JSON 到 PHP 数组使用 file_get_contents
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【中文标题】JSON 到 PHP 数组使用 file_get_contents【英文标题】：JSON to PHP Array using file_get_contents 【发布时间】：2012-03-20 17:01:19 【问题描述】：我正在尝试使用杂志 api 获取以下 json 内容。 json的输出是这样的。我希望下面的 json 转换为 php 数组。
"bpath": "http://www.sampledomain.com/",
"clist": [
"cid": "11",
"display_type": "grid",
"ctitle": "abc",
"acount": "71",
"alist": [
"aid": "6865",
"adate": "2 Hours ago",
"atitle": "test",
"adesc": "test desc",
"aimg": "",
"aurl": "?nid=6865",
"weburl": "news.php?nid=6865",
"cmtcount": "0"
,
"aid": "6857",
"adate": "20 Hours ago",
"atitle": "test1",
"adesc": "test desc1",
"aimg": "",
"aurl": "?nid=6857",
"weburl": "news.php?nid=6857",
"cmtcount": "0"
]
,
"cid": "1",
"display_type": "grid",
"ctitle": "test1",
"acount": "2354",
"alist": [
"aid": "6851",
"adate": "1 Days ago",
"atitle": "test123",
"adesc": "test123 desc",
"aimg": "",
"aurl": "?nid=6851",
"weburl": "news.php?nid=6851",
"cmtcount": "7"
,
"aid": "6847",
"adate": "2 Days ago",
"atitle": "test12345",
"adesc": "test12345 desc",
"aimg": "",
"aurl": "?nid=6847",
"weburl": "news.php?nid=6847",
"cmtcount": "7"
]
,
]
我的代码如下所示。
<?php
$json_url = "http://api.testmagazine.com/test.php?type=menu";
$json = file_get_contents($json_url);
$data = json_decode($json, TRUE);
echo "<pre>";
print_r($data);
echo "</pre>";
?>
上面的代码返回一个空数组。 :( 如何将上述 JSON 转换为 php 对象数组。 我很无奈。
谢谢 韩
【问题讨论】：
您可能会得到一个空数组，因为 JSON 对象格式不正确。尝试通过JSONLint 验证它。 print_r($data) 的输出是什么？ @Jack & Aldo - 解决了我的问题。 Shivam S.Kara 代码，终于成功了。 【参考方案1】：检查一些错字','
<?php
//file_get_content(url);
$jsonD = '
"bpath":"http://www.sampledomain.com/",
"clist":[
"cid":"11",
"display_type":"grid",
"ctitle":"abc",
"acount":"71",
"alist":[
"aid":"6865",
"adate":"2 Hours ago",
"atitle":"test",
"adesc":"test desc",
"aimg":"",
"aurl":"?nid=6865",
"weburl":"news.php?nid=6865",
"cmtcount":"0"
,
"aid":"6857",
"adate":"20 Hours ago",
"atitle":"test1",
"adesc":"test desc1",
"aimg":"",
"aurl":"?nid=6857",
"weburl":"news.php?nid=6857",
"cmtcount":"0"
]
,
"cid":"1",
"display_type":"grid",
"ctitle":"test1",
"acount":"2354",
"alist":[
"aid":"6851",
"adate":"1 Days ago",
"atitle":"test123",
"adesc":"test123 desc",
"aimg":"",
"aurl":"?nid=6851",
"weburl":"news.php?nid=6851",
"cmtcount":"7"
,
"aid":"6847",
"adate":"2 Days ago",
"atitle":"test12345",
"adesc":"test12345 desc",
"aimg":"",
"aurl":"?nid=6847",
"weburl":"news.php?nid=6847",
"cmtcount":"7"
]
]
';
$parseJ = json_decode($jsonD,true);
print_r($parseJ);
?>
【讨论】：
【参考方案2】：您的 JSON 不是 P. Galbraith 上面告诉您的有效字符串。
这是解决方案。
<?php
$json_url = "http://api.testmagazine.com/test.php?type=menu";
$json = file_get_contents($json_url);
$json=str_replace(',
]',"
]",$json);
$data = json_decode($json);
echo "<pre>";
print_r($data);
echo "</pre>";
?>
使用此代码，它将为您工作。
【讨论】：
我知道您正在从 API 获取数据，因此您无法验证它。但是我给定的代码对你有用 能否请您选择我的答案作为正确答案，以便其他人使用。【参考方案3】：您提供的 JSON 示例无效。使用此 JSON 验证器 http://jsonlint.com/ 在线检查。您需要删除第 59 行多余的逗号。
如果您拥有有效的 json，您可以使用此代码将其转换为数组。
json_decode($json, true);
Array
(
[bpath] => http://www.sampledomain.com/
[clist] => Array
(
[0] => Array
(
[cid] => 11
[display_type] => grid
[ctitle] => abc
[acount] => 71
[alist] => Array
(
[0] => Array
(
[aid] => 6865
[adate] => 2 Hours ago
[atitle] => test
[adesc] => test desc
[aimg] =>
[aurl] => ?nid=6865
[weburl] => news.php?nid=6865
[cmtcount] => 0
)
[1] => Array
(
[aid] => 6857
[adate] => 20 Hours ago
[atitle] => test1
[adesc] => test desc1
[aimg] =>
[aurl] => ?nid=6857
[weburl] => news.php?nid=6857
[cmtcount] => 0
)
)
)
[1] => Array
(
[cid] => 1
[display_type] => grid
[ctitle] => test1
[acount] => 2354
[alist] => Array
(
[0] => Array
(
[aid] => 6851
[adate] => 1 Days ago
[atitle] => test123
[adesc] => test123 desc
[aimg] =>
[aurl] => ?nid=6851
[weburl] => news.php?nid=6851
[cmtcount] => 7
)
[1] => Array
(
[aid] => 6847
[adate] => 2 Days ago
[atitle] => test12345
[adesc] => test12345 desc
[aimg] =>
[aurl] => ?nid=6847
[weburl] => news.php?nid=6847
[cmtcount] => 7
)
)
)
)
)
【讨论】：
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