我在php中有这部分代码。当玩家按下按钮在客户端(使用ajax)我想我的数据库显示下一个记录。但我不会。
if(isset($_POST['req'])){
$counter++;
$sql = "SELECT question FROM mytable WHERE id = $counter";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
while($row = $result->fetch_assoc()) {
echo $row["id"]." ". $row["question"]. " " . "<br>";
}
} else {
echo "0 results";
}
}
我建议将计数器存储在会话中。然后每次玩家做这个动作时,你可以给他们下一行,像这样:-
session_start();
if(isset($_POST['req'])){
if ( ! isset($_SESSION['counter']) ) {
$_SESSION['counter'] = 1;
} else {
$_SESSION['counter'] = $_SESSION['counter'] + 1;
}
$sql = "SELECT question FROM mytable WHERE id = {$_SESSION['counter']}";
$result = $conn->query($sql);
if ( ! $result ) {
// log error to error log
error_log(print_r($conn->errorinfo(),true), 3, 'app_error.log');
echo "Temporary database issues, please try again later";
header('Location: error_page.php');
exit;
}
if ($result->num_rows > 0) {
$row = $result->fetch_assoc();
echo $row["id"]." ". $row["question"]. " " . "<br>";
} else {
echo "0 results";
}
}
简单的方法是在Ajax请求中发送当前id。增加它,然后使用它从数据库中提取下一个问题