一、json_encode()
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<?php
$arr =array ('a'=>1,'b'=>2,'c'=>3,'d'=>4,'e'=>5);
echo json_encode($arr);
?>
输出
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{"a":1,"b":2,"c":3,"d":4,"e":5}
再看一个对象转换的例子:
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$obj->body = 'another post';
$obj->id = 21;
$obj->approved = true;
$obj->favorite_count = 1;
$obj->status = NULL;
echo json_encode($obj);
输出
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{
"body":"another post",
"id":21,
"approved":true,
"favorite_count":1,
"status":null
}
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
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$arr = Array('one','two', 'three');
echo json_encode($arr);
输出
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["one","two","three"]
如果将它改为关联数组:
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$arr = Array('1'=>'one','2'=>'two','3'=>'three');
echo json_encode($arr);
输出变为
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{"1":"one","2":"two","3":"three"}
如果你需要将"索引数组"强制转化成"对象",可以这样写
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json_encode( (object)$arr);
或者
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json_encode ( $arr, JSON_FORCE_OBJECT );
下面是一个PHP的类:
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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';
publicfunction getErrorCode() {
returnself::ERROR_CODE;
}
}
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$foo =new Foo;
$foo_json = json_encode($foo);
echo $foo_json;
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{"public_ex":"this is public"}
该函数用于将json文本转换为相应的PHP数据结构。下面是一个例子:
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$json ='{"foo": 12345}';
$obj = json_decode($json);
print $obj->{'foo'};// 12345
通常情况下,json_decode()总是返回一个PHP对象,而不是数组。比如:
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$json ='{"a":1,"b":2,"c":3,"d":4,"e":5}';
var_dump(json_decode($json));
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object(stdClass)#1 (5) {
["a"] => int(1)
["b"] => int(2)
["c"] => int(3)
["d"] => int(4)
["e"] => int(5)
}
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$json ='{"a":1,"b":2,"c":3,"d":4,"e":5}';
var_dump(json_decode($json,true));
结果就生成了一个关联数组:
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array(5) {
["a"] => int(1)
["b"] => int(2)
["c"] => int(3)
["d"] => int(4)
["e"] => int(5)
}
下面三种json写法都是错的,你能看出错在哪里吗?五、json_decode()的常见错误
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$bad_json ="{ 'bar': 'baz' }";
$bad_json ='{ bar: "baz" }';
$bad_json ='{ "bar": "baz", }';
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。
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var_dump(json_decode("Hello World"));//null