如何執行SQL並從PHP獲取結果？

Question

我有這樣的代碼

$tnid = mysql_query("SELECT TrainingID FROM training ORDER BY TrainingID DESC LIMIT 1"); 

上面的查詢將從訓練表返回的最後一行。 如果我回顯$ tnid它會顯示'資源ID＃5'。 如果我添加

$d = mysql_fetch_array($tnid); 

那麼我回聲$ d，它會在 U顯示錯誤消息

數組字符串轉換：\ XAMPP \ htdocs中\ PDS \行動\ doInsertSchedule.php在線32

如何顯示查詢的確切結果？ 有人請幫忙。

Answer 1

<?php 
    $con=mysqli_connect("example.com","peter","abc123","my_db"); 
    // Check connection 
    if (mysqli_connect_errno()) { 
     echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
    } 
    $result = mysqli_query($con,"SELECT TrainingID FROM training ORDER BY TrainingID DESC LIMIT 1"); 
    while($row = mysqli_fetch_array($result)) { 
     echo " Training ID - " . $row['TrainingID '] .; 
     echo "<br>"; 
    } 
    mysqli_close($con); 
?> 

Answer 2

嘗試......

$con = mysql_connect("host_name", "user_name", "password"); 
mysql_select_db("database_name", $con); 
$query="SELECT TrainingID FROM training ORDER BY TrainingID DESC LIMIT 1"; 
$res=mysql_query($query,$con); 
$farow=mysql_fetch_array($res); 
$answer=$farow['TrainingID'];