【发布时间】:2014-05-19 09:37:11
【问题描述】:
我是 PHP 新手,我这样做是为了向我的网站显示 fdpp 门户中上传的内容
<?php $resp = file_get_contents("http://fdpp.blgs.gov.ph/api/documents?source=datatable&sSearch=kalinga");
$clean = json_decode($resp);
print_r($clean); ?>
这是结果:
stdClass Object
(
[iTotalRecords] => 130035
[iTotalDisplayRecords] => 879
[sEcho] => 0
[aaData] => Array
(
[0] => stdClass Object
(
[lgu] => <a href=' http://fdpp.blgs.gov.ph/documents/view/129293'>CAR<br/>Kalinga<br />Balbalan</a>
[document] => <a href=' http://fdpp.blgs.gov.ph/documents/view/129293'>Local Disaster Risk Reduction and Management Fund Utilization (LDRRMF)</a>
[period] => Quarter 1 2014
[status] => Required • SUBMITTED
[desc] => The atng.
)
[1] => stdClass Object
(
[lgu] => <a href=' http://fdpp.blgs.gov.ph/documents/view/129188'>CAR<br/>Kalinga<br />Balbalan</a>
[document] => <a href=' http://fdpp.blgs.gov.ph/documents/view/129188'>Manpower Complement</a>
[period] => Quarter 1 2014
[status] => Required • SUBMITTED
[desc] => The file ag.
)
我应该在我的代码中添加什么以将其放入列名为 lgu、文档、句点的表中?我尝试阅读 foreach 手册,但我不知道有人可以帮助我吗?
【问题讨论】:
-
如果没有您的尝试并告诉我们您遇到的问题,几乎不可能帮助您。
-
澄清一下,他似乎正在尝试将 jQuery DataTables 库与服务器端处理一起使用。 Here's an example.