我正在通过 lynda.com 的视频学习 PHP。我在本地主机 phpmyadmin 面板中创建了一个名为 widget_corp 的数据库，并编写了这些代码块
<?php
/* 1.Create a database connection */
$connection = mysql_connect("localhost", "root", "*");
if(!$connection){
die("Database connection failed: " .mysql_error());
}
/* 2. Select a database to use */
$db_select = mysql_select_db("widget_corp", $connection);
if(!$db_select)
{
die("Database selection failed: " . mysql_error());
}
?>
<html>
<head>
<title> Connection To the Database </title>
</head>
<body>
<?php
//3. Perform database query
$result = mysql_query("SELECT * FROM subjects",$connection);
if(!$result)
{
die("Database query failed: " .mysql_error());
}
//4. Use returned data
while($row = mysql_fetch_array($result));
{
echo $row["menu_name"]." ".$row["position"]."<br/>";
}
?>
</body>
</html>
<?php
//5. Close connection
mysql_close($connection);
?>
我总是得到这种错误类型:
Object not found!
The requested URL was not found on this server. If you entered the URL manually please check your spelling and try again.
If you think this is a server error, please contact the webmaster.
Error 404
localhost
Apache/2.4.4 (Unix) PHP/5.5.3 OpenSSL/1.0.1e mod_perl/2.0.8-dev Perl/v5.16.3
我该如何克服这个问题？谢谢大家
最佳答案
你犯了一个非常小的错误。您在 while 语句中使用了分号，因此只需将其删除即可。然后它将正常工作。
使用返回的数据:
while($row = mysql_fetch_array($result))
不是:
while($row = mysql_fetch_array($result));
关于PHP-MySQL数据库连接，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/20172805/