{
"categories": [
{
"necklaces": [
{ title: "Outlet", url: "W_Outlet" },
{ title: "Baby-G", url: "W_BabyG" },
{ title: "Bulova", url: "W_Bulova" },
{ title: "Children_s", url: "W_Childrens" }
]
}
]
}
当我在PHP中使用json_decode
时,我得到一个空结果,这让我相信JSON格式不正确。
我需要的是:一个名为“categories”的对象,在该对象内部将有一个对象数组,其中包含一组对象。
即。类别 - > [项链 - > {title:'Gold',url:'Gold'}]
等等,但它似乎没有起作用。有什么想法吗?
答案 0 :(得分:1)
{
"categories": [
{
"necklaces": [
{
"title": "Outlet",
"url": "W_Outlet"
},
{
"title": "Baby-G",
"url": "W_BabyG"
},
{
"title": "Bulova",
"url": "W_Bulova"
},
{
"title": "Children_s",
"url": "W_Childrens"
}
]
}
]
}
答案 1 :(得分:1)
您的字段名称周围缺少引号。
{
"categories": [
{
"necklaces": [
{ "title": "Outlet", "url": "W_Outlet" },
{ "title": "Baby-G", "url": "W_BabyG" },
{ "title": "Bulova", "url": "W_Bulova" },
{ "title": "Children_s", "url": "W_Childrens" }
]
}
]
}
答案 2 :(得分:1)
必须引用title
和url
:
<?php
$json = '{
"categories": [
{
"necklaces": [
{ "title": "Outlet", "url": "W_Outlet" },
{ "title": "Baby-G", "url": "W_BabyG" },
{ "title": "Bulova", "url": "W_Bulova" },
{ "title": "Children_s", "url": "W_Childrens" }
]
}
]
}';
$b = json_decode($json);
print_r($b);
?>
结果将是:
stdClass Object
(
[categories] => Array
(
[0] => stdClass Object
(
[necklaces] => Array
(
[0] => stdClass Object
(
[title] => Outlet
[url] => W_Outlet
)
[1] => stdClass Object
(
[title] => Baby-G
[url] => W_BabyG
)
[2] => stdClass Object
(
[title] => Bulova
[url] => W_Bulova
)
[3] => stdClass Object
(
[title] => Children_s
[url] => W_Childrens
)
)
)
)
)
答案 3 :(得分:1)
所有键和值必须是JSON中的字符串,您缺少引号。
{
"categories": [
{
"necklaces": [
{
"title": "Outlet",
"url": "W_Outlet"
},
{
"title": "Baby-G",
"url": "W_BabyG"
},
{
"title": "Bulova",
"url": "W_Bulova"
},
{
"title": "Children_s",
"url": "W_Childrens"
}
]
}
]
}
您可以尝试JSONLint来测试JSON。