Question

如何将我的$url和$name变量用作html代码中的值，因为现在它们只是字符串$url和$ name？

$curnp = $_COOKIE["username"];
$query = mysqli_query($con,"SELECT clientname FROM npclients WHERE npname='$curnp'");
$data = $query->fetch_array(MYSQLI_ASSOC);

foreach ($data as $name): 
    $url = "http://www.dev.healthfullu.com/clients.php?name=".urlencode($name);
    echo '<li>';
    echo '<div class="news-item-detail"><a href="$url" class="news-item-title" target="_blank">$name</a>';
    echo '</div>';
    echo '</li>';
endforeach ?>

Answer 1

变化：

echo '<div class="news-item-detail"><a href="$url" class="news-item-title" target="_blank">$name</a>';

要么：

echo "<div class='news-item-detail'><a href='$url' class='news-item-title' target='_blank'>$name</a>";

或

echo '<div class="news-item-detail"><a href=' . $url . ' class="news-item-title" target="_blank">' . $name . '</a>';

在http://php.net/manual/en/language.types.string.php

阅读有关报价样式和字符串的区别

如何正确打印html代码中的PHP变量？

1 个答案: