如何执行SQL并从PHP获取结果？

Question

我有这样的代码

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

上面的查询将从训练表返回的最后一行。 如果我回显$ tnid它会显示'资源ID＃5'。 如果我添加

$d = mysql_fetch_array($tnid); 

那么我回声$ d，它会在 U显示错误消息

数组字符串转换：\ XAMPP \ htdocs中\ PDS \行动\ doInsertSchedule.php在线32

如何显示查询的确切结果？ 有人请帮忙。

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

尝试......

$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'];