PHP簡單的HTML DOM解析器不會返回任何東西

Question

爲什麼我的腳本不會返回ID爲「pp-featured」的div？

<?php 
# create and load the HTML 
include('lib/simple_html_dom.php'); 
$html = new simple_html_dom(); 
$html->load("http://maps.google.com/maps/place?cid=6703996311168776503&q=hills+garage&hl=en&view=feature&mcsrc=google_reviews&num=20&start=0&ved=0CFUQtQU&sa=X&ei=sCq_Tr3mJZToygTOmuCGCg"); 

$ret = $html->find('div[id=pp-featured]'); 

# output it! 
echo $ret->save(); 
?> 

Answer 1

這讓我在路上。謝謝你的幫助。

<?php 

include_once 'lib/simple_html_dom.php'; 

$url = "http://maps.google.com/maps/place?cid=6703996311168776503&q=hills+garage&hl=en&view=feature&mcsrc=google_reviews&num=20&start=0&ved=0CFUQtQU&sa=X&ei=sCq_Tr3mJZToygTOmuCGCg"; 

$html = file_get_html($url); 

$ret = $html->find('div[id=pp-reviews]'); 

foreach($ret as $story) 
    echo $story; 

?> 

Answer 2

庫總是返回一個數組，因爲可能有多個項目與選擇器匹配。

如果您希望只有一個您應該檢查以確保您分析的頁面按預期行事。

建議解決方案：

<?php 

include_once 'lib/simple_html_dom.php'; 

$url = "http://maps.google.com/maps/place?cid=6703996311168776503&q=hills+garage&hl=en&view=feature&mcsrc=google_reviews&num=20&start=0&ved=0CFUQtQU&sa=X&ei=sCq_Tr3mJZToygTOmuCGCg"; 

$html = file_get_html($url); 

$ret = $html->find('div[id=pp-reviews]'); 
if(count($ret)==1){ 
echo $ret[0]->save(); 
} 
else{ 
echo "Something went wrong"; 
}