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php中json的使用

Jun 23, 2016 pm 02:30 PM

 一、json_encode()

  该函数主要用来将数组和对象,转换为json格式。先看一个数组转换的例子:

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

echo json_encode($arr);

  结果为

{"a":1,"b":2,"c":3,"d":4,"e":5}

  再看一个对象转换的例子:

$obj->body = 'another post';

$obj->id = 21;

$obj->approved = true;

$obj->favorite_count = 1;

$obj->status = NULL;

echo json_encode($obj);

  结果为

{
    "body":"another post",

"id":21,

"approved":true,

"favorite_count":1,

"status":null
  }

  由于json只接受utf-8编码的字符,所以json_encode()的参数必须是utf-8编码,否则会得到空字符或者null。当中文使用GB2312编码,或者外文使用ISO-8859-1编码的时候,这一点要特别注意。

  二、索引数组和关联数组

  PHP支持两种数组,一种是只保存"值"(value)的索引数组(indexed array),另一种是保存"名值对"(name/value)的关联数组(associative array)。

  由于javascript不支持关联数组,所以json_encode()只将索引数组(indexed array)转为数组格式,而将关联数组(associative array)转为对象格式。

  比如,现在有一个索引数组

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

echo json_encode($arr);

  结果为:

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

  如果将它改为关联数组:

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

echo json_encode($arr);

  结果就变了:

{"1":"one","2":"two","3":"three"}

  注意,数据格式从"[]"(数组)变成了"{}"(对象)。

  如果你需要将"索引数组"强制转化成"对象",可以这样写

json_encode( (object)$arr );

  或者

json_encode ( $arr, JSON_FORCE_OBJECT );

  三、类(class)的转换

  下面是一个PHP的类:

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;

}

}

  现在,对这个类的实例进行json转换:

$foo = new Foo;

$foo_json = json_encode($foo);

echo $foo_json;

  输出结果是

{"public_ex":"this is public"}

  可以看到,除了公开变量(public),其他东西(常量、私有变量、方法等等)都遗失了。

  四、json_decode()

  该函数用于将json文本转换为相应的PHP数据结构。下面是一个例子:

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

$obj = json_decode($json);

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

  通常情况下,json_decode()总是返回一个PHP对象,而不是数组。比如:

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

var_dump(json_decode($json));

  结果就是生成一个PHP对象:

object(stdClass)#1 (5) {

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

}

  如果想要强制生成PHP关联数组,json_decode()需要加一个参数true:

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

var_dump(json_decode($json),true);

  结果就生成了一个关联数组:

array(5) {

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

}

  五、json_decode()的常见错误

  下面三种json写法都是错的,你能看出错在哪里吗?

$bad_json = "{ 'bar': 'baz' }";

$bad_json = '{ bar: "baz" }';

$bad_json = '{ "bar": "baz", }';

  对这三个字符串执行json_decode()都将返回null,并且报错。

  第一个的错误是,json的分隔符(delimiter)只允许使用双引号,不能使用单引号。第二个的错误是,json名值对的"名"(冒号左边的部分),任何情况下都必须使用双引号。第三个的错误是,最后一个值之后不能添加逗号(trailing comma)。

  另外,json只能用来表示对象(object)和数组(array),如果对一个字符串或数值使用json_decode(),将会返回null。

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

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