使用PHP＆amp; MySQLi：我正在尝试使用选择列表，当用户使用PHP点击提交按钮时，我试图在用户选择特定选择时显示该列数据：
//database set up information to connect to SQLi:
$servername = "*******";
$username = "********";
$password = "******";
$dbname = "********";
`// Create connection`
`$conn = new mysqli($servername, $username, $password, $dbname);`
`// Check connection`
`if ($conn->connect_error) {`
`die("Connection failed: " . $conn->connect_error);
}`
`//To get value of a selected option from select tag:
if(isset($_POST['submit'])){
$selected_val = $_POST['selection']; // Storing Selected Value In Variable
echo " You have selected :" .$selected_val; // Displaying Selected Value
}`
`//a string was submitted:`
`$selection = $_POST['selection'];`
`$selection = intval($_POST['selection']);`
`//selecting data from database:`
`$sql = "SELECT * FROM teams WHERE teamname = '$selection'";
$sql = "SELECT * FROM teams WHERE city = '$selection'";
$sql = "SELECT * FROM teams WHERE bestplayer = '$selection'";
$sql = "SELECT * FROM teams WHERE yearformed = $selection";
$sql = "SELECT * FROM teams WHERE website = '$selection'";`
`$conn->close();
?>`
`<form action=<?php echo htmlspecialchars($_SERVER["PHP_SELF"]); ?> method="post">
<select name="selection">
<option value="teamname">Team Name</option>
<option value="city">City</option>
<option value="bestplayer">Best Player</option>
<option value="yearformed">Year Formed</option>
<option value="website">Website</option>
</select>
<input type="submit" name="submit" value="Get Selected Values" />
</form>`