Question

我有一个链接到一个页面的链接，该页面根据链接的ID回显信息，但它没有回显信息。在此先感谢您的帮助。

if($connect) {
            mysql_selectdb('phplogin');
            $id = $_GET['id'];
            $query = mysql_query("SELECT * FROM forum WHERE id = '$id'");
            $data = mysql_fetch_array($query);

           echo "<div class='post'>" . "<div class='leftside'>" . "<h3 class='by'>" . $data['user'] . "</h3>" . "<h5 class='date'>" . $data['time'] . "</h5>" . "</div>" . "<div class='after'>" . "</div>" . "<div class='rightside'>" . "<h2 class='title'>" . htmlspecialchars($data['title']) . "</h2>" . "<p class='description'>" . htmlspecialchars($data['description']) . "</p>"  . "</div>" . "<div class='clear'>" . "</div>" . "</div>";

        } else {
            die ('failed to connect to database');
        }

Answer 1

在此查询字符串下：localhost / Website / HTML / Post.php？id = 7

试试这个：

if(isset($_GET['id']))
{
     $id       = $_GET['id'];
     $query    = mysql_query("SELECT * FROM forum WHERE id = '$id'");
     $countQry = mysql_num_rows($query);
     if($countQry>0)
     {
          $data     = mysql_fetch_array($query);
          //-- Fetch your Data here ---//
     }
     else
     {
          echo "No record found.";
     }
}
else
{
     echo "Invalid Id";
}

你的MYSQL连接应该是这样的：

<?php
$link = mysql_connect('localhost', 'mysql_user', 'mysql_password');
if (!$link) {
    die('Could not connect: ' . mysql_error());
}
echo 'Connected successfully';
mysql_close($link);
?>

数据不显示

1 个答案: