我有一个链接到一个页面,该页面根据链接的 id 回显信息,但它没有回显信息。预先感谢您的帮助。
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($data['description']) . "</p>" . "</div>" . "<div class='clear'>" . "</div>" . "</div>";
} else {
die ('failed to connect to database');
}
最佳答案
在此查询字符串下:localhost/Website/HTML/Post.php?id=7
试试这个:
if(isset($_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);
?>
关于php - 数据不显示,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/22591533/