从 SQL 和 PHP 创建超链接

Question

我有几个“供应商”，其网站网址列在表格中。我想知道如何链接表中列出的他们的网站以使其成为实时网址。

<?php if($Website){
  echo "<div class='providerData1'>Website: </div> <div class='providerData1 providerData2'>" . $Website .  "</div><br />"; }
?> 

有什么想法吗？谢谢你。 :)

Answer 1

立足于以下几点：（详情可在 cmets 中找到）

<?php

$DB_HOST = "xxx";
$DB_NAME = "xxx";
$DB_PASS = "xxx";
$DB_USER = "xxx";

$db = new mysqli($DB_HOST, $DB_USER, $DB_PASS, $DB_NAME);
if($db->connect_errno > 0) {
  die('Connection failed [' . $db->connect_error . ']');
}

if($db->connect_errno > 0){
    die('Unable to connect [' . $db->connect_errno . ']');
}

// Use
// $sql = mysqli_query($db, "select * from tablename ");
// Or select particular columns
$sql = mysqli_query($db, "select link,name from tablename LIMIT 1");
// You can remove LIMIT 1 if you want to show them all

    echo "Pages (found):";
    echo "<hr>";

while ($row = mysqli_fetch_array($sql)){

    $name= $row['name'];
    $Website = $row['link'];

    echo $name;
    echo "<br>";

    echo "<div class='providerData1'>Website: </div> <div class='providerData1 providerData2'><a href='$Website'>" . $name .  "</a></div><br />";

    }

    echo "<br>";

?>

---

代替整个 while 循环，您可以将其替换为：

旁注（具有link 列的 URL）

echo '<table><tr><th>Name</th><th>Page</th><th>Link</th></tr>';
while ($row = $result->fetch_assoc()){
  echo '<tr><td>'.$row["name"].'</td>';
  echo '<td>'.$row["link"].'</td>';
  echo '<td><a href="'.$row["link"].'">Link</a></td></tr>';

}
echo '</table>';

并在 HTML 表格中整齐地显示您的链接。