在PHP对象的JSON数组 [英] Json array of objects in php
本文介绍了在PHP对象的JSON数组的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
这是我的code生产
{
"aaa":1,
"b":2,
"c":3,
"d":4,
"e":5,
"fff":{"a":11111,"b":222222,"c":33333,"d":444454,"e":55555555}
}
这就是code
and this is the code
<?php
$c = array('a' => 11111, 'b' => 222222, 'c' => 33333, 'd' => 444454, 'e' => 55555555 );
$arr = array('aaa' => 1, 'b' => 2, 'c' => 3, 'd' => 4, 'e' => 5 , 'fff'=>$c);
echo json_encode($arr);
?>
但我想有一些像这样的结构
but I want to have some structure like this
{
"aaa":1,
"b":2,
"c":3,
"d":4,
"e":5,
"fff":{"a":11111,"b":222222,"c":33333,"d":444454,"e":55555555},
"last":[
{
"id": 8817,
"loc": "NEW YORK CITY"
},
{
"id": 2873,
"loc": "UNITED STATES"
},
{
"id": 1501,
"loc": "NEW YORK STATE"
}
]
}
我在JSON和PHP新我需要这个快,所以我没有时间去阅读有关此JSON结构......所以,请,如果有人知道如何添加这最后一个元素,请提供一些PHP code。
I am new in json and php and I need this fast so I do not have time to read about this json structure... So please if someone know how to add this last element please provide some php code.
谢谢,
推荐答案
- 以JSON-CN codeD字符串,并把它传递给json_de code()
- 的返回值赋值给一个变量
- 通过该变量var_export()得到一个PHP-CN codeD重新串数据的presentation。
例如
<?php
$json = '{
"aaa":1,
"b":2,
"c":3,
"d":4,
"e":5,
"fff":{"a":11111,"b":222222,"c":33333,"d":444454,"e":55555555},
"last":[
{
"id": 8817,
"loc": "NEW YORK CITY"
},
{
"id": 2873,
"loc": "UNITED STATES"
},
{
"id": 1501,
"loc": "NEW YORK STATE"
}
]
}';
$php = json_decode($json, true);
echo var_export($php);
打印
array (
'aaa' => 1,
'b' => 2,
'c' => 3,
'd' => 4,
'e' => 5,
'fff' =>
array (
'a' => 11111,
'b' => 222222,
'c' => 33333,
'd' => 444454,
'e' => 55555555,
),
'last' =>
array (
0 =>
array (
'id' => 8817,
'loc' => 'NEW YORK CITY',
),
1 =>
array (
'id' => 2873,
'loc' => 'UNITED STATES',
),
2 =>
array (
'id' => 1501,
'loc' => 'NEW YORK STATE',
),
),
)
这篇关于在PHP对象的JSON数组的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!
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