得到錯誤試圖讓非對象的屬性在簡單的HTML DOM

Question

這裏是我用來提取數據的代碼：

ini_set("user_agent","Opera/9.80 (Windows NT 6.1; U; Edition Campaign 21; en-GB) Presto/2.7.62 Version/11.00"); 

$html14=file_get_html("https://www.gamesrocket.com/battlefield-1.html"); 

$title14 = $html14->find('span.price',0)->children(0)->plaintext; 

for ($i=0; $i <sizeof($title14) ; $i++) { 
    # code... 

    if($title14[$i]){ 
     echo $title14[$i]->plaintext; 
     $a13=$title14[$i]->plaintext; 
    } 
    else{ 

     echo "problem"; 
    } 

} 

在這裏，我想從給定的頁面，但我所提取的價格我得到這個錯誤：Trying to get property of non-object。我使用簡單的HTML DOM來提取數據

Answer 1

因爲$ title14是字符串，而不是對象， 你可以var_dump($title14);知道它的類型。

ini_set("user_agent","Opera/9.80 (Windows NT 6.1; U; Edition Campaign 21; en-GB) Presto/2.7.62 Version/11.00"); 



$html14=file_get_html("https://www.gamesrocket.com/battlefield-1.html"); 

$title14 = $html14->find('span.price',0)->children(0)->plaintext; 
    for ($i=0; $i <sizeof($title14) ; $i++) { 

    if($title14[$i]){ 
    echo $title14[$i]; 
    $a13=$title14[$i]; 
    } 
    else{ 

    echo "problem"; 
} 

}