我使用ajax post方法接收了json数据。
我从tomorrow.php
收到的示例格式:
POSTDATA = country = America& competition = Copa + Libertado& team = Boca + Juniors,content type = application / json,charset utf-8)via tomorrow.php at my simple html dom query file。我收到多个json数据,所以不只是一个POSTDATA。
我想将所有数据转换为php变量。我怎样才能做到这一点?我如何与现有的simple_html_dom
查询结合使用?
我现有的简单html dom查询(我收录了bright.php,它发送了json数据与郡,竞争和俱乐部):
<?php
header('Content-Type: text/html; charset=utf-8');
ini_set("user_agent", "User-Agent: Mozilla/5.0 (Windows NT 6.1; WOW64) AppleWebKit/537.17 (KHTML, like Gecko) Chrome/24.0.1312.60 Safari/537.17");
include ("tomorrow.php");
include ("mysql.php");
include_once('simple_html_dom.php');
ini_set('display_errors', true);
error_reporting(E_ALL);
$filename = "footballfixtures.html";
$html = file_get_html($filename);
class matches {
var $day;
var $kickofftime;
var $title;
function matches ($day, $kickofftime, $title){
$this->day=$day;
$this->kickofftime=$kickofftime;
$this->title=$title;
return $this;
}
}
$i=0;
$day=$html->find('h1',0);
$day->plaintext;
$day=str_replace("<h1>TODAY FOOTBALL FIXTURES: ","", $day);
$day=str_replace("</h1>","", $day);
$matchday = str_replace(array('MONDAY ', 'TUESDAY ', 'WEDNESDAY ', 'THURSDAY ', 'FRIDAY ', 'SATURDAY ', 'SUNDAY '), '', $day);
$matchday=str_replace(" ","-", $matchday);
$matchday=date('Y-m-d', strtotime($matchday));
foreach($html->find('table.fixtures') as $matches)
{
foreach ($matches->find('tr[class=a1],tr[class=a2]') as $matchesTR) {
$kickofftime=$matchesTR->find('td[matches]',0)->plaintext;
$kodate = date('Y-m-d H:i:s', strtotime("$matchday $kickofftime +1 hour"));
$title=$matchesTR->find('pre[class=title]',0)->plaintext;
$i++;
$title = ucwords(strtolower($title));
echo $kodate;
echo $title;
}
}
// clean up memory
$html->clear();
unset($html);
?>
我想将我从footballfixtures.html
(启动时间和标题)收到的数据加上我通过json数据收到的数据(通过tomorrow.php)。
我希望每一行都有这些结果:
kodate,职称,比赛,国家,球队
我希望有人可以帮助我,因为我不知道。