人們上網並提交一篇文章,標題和說明。人們通常會包含鏈接,但它們顯示爲基本文本。我想要一種方式,當人們包含一個url時,它被認爲是一個工作鏈接。我寫了一段代碼,但它只掃描一行,似乎在實際的表中不起作用,因爲它在表外回顯。從Mysql行中提取網址
基本上...我想要一個表格,當鏈接被提交時,超鏈接被創建。有任何想法嗎? 下面更新後的代碼會保持相同的狀態。
我的代碼如下:
$query = "SELECT * FROM rumours ORDER BY id DESC";
$query = mysql_query($query) or die('MySQL Query Error: ' . mysql_error($connect));
while ($row = mysql_fetch_assoc($query)) {
$id = $row['id'];
$band = $row['band'];
$title = $row['Title'];
$description = $row['description'];
$reg_exUrl = "/(http|https|ftp|ftps)\:\/\/[a-zA-Z0-9\-\.]+\.[a-zA-Z]{2,3}(\/\S*)?/";
// The Text you want to filter for urls
// Check if there is a url in the text
if(preg_match($reg_exUrl, $description, $url)) {
// make the urls hyper links
preg_replace($reg_exUrl, '<a href="'.$url[0].'">'.$url[0].'</a>', $description);
}
echo "<table border='1'>";
echo "<tr>";
echo "<td> $title </td>";
echo "</tr>";
echo "<tr>";
echo "<td class = 'td1'> $description </td>";
echo "</tr>";
echo "</table>";
}
看到這個問題:http://stackoverflow.com/questions/1188129/replace-urls-in-text-with-html-links – Ollie