尝试回显元素的属性时尝试获取非对象的属性

Question

我不断收到错误 Trying to get 属性 of non-object on the line $title->$my_xpath->query 当运行脚本。我有正确的节点和路径，但直到无法正常工作

    $nodes = $my_xpath->query( '//div[@class="info_coupon"]' );       
    foreach( $nodes as $node )
    {
        $title  = $my_xpath->query( 'a', $node )->item(0)->nodeValue;
        echo $title;
        $code = $my_xpath->query( 'a/@data-code', $node );
        if( $code->length>0 ) {
        $coupon = $code->item(0)->nodeValue ;
        echo $coupon;        
        }
    }

Answer 1

该页面上有一些 javascript 操作生成了 <a>。但本质上（当你得到初始$html）时，有none.

这是您从代码的初始行中得到的片段：

$url = "http://zoutons.com/stores/paytm-coupons/";    
$html = file_get_contents($url); <-- this one contains

这个：

<div class="info_coupon"> 
    <span rel="nofollow" data-lnu="aHR0cDovL3RyYWNraW5nLnZjb21taXNzaW9uLmNvbS9hZmZfYz9vZmZlcl9pZD0xMDIyJmFmZl9pZD0yMDYwJnVybD1odHRwcyUzQSUyRiUyRnBheXRtLmNvbSUyRiUzRnV0bV90ZXJtJTNEe2FmZmlsaWF0ZV9pZH0=" href="http://zoutons.com/stores/paytm-coupons/?#cid=31215" class="heading affiliate affiliate_map c_data_31215" data-affiliate="aHR0cDovL3RyYWNraW5nLnZjb21taXNzaW9uLmNvbS9hZmZfYz9vZmZlcl9pZD0xMDIyJmFmZl9pZD0yMDYwJnVybD1odHRwcyUzQSUyRiUyRnBheXRtLmNvbSUyRiUzRnV0bV90ZXJtJTNEe2FmZmlsaWF0ZV9pZH0="  data-id="31215" data-code="NEW50" data-link_type="text" store="Paytm">GET FREE Rs.50/- ON RECHARGE  (VALID TILL &#8211; APRIL 27)
    </span>

原来真的没有<a>

但您要查找的数据实际上在 <span> 中：

href="http://zoutons.com/stores/paytm-coupons/?#cid=31215"
data-code="NEW50"

所以就把它放在那里：

$nodes = $my_xpath->query( '//div[@class="info_coupon"]' );       
foreach( $nodes as $node )
{
    $title = $my_xpath->evaluate('string(./span/@href)', $node);
    $code = $my_xpath->evaluate('string(./span/@data-code)', $node);
    echo $title;
    echo $code;
}

尝试回显元素的属性时尝试获取非对象的属性

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尝试回显元素的属性时尝试获取非对象的 属性

Trying to get property of non-object when trying to echo the attribute of element

html

php

xpath

domdocument

尝试回显元素的属性时尝试获取非对象的属性