Question

我是编程特别laravel的新手

这是我想从中获取数据的链接.... 这是一个体育名单

[http://www.lionpubpattaya.com/index.php/tv-guide][1]

我想在我的应用程序中显示该数据，并且应该显示如下

JSON
{

"games":[

{

"gameTitle":"PREMIER LEAGUE FOOTBALL",

"gameList" : {

"0" : "Sat 15 - 1845 SOUTHAMPTON VS EVERTON",

"1" : "Sat 15 - 2100 SUNDERLAND VS NORWICH",

"2" : "Sat 15 - 2100 SWANSEA VS NEWCASTLE",

"3" : "Sat 15 - 2100 TOTTENHAM VS STOKE",

"4" : "Sat 15 - 2100 WATFORD VS WEST BROM",

"5" : "Sat 15 - 2100 WEST HAM VS LEICESTER",

"6" : "Sun 16 - 1930 CRYSTAL PALACE VS ARSENAL"

}

},

{

"gameTitle":"NRL",

"gameList" : {

"0" : "Thurs 13 - 1630 COWBOYS VS RABBITOHS",

"1" : "Fri 14 - 1630 BRONCOS VS DRAGONS",

"2" : "Sat 15 - 1155 WEST TIGERS VS KNIGHTS",

"3" : "Sat 15 - 1425 PANTHERS VS WARRIORS",

"4" : "Sat 15 - 1625 ROOSTERS VS EELS",

"5" : "Sun 16 - 1100 RAIDERS VS SEA EAGLES",

"6" : "Sun 16 - 1300 BULLDOGS VS TITANS",

"7" : "Mon 17 - 1600 SHARKS VS STORM"

}

},

{

"gameTitle":"AFL",

"gameList" : {

"0" : "Fri 14 - 1630 SYDNEY VS COLLINGWOOD",

"1" : "Sat 15 - 1030 ESSENDON VS ADELAIDE",

"2" : "Sat 15 - 1110 KANGAROOS VS ST KILDA",

"3" : "Sat 15 - 1330 PT ADELAIDE VS GW SYDNEY",

"4" : "Sat 15 - 1620 GEELONG VS HAWTHORN",

"5" : "Sat 15 - 1620 BRISBANE VS CARLTON",

"6" : "Sun 16 - 1000 RICHMOND VS GOLD COAST",

"7" : "Sun 16 - 1220 BULLDOGS VS MELBOURNE",

"8" : "Sun 16 - 1330 FREMANTLE VS WEST COAST"

}

}




this is my code in my laravel controller

   public function index()
    {

       // $file = file_get_contents("excel/sportstv.txt");
       // return $file;
        libxml_use_internal_errors(true);
        $html = file_get_contents("http://www.lionpubpattaya.com/index.php/tv-guide");

        $dom = new \DOMDocument();
        $dom->loadHTML($html);
        $nodes = $dom->document.getElementById('<h3>');
        $result = array();
        foreach ($nodes as $node) {
            $result = $node->nodeValue; // return <h3> tag data
        } 
        $response = response()->json(array("Title",$result));

        return $response;
        libxml_use_internal_errors(true);
    }

Answer 1

我已经解决了我的问题。这就是答案。

$url = "http://www.lionpubpattaya.com/index.php/tv-guide";
$html = file_get_html ( $url );

$games = [];
$i = 0;
foreach ( $html->find ( 'h2' ) as $element ) {
    $games[$i]['gameTitle'] = trim( str_replace("\r\n","", $element->plaintext ) );  
    $games[$i]['gameList'] = explode("\r\n", $element->next_sibling()->plaintext);
    $i++;           
    flush();
}

unset($games[15]);
unset($games[16]);

$games = array_filter($games);

return $games;

如何从其他网站获取数据？

1 个答案: