如何使用此php代码从wamp数据库访问mysql表？

Question

如何通过在Wamp Server中使用php从数据库访问表。我已经完成了以下代码，但由于某些原因而无法正常工作。是否有任何内容要放入“ action =“” page.i想在下拉菜单中的任何其他条目上并按搜索按钮显示数据库中的表。

<p class="h2">Quick Search</p>
    <div class="sb2_opts">
     <p>
   </p>
<form method="post" action="" >
 <p>Enter your source and destination.</p>
<p>
    From:</p>
<select name="from">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<p>
    To:</p>
   <select name="To">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<input type="submit" value="search" /> 
</form>
</form> </table>

<?php


if(isset($_POST['from']) and isset($_POST['To'])) {
$from = $_POST['from'] ;
$to = $_POST['To'] ;
$table = array($from, $to);
$con=mysqli_connect("localhost");
$mydb=mysql_select_db("homedb"); 
if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}


switch ($table) {
  case array ("Islamabad", "Lahore") :
$result = mysqli_query($con,"SELECT * FROM flights");
echo "</flights>";                                    //table name is flights


 break;
  case array ("Islamabad", "Murree") :

$result = mysqli_query($con,"SELECT * FROM `isb to murree`");
echo "</`isb to murree`>";                                     //table name isb to murree
  ;
 break;
  case array ("Islamabad", "Muzaffarabad") :

$result = mysqli_query($con,"SELECT * FROM `isb to muzz`");
echo "</`isb to muzz`>";
 break;
//.....
//......
default:
echo "Your choice is nor valid !!";
}

}
mysqli_close($con);
?>

Answer 1

检查Php手册中的mysqli_connect => 程序样式

mysqli mysqli_connect（[字符串$ host = ini_get（“ mysqli.default_host”）[，字符串$ username = ini_get（“ mysqli.default_user”）[，字符串$ passwd = ini_get（“ mysqli.default_pw”）[，字符串$ dbname = “” [，int $ port = ini_get（“ mysqli.default_port”）[，字符串$ socket = ini_get（“ mysqli.default_socket”）]]]]]]]）

例如 像下面

$con = mysqli_connect("myhost","myusername","mypassword","mydatabase") or die("Error " . mysqli_error($link)); 

在您的代码中，您将打开与mysqli_connect连接，并通过mysql_select_db选择数据库。 那是错的..

您可以像下面这样重写它：

<p class="h2">Quick Search</p>
    <div class="sb2_opts">
     <p>
   </p>
<form method="post" action="" >
 <p>Enter your source and destination.</p>
<p>
    From:</p>
<select name="from">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<p>
    To:</p>
   <select name="To">
<option value="Islamabad">Islamabad</option>
<option value="Lahore">Lahore</option>
<option value="murree">Murree</option>
<option value="Muzaffarabad">Muzaffarabad</option>
</select>
<input type="submit" value="search" /> 
</form>
</form> </table>

<?php


if(isset($_POST['from']) and isset($_POST['To'])) {
$from = $_POST['from'] ;
$to = $_POST['To'] ;
$table = array($from, $to);
$con=mysqli_connect("localhost", "yourusername", "yourpassword", "yourdatabasename");

if (mysqli_connect_errno()) 
{ 
echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
}


switch ($table) {
  case array ("Islamabad", "Lahore") :
$result = mysqli_query($con,"SELECT * FROM flights");
echo "flights";                                    //table name is flights


 break;
  case array ("Islamabad", "Murree") :

$result = mysqli_query($con,"SELECT * FROM `isb to murree`");
echo "</`isb to murree`>";                                     //table name isb to murree
  ;
 break;
  case array ("Islamabad", "Muzaffarabad") :

$result = mysqli_query($con,"SELECT * FROM `isb to muzz`");
echo "isb to muzz";
 break;
//.....
//......
default:
echo "Your choice is nor valid !!";
}

}
mysqli_close($con);
?>

Answer 2

应该提供用户名和密码。

$con = mysqli_connect("localhost","root","","mydatabase") or die("Error " . mysqli_error($link));