因此我创建了此表，该表显示了来自phpmyadmin的数据表。当我单击编辑游戏以说出ID 9时，它仅显示第8行而不是第9行的数据。我相信这与这段代码有关，因为这是我看到错误发生的唯一位置。有人可以帮忙吗？如果您需要查看我的另一段代码，可以的，谢谢！
<?php
$dbServername = "localhost";
$dbUsername = "root";
$dbPassword = "";
$dbName = "games";
$conn = mysqli_connect($dbServername,$dbUsername,$dbPassword,$dbName);
$sql = "SELECT Id,Dates,GameType,SubType,GameName,FeeType,EntryFee,MinPlayers,MaxPlayers,MaxEn
tries,PrizeDetails,GameType2 FROM games";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
$Id = $row["Id"];
$Dates = $row["Dates"];
$GameType = $row["GameType"];
$SubType = $row["SubType"];
$GameName = $row["GameName"];
$FeeType = $row["FeeType"];
$EntryFee = $row["EntryFee"];
$MinPlayers = $row["MinPlayers"];
$MaxPlayers = $row["MaxPlayers"];
$MaxEntries = $row["MaxEntries"];
$PrizeDetails = $row["PrizeDetails"];
$GameType2 = $row["GameType2"];
echo "<br/>
<h1><center> Edit Game </h1> <br/><br/>