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.

3. JavaScript uses json string

Method to convert JSON string into JSON object

Use the following method to convert into JSON object first:

//由JSON字符串转换为JSON对象var obj = eval('(' + str + ')');

Or

var obj = str.parseJSON(); //由JSON字符串转换为JSON对象

or

var obj = JSON.parse(str); //由JSON字符串转换为JSON对象

Then, you can read it like this:

Alert(obj.name);Alert(obj.sex);

Special note: If obj is originally a JSON object, use the eval() function to convert it Finally (even after multiple conversions) it is still a JSON object, but there will be problems (throwing a syntax exception) after using the parseJSON() function.

Related recommendations:

Solution to the inability of json_decode to parse the special question mark character in PHP

How does PHP implement the json_decode non-escaping Chinese method Introduction

Compare the difference between json_encode and json_decode

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