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我用我的數據庫中的值做了一個下拉列表。現在,當我在下拉列表中選擇另一個值時,我想顯示來自同一數據庫的信息。例如,我在我的下拉列表中選擇了天使隊,然後我想在數據庫中顯示關於天使隊的信息。我真的很想知道如何做到這一點。 這裏是我的代碼:如何顯示來自數據庫的信息onchange
<?php
$servername = "localhost";
$username = "id1419279_root";
$password = "******";
$dbname = "id1419279_mlb";
// Create connection
$conn = new mysqli($localhost, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT * FROM teams";
$result = $conn->query($sql);
$dropdownlist = '';
while($row = mysqli_fetch_array($result)) {
$teamnaam = $row['teamnaam'];
$dropdownlist .="<option value='" . $teamnaam . "'>" . $teamnaam . "
</option>";
}
if(isset($dropdownlist)){
echo "<select name='naamteam' onchange='submit()'>";
echo $dropdownlist;
echo "</select>";
}
?>
<?php
if(isset($_POST['naamteam'])) {
$naamteam = $_POST['teamnaam'];
$query = "SELECT * FROM spelers WHERE naamteam = '$teamnaam'";
$result = mysqli_query($link, $query);
while ($row = mysqli_fetch_array($result)) {
echo "{$row['teamnaam']} <br>";
echo "{$row['coach']} <br>";
echo "{$row['info']} <br>";
}
}
?>
那麼我真的沒有任何AJAX的經驗,所以可能不會很好,我明天的最後期限hahaha – NivardJ