doubi8512
2015-06-03 19:34
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从SELECT语句中回显变量

i'm trying to make a select query and output this information into a html tags, however it keep converting the php code to this:

<!--? echo $news_text; ?-->

What am i doing wrong? and why isnt it returning error instead of converting it to above?

code

$con = mysqli_connect('localhost','root','pass','db');
$theId = $_GET['id'];
<?

$all_news = $con->prepare("SELECT 
n.id, 
n.title, 
n.url, 
n.image_url, 
n.date, 
n.news_text, 
w.url as website_url, 
w.image as website_image 
from news n
join website w on w.id = n.website_id
where  n.id = ? limit 1"); 
$all_news->bind_param("i", $theId);
$all_news->execute();
$all_news->bind_result($id, $title, $url, $image_url, $date, $news_text,     $url, $referer_img);
$all_news->fetch();

  ?>

<div class='main_image' style='background-image:url(<?php echo $image_url; ?>)' title=''></div>
   <p class="title">
    <? echo $title; ?>
  </p>
   <img class="referer" src="<?php echo $referer_img ?>" alt="">
   <div class="text_div">
        <p>
          <? echo $news_text; ?>
        </p>
   </div>
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2条回答 默认 最新

  • dongtao6842 2015-06-03 19:42
    最佳回答

    Is the part of the code where you echo the information out at contained in '<?php ... ?>'

    EDIT I can't respond to other posts yet (need 50 rep) but the reason you can't place the echo below the <p> as opposed to the $all_news->fetch(); tag is because it is below the ?> tag. Just place it wherever you want but surround it in <?php echo $news_text; ?> and done.

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