Question

下面的代码应该创建超链接到它的位置的医院名称，但是当我输出$ hyperlink变量发送时，它只是输出医院位置和医院名称。我正在使用notepad ++并在google chrome中运行我的代码。可能是PHP没有正确处理HTML。顺便说一句，它还连接html标签（在$ hyperlink内）作为输出的一部分。请帮忙

    while($processtable=mysqli_fetch_assoc($fetchospitalperson)){
       // $messagebody= "Patient Name: ".$processtable['firstname']." ".$processtable['fathername']." ".$processtable['lastname']."  "."Hospital Name:"; 
        $messagebody = "Patient Name: ".$processtable['firstname']." ".$processtable['fathername']." ".$processtable['lastname']."  "."Hospital Name:".PHP_EOL; 
        //$hyperlink= new DOMDocument();
        //$hyperlink->loadHTML("<html><body><a href='<".$processtable['geolocation']."'>" .$processtable['hospitalname']."</a></body></html>"); 
        $hyperlink = "<html><body><a href='<".$processtable['geolocation']."'>".$processtable['hospitalname']."</a></body></html>".PHP_EOL;
         //Name of the person that needs the blood transfusion along with hospital he is staying at,hyperlinked to its location
         $string .= $messagebody."  ".$hyperlink.PHP_EOL ;
        }
    $message= "Dear"." ".$firstname." ".$lastname.",".PHP_EOL .$string; 
    if(isset($sendtoperson)){   
        if(mail($to,$header,$message)){
            echo "Sent";
        }
        else{ echo "Not sent";}
    }

html标签输出作为变量的一部分

0 个答案: